Is this the correct way to measure the Back EMF Voltage? - Page 1
Forum section: electronics / beginners

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 01:26:55 am
content: Hello, I measured the Back EMF Voltage of a DC motor by connecting the shaft of two motors together. Then, applied 11.1V to the driving motor (called Motor A). While the motors were spinning, I measured the voltage across Motor A and also the voltage across Motor B (the one being driven) at the same time. Meanwhile, I measured the spinning speed. I obtained the following data: Voltage across Motor A: 9.37V. Voltage across Motor B: 8.15V. Am I correct that the Back EMF Voltage is: 8.15V? How come the voltage across Motor A was 9.37V rather than the applied voltage 11.1V?

author: neko efecktz
author_rank: Regular Contributor
timestamp: May 01, 2017, 01:53:12 am
content: NO...You are using motor A to drive motor B.DC motors are also generators.You are simply measuring the voltage that motor 2 is producing.this has nothing to do with back EMF.To learn more on Back EMF there is an enless supply of information on line .wiki pedia has some interesting informationQuote

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 01:56:53 am
content: I used the method from a book recommended by a forum user. See 2-77 Under Voltage Constant and Fig. 2.10.13. I think the Eg mentioned there is the Back EMF voltage. Perhaps I read the method incorrectly? Could anybody please double check?https://www.elsevier.com/books/dc-motors-speed-controls-servo-systems/zhou/978-0-08-021714-7

author: rstofer
author_rank: Super Contributor
timestamp: May 01, 2017, 03:55:26 am
content: re: link - TLDR;You are making this harder than it is. You apply a voltage to the terminals and, given the measured resistance, you know what current should flow. But you measure the current of an unloaded motor and you find it is far less than what you calculated for the winding resistance and applied voltage. The difference is the back EMF. What you really have is the applied voltage minus the back EMF (producing a net lower voltage) divided by the measured resistance. Ohm's Law...As you load the motor, the back EMF is reduced and the current increases accordingly.So, (Applied Voltage - Back EMF) / Measured Resistance => Input CurrentBack EMF = Applied Voltage minus (Input Current times Measured Resistance) <= this is the equation you want!

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2017, 04:13:03 am
content: Identical DC motors with the same field current (or identical field magnets) spinning in the same direction at the same speed will have the same back-EMF. However if you couple the two motors shaft to shaft, they are spinning in opposite directions and asymmetries in the brush positions will result in non-equal back-EMFs. Many DC motors have asymmetric brush positions for improved high speed performance in one direction at the expense of performance in reverse.If you need to actually measure*the back-EMF of a permanent magnet motor, the easiest is to PWM the motor supply with a high voltage MOSFET (to withstand the inductive kickback with only a minimal snubber) and measure the generated voltage during the off time (discarding the initial inductive transient). As the duty cycle approaches 100% the measured back-EMF will approach its non-PWMed limit for that supply voltage. Shunt wound motors can be handled similarly, but you need to maintain power to the field circuit. Series or compound wound motors are not amenable to such methods of measurement.* Directly measure, rather than calculate it from other measurements.

author: Zero999
author_rank: Super Contributor
timestamp: May 01, 2017, 10:21:45 am
content: Quote from: neko efecktz on May 01, 2017, 01:53:12 amThe voltage generated by motor 2 spinning is the back EMF.Quote from: fishandchips on May 01, 2017, 01:26:55 amIt could be the resistance of the wiring, power supply or PSU current limit.Quote from: rstofer on May 01, 2017, 03:55:26 amYes, that's the easiest way to do it:BackEMF = VIN- R×I

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 11:10:38 am
content: About "Back EMF = Applied Voltage minus (Input Current times Measured Resistance)"I conducted another set of 10 experiments. I set the applied voltage from a power supply to 11.1V and measured the voltage across the two terminals of the DC motor as well as the current in series with the motor. Meanwhile, I measured the turning speed using a tachometer.The averaged voltage measured across the motor was 10.18V.The averaged current measured in series with the motor was 1.45ANote that the measured voltages across the motor of all 10 runs were about 1V less than the supplied voltage of 11.1V.I calculated the resistance by stalling the motor and measured the voltage across the motor and the current in series with the motor. Then, calculated the resistance as averaged voltage of 10 runs divided by averaged current of 10 runs. I obtained 0.1 Ohm.Is the Back EMF = 11.1V - (1.45A * 0.1Ohm) = 10.96V?

author: max_torque
author_rank: Super Contributor
timestamp: May 01, 2017, 11:27:08 am
content: The driving motor is doing useful work (spinning itself and the "load" motor) and this takes power, that means a non zero phase current is flowing in the driving motor. And, as the driving motor cannot have a zero phase resistance, than means there is a voltage drop across the phases.The "Load" motor is, in this case, effectively unconnected (assuming you were measuring the voltage with a high impedance multimeter!) and hence has zero phase current, and so no voltage drop.So, the Back EMF measured from the second motor is of course lower than the voltage supplied to the first motor!For example, lets say at speed X, with both motors being identical (and ignoring any unsymmetrical brush effects) the KE (the motors back emf vs rotational speed) of each motor generates a back emf of 10 volts. In the driving motor, there is a positive phase current, of say 1 amp, and if the winding impedance were 1 ohm, then 1 volt would be required to push that current through the phase windings, and so you'd have to apply 11 volts to the driving motor, yet get just 10V out of the load motor.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 11:44:51 am
content: I am using the following instruments. Are they OK for the kind of experiments I am doing to characterize the motor?To measure the voltage across the DC motor:http://overseas.sanwa-meter.co.jp/items/detail.php?id=30To measure the current in series with the motor:Mastercraft 052-0052-2 (the version that could measure 20A)https://www.eevblog.com/forum/testgear/canadian-tire-mastercraft-dmm-new-and-old-revision-teardown/Power Supply (I used the front outputs to supply voltage to the motor)http://www.alinco.com/Products/ps/DM-330/

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2017, 11:57:08 am
content: Max_torque's explanation has a further implication: Because the driving motor has to provide power to overcome twice the friction and windage losses of an unloaded motor, you will need more voltage to match the speed of an unloaded motor. You therefore cant get accurate results from the two motor method without a tachometer and an adjustable power supply.Asymmetric brush effects are fairly easy to test for - if, for both directions at the same voltage, you get near enough the same unloaded speed and current, the brushes must be symmetric. As the brushes may cock slightly in their holders, if the motor has been running in one direction for a significant period (as a part of the brush lifespan), its likely to require running in in the other direction before frictional losses become equal, which may confuse the issue. However as there would be no point in deliberately building a motor with very small brush asymmetry, one would expect to see a gross difference for a motor constructed with deliberate asymmetry.One thing to note: Copper has a notable positive temperature coefficient of resistivity of approximately 0.4%/°C, so if you run the motor long enough to heat up significantly, you will get different results as the winding resistance increases. It may be useful to measure the motor temperature, and directly measure the DC resistance between runs.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 12:15:45 pm
content: Quote from: Ian.M on May 01, 2017, 11:57:08 amI tried to measure the resistance directly by connecting the probes from the multi-meter to the motor (see the two areas pointed by the arrows in the photo). Unfortunately, even I have tried three different multi-meters, I got 0 Ohm in each case. So, I used the calculated resistance of 0.1 Ohm.https://bbqbbq2bbq.smugmug.com/My-First-Gallery/i-fpFQz8H/A

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 12:31:00 pm
content: Quote from: fishandchips on May 01, 2017, 11:10:38 amFor your reference, if I put back the gearhead to the motor and repeated the experiment, I got the following data:The averaged voltage measured across the motor was 10.2V (also less than the supplied voltage of 11.1V).The averaged current measured in series with the motor was 1.2A.The Back EMF = 11.1V-(1.2A*0.1 Ohm) = 10.98V.Are these calculated Back EMF values correct? I need to enter a Back EMF value to simulate the motor in Simulink. Are the instruments I used to do the measurements and to supply the voltage good enough to provide valid results?

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2017, 12:52:05 pm
content: Quote from: fishandchips on May 01, 2017, 12:15:45 pmMost multimeters cant perform useful measurements down in the milliohms. You need a good bench milliohmmeter that uses Kelvin connections for that sort of stuff.Otherwise, its apply a forcing current and measure the voltage, as you did, but that has complications from self-heating unless you keep the duty cycle very low.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 01:04:43 pm
content: Thanks Ian. Do you mean to get an accurate resistance, I should get a better bench milliohmeter that uses Kelvin connections and measure the resistance directly? From the manufacturer, the "dynamic" resistance at a lower test voltage is also about 0.1 Ohm. In this case, can I just use 0.1 Ohm to save some money? The kind of meter you mentioned costs hundreds of dollars.How about the measured voltage across the motor and the current I measured. Are those data valid?If I use the measured voltage across the motor (10.18V) instead of the 11.1V I measured before I connected the motor to the power supply, the Back EMF (without gearhead installed) becomes:10.18V - (1.45A * 0.1 Ohm) = 10VWith the gearhead installed and using the measured voltage across the motor (10.2V) instead of the 11.1V I measured before I connected the motor to the power supply, the Back EMF becomes:10.2V - (1.2A* 0.1 Ohm) = 10.08VIf the 0.1 Ohm is acceptable, which calculated Back EMF value is the correct one to use?

author: Ian.M
author_rank: Super Contributor
timestamp: May 01, 2017, 01:21:34 pm
content: It depends on how much accuracy you need. If you only need four digits, its possible to home-brew one for an affordable price. e.g.http://hackaday.com/2017/01/24/milliohm-meter-version-1-5/Settle for three digits and you could cheapen it further e.g. set up a 100mA LM317 current source and a x10 differential amplifier using an ordinary singe supply OPAMP with a common mode input and output range that goes down to ground, and 0.1% resistors and use an ordinary DMM on mV for readout.Without having measured the temperature, you could easily have had a 10% change in resistance between your first and last run in a series - that's only a 25°C wiinding temperature change.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 01:45:58 pm
content: Quote from: Ian.M on May 01, 2017, 01:21:34 pmThanks. As I recall, the motor was not even warm. I paused between runs.

author: rstofer
author_rank: Super Contributor
timestamp: May 01, 2017, 02:09:09 pm
content: Quote from: fishandchips on May 01, 2017, 12:31:00 pmAgain, you're making this harder than it is. External to the motor, you have absolutely no way to directly measure Back EMF. The reduced voltage you are supplying to the terminals is caused by factors external to the motor - poor PS regulation, excessive voltage drop in the cables, etc. In no way does it have anything to do with Back EMF. The only voltage number you care about is measured at the terminals and this is the applied voltage. Measured at the terminals...Running a second motor as a generator provides no useful information because there is no load and Back EMF is always a function of load.You did a locked rotor experiment and that gave you a measure of the winding resistance. Use that value. It was derived from Applied Voltage divided by Input Current with the rotor locked.All you need to do is measure the terminal voltage and input current then use the formula for Back EMF. At no load the current will be less than the simple Applied Voltage divided by Measured Resistance would suggest. This is because the windings are generating a voltage internal to the motor which is the Back EMF.The degenerate case of zero Back EMF occurs at locked rotor where that voltage is 0. This is where you can get a measure of the winding resistance by working through Ohm's Law for the terminal voltage and input current.

author: rstofer
author_rank: Super Contributor
timestamp: May 01, 2017, 02:22:43 pm
content: Your value for Back EMF above seems mathematically correct. What is the full load current of the motor? Given your 0.1 Ohm value and 12V supplied, it would seem like the locked rotor current would be 120 Amps. That's a big honking motor! I don't recall ever seeing any nameplate data discussed in the thread. I have been thinking about toy motors where it seems we're talking about golf cart motors.That would explain why your unloaded run current is so high. Maybe a snapshot of the motor so we can get an idea what you're working with?If the locked rotor current IS 120 Amps then it's no wonder you are dropping voltage in the power leads.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 02:40:48 pm
content: Quote from: rstofer on May 01, 2017, 02:22:43 pmThanks rstofer. So, forget about the method of using two motors to get the Back EMF and also forget about using 11.1V in the calculation.When I stalled the motor to calculate the resistance, I did 10 runs. At 4V (again I got this measured value when I measured it directly from the output of the power supply before it was connected to the motor), the averaged measured voltage was 0.55V and the averaged measured current was 5.83A. At 5V, the averaged measured voltage was 0.64V and the averaged measured current was 7.35A. At 6V, smoke started to come out and I stopped the experiment immediately.Photo of the disassembled motor is as follows:https://bbqbbq2bbq.smugmug.com/My-First-Gallery/i-cbVKK5j/A

author: rstofer
author_rank: Super Contributor
timestamp: May 01, 2017, 04:29:05 pm
content: You have some massive voltage drop in your setup. Only terminal voltage counts but you are losing a lot of voltage in the PS and cables. Wimpy! But that's good! If you didn't have that voltage drop, the motor would have gone up in smoke a long time back.You don't need very much voltage to measure the internal resistance and after accounting for voltage drop, you don't have much.It seems to me that your measured voltages and currents are directly related.0.55V/5.83A -> 0.094 Ohms0.64V/7.35A -> 0.087 OhmsSo, the winding resistance (including the brushes) is about 0.09 Ohms. Given that number, you can get your Back EMF versus load (or RPM to the extent that RPM is a function of load) using the formula above. But, again, it is a function of load. If you could lash up something like a Prony Brake you could get a nice graph of torque versus current and hence Back EMF.http://www.instructables.com/id/Bench-top-dynomometer/

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 05:43:06 pm
content: Thanks rstofer. So, the Back EMF will be 10.1V becauseWithout gearhead case: 10.18V - (1.45A * 0.09 Ohm) = 10.05VWith gearhead case: 10.2V - (1.2A* 0.09 Ohm) = 10.1VHow come having the gearhead installed does not affect the Back EMF much. I thought having gears means more friction.In the case without the gearhead, with the averaged measured speed of 2358.55 rpm from the tachometer, I could calculate the Back EMF constant as:10.1V/23583.55 rpm = 10.1V/247 rad/s = 0.04 V-s/radSince Kt = Ke, the Torque Constant is also 0.04 but in N.m/Amp?

author: rstofer
author_rank: Super Contributor
timestamp: May 01, 2017, 07:55:05 pm
content: It doesn't make sense that the current goes down when the gearhead is installed. Maybe it doesn't have to...The voltmeter is 1.3%+3 counts and the ammeter is 1.2%+5 counts. The readings aren't far enough apart to say the problem isn't just the accuracy moving around. Separating 1/10 volt or 1/10 amp isn't something you can count on with these meters.My HP3478A bench multimeter is 0.004% of reading + 5 counts in 5-1/2 digit mode for DC voltage and 1% plus 30 counts for current (not all that good). Bought used from eBay.The EEVblog Brymen is 0.3% + 2 digits for voltage and 0.7% plus 3 digits for current on a 3-5/6 digit meter.The GW Instek GDM8251A bench meter is 0.012% plus 5 digits for voltage and 0.2% plus 5 digits with a 20,000 count display (5-1/2 digits). Also bought used and calibrated from eBay for about $100.So, the HP for voltage and the Instek for current.Maybe a better way to measure current is with a shunt resistor. These usually have a full scale voltage drop of 50 mV for industrial units and are measured in milliohms for current sense resistors:http://www.mouser.com/Passive-Components/Resistors/_/N-5g9n?Keyword=shunt&FS=True0.001 Ohms times 10A is 10 mV and good meters have excellent resolution at lower voltages. Fifty milliohms is also available but it will drop 0.5V. Either way, the resistor needs to be upstream of the voltmeter at the motor termnals.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 08:16:03 pm
content: Thanks. Do you think having a clamp meter such as the Klein Tools's CL600 sold in Home Depot would help to measure the current more accurately? I double checked the output of the power supply. The max amp from the ports at the front is 5A. Should be sufficient?I used very similar kind of hooked wires to connect the motor to the multi-meters and power supply. Could such thin wires be the source of the problem? I suppose thin wires do not allow much current to pass through. As a result, the voltage measured across the motor might be lower than the 11.1V I got when measuring the voltage directly from the power supply.https://picclick.com/5-pairs-Wire-Kit-Test-Hook-to-Hook-201502669119.html

author: Zero999
author_rank: Super Contributor
timestamp: May 01, 2017, 08:35:49 pm
content: Quote from: fishandchips on May 01, 2017, 08:16:03 pmIt's completely unsuitable for what you're doing. It can only measure AC current. You need something capable for measuring DC current.Quotehttp://www.kleintools.com/catalog/clamp-meters/digital-clamp-meter-ac-auto-ranging-600a

author: rstofer
author_rank: Super Contributor
timestamp: May 01, 2017, 10:28:29 pm
content: Quote from: fishandchips on May 01, 2017, 08:16:03 pmMeasure the voltage at the PS and again at the motor. Every little bit that is missing is being dropped in the cables. Those cables are totally unsuitable for higher currents. Just measure the voltage drop... Zero is a good answer, maybe 0.5V could be tolerated but you have several volts lost. Those cables have to be getting hot.You need precision measurements, not more of the +- several percent variety. If you use the shunt resistors, you get your meter into something it does fairly well, measuring DC voltage, not current. And if you want the right answer, you'll probably need to buy a meter with a whole lot more digits and accuracy.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 01, 2017, 10:38:44 pm
content: Thanks. I am a bit confused. Before I connected the motor to the power supply, I plugged in a multi-meter to the outputs of the power supply to get a measured voltage of 11.1V. In the without gearhead case, the measured voltage across the terminals of the DC motor was 10.18V. In the with gearhead installed case, the measured voltage across the terminals was 10.2V. So, about 1V was dropped mysteriously perhaps due to my use of the very thin wires. Why you mentioned that I have several volts lost? Am I missing something?Do you think the following kind of large alligator clip test lead with 22 gauge heavy insulation copper wires would make the voltage drop less?https://www.adafruit.com/product/321So, from your post, it seems that the method is correct but the wires and instruments are not good enough to get accurate measurements.Do Ke=0.04 V-s/rad and Kt = 0.04 Nm/A look reasonable?

author: rstofer
author_rank: Super Contributor
timestamp: May 02, 2017, 02:06:19 am
content: Quote from: fishandchips on May 01, 2017, 10:38:44 pmSee your reply #18 where you did some locked rotor tests.QuoteThe resistance of #22 AWG is 0.05 Ohms per meter so if you had 2 19" leads, the resistance of the wire would be about half that of the motor. You can work out the voltage drops versus running current. 0.05 Ohms times 2 Amps will only drop 0.1V in the leads.QuoteIt depends on how accurate your simulation needs to be. I would probably settle for 10% accuracy for the overall project.QuoteI really don't know. I haven't thought about DC motors since I took the course back around '72.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 02, 2017, 02:56:16 am
content: The very small measured voltages in that experiment is indeed very strange. Maybe by stalling the motor, the system tried to get more current but the thin wires prevented it from happening. Then, strange thing happened. Let me get those thicker wires and retest.Usually do people use those #22 AWG wires with alligator clips in this kind of experiments? Given that I can choose small, medium or large alligator clips, which one do you recommend?

author: Ian.M
author_rank: Super Contributor
timestamp: May 02, 2017, 03:55:35 am
content: If you are messing around with high current circuits, you shouldn't be using croc clips and 22AWG cable. The leads used need to be made of wire with an ampacity rating sufficient for the maximum output of the bench PSU. 22AWG is only good for 7A unless it has high temperature insulation. You should probably be using 18AWG or even heavier with crimp spade terminals at the PSU end that can be securely clamped by the PSU terminal binding posts, and at the motor end, either soldered, or crimped and bolted connections. If you use 4mm plugs anywhere (e.g. to insert an ammeter) they need to be good quality brand name ones with sufficient current rating.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 02, 2017, 11:04:13 am
content: Thanks. In the real circuit, I use thick wires and solder the wires directly to the motor driver. However, in the test, I need a temporary solution. As I only got two hands, I used clips cables. What good quality brand name with sufficient current rating do you have in mind? In some stores, the products have no name nor specifications provided. In general, is the larger the croc clips the better? If I have no helping hands, how to I connect the two multi-meters to the circuit?When I repeat the Back EMF experiments, I guess I better adjust the power supplying output so that the to-be-measured voltage across the motor, rather than the power supply, is 11.1V. Am I right?I recall reading that in the resistance calculation experiment, one should set the current output from the power supply to 25% of the rated current. Why 25%? I am characterizing the motor at a higher voltage than the recommended one. I can directly adjust only the voltage output of my power supply. Changing the voltage output changes the current though.

author: Ian.M
author_rank: Super Contributor
timestamp: May 02, 2017, 11:35:48 am
content: The problem with clip leads is the small contact area and limited contact pressure. You aren't going to find heavy duty 20A copper clips like Mueller BU-51C (http://www.alliedelec.com/mueller-bu-51c/70188674/on any affordable off-the-shelf lead. Get something similar with copper contacts not plated steel and an actual current rating and make up your own leads. You'll also need some 20A rated 4mm plugs, or if you are cheap-skating it, use the crimp spades I previously mentioned for any end that goes onto a binding post.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 02, 2017, 12:22:56 pm
content: Thanks. rstofer recommended two multi-meters in Post #21. The HP/Keysight is not in production and both units are quite expensive for me. For the Mastercraft 052-0052-2 and Sanwa meters I mentioned in Post #8, are they generally sufficient for most uses? I chose the Mastercraft 052-0052-2 as it was the only one in nearby stores that could measure 20A at affordable price. If it is not a good meter, may be I should return it to the store.

author: Ian.M
author_rank: Super Contributor
timestamp: May 02, 2017, 12:50:53 pm
content: I wouldn't trust the Mastercraft DMM for high voltage or high current work, its perfectly adequate for general electronics use and limited line voltage work - fused plug-in appliances only. Its 20A current rating is fairly dubious - seehttps://www.eevblog.com/forum/testgear/canadian-tire-mastercraft-dmm-new-and-old-revision-teardown/If you are measuring high currents you should probably be using an external precalibrated current shunt. Typically current shunts are designed to have a burden voltage of 50mV, 75mV or 100mV full scale, so if you want direct read-out on a cheap DMM's 1999mV range, with only a power of ten scaling to do in your head, you'll need to add a precision amplifier.

author: Zero999
author_rank: Super Contributor
timestamp: May 02, 2017, 01:52:06 pm
content: Come to think of it. You don't need over-rated, chunky cable or a clamp meter. The cable just needs to be able to carry the required current, without overheating.All you need is a separate amp and volt meter. Put the amp meter anywhere in series with the power supply and the motor. Measure the voltage, as near as the motor's terminals, as you possibly can. Adjust the power supply voltage, until the desired voltage appears across the motor and stabilises. The calculations can then be performed, using the actual motor voltage and current.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 02, 2017, 02:05:24 pm
content: I recall borrowing two other multi-meters to do the resistance-related, stall motor experiment. Both gave me zero amp. Then, I bought the Mastercraft DMM and it gave me some non-zero amp values.

author: rstofer
author_rank: Super Contributor
timestamp: May 02, 2017, 02:06:57 pm
content: The issue with the meters boils down to trying to compare measurements that vary by less than the inherent accuracy of the meter.These shunts are 5 times better than the meter on DCVhttps://www.digikey.com/product-detail/en/riedon/RSA-10-100/696-1598-ND/4967075Maybe the 10 mOhm would work since 1A would produce 10 mV which would read 10.00 mV on the EEVblog Brymen BM235.It would be nice to have a meter that has averaging because the readings are going to be bouncing around quite a bit. This would be a great time to look into high side current sense amplifiers and scale the voltage up to something that can be handled with an Arduino.http://cds.linear.com/docs/en/product-selector-card/2PB_currentsense_RevA.pdf

author: fishandchips
author_rank: Regular Contributor
timestamp: May 04, 2017, 03:30:48 am
content: Thanks. Are shunts some kind of connection hub?What is "EEVblog Brymen BM235"? Is it invented by a member of EEVblog?http://www.eevblog.com/product/bm235-multimeter/

author: Ian.M
author_rank: Super Contributor
timestamp: May 04, 2017, 04:22:21 am
content: Q1. No. Seehttp://www.rc-electronics-usa.com/current-shunt.htmlQ2. Seehttps://www.eevblog.com/forum/testgear/eevblog-bm235-multimeter-dilemma/msg856498/#msg856498. TLDR: Dave Jones (EEVblog founder/owner) saw a gap in the Australian market for an affordable high quality multimeter. Brymen make multimeters for resellers under many different brands and sell some under their own brand. The BM235 was developed by Brymen in consultation with Dave to meet his exacting specs and the Brymen/EEVblog joint branded edition of it showcases that collaboration while helping Dave fund EEVblog and put bread on his own table.

author: rstofer
author_rank: Super Contributor
timestamp: May 04, 2017, 02:42:32 pm
content: Current shunts are just high wattage low value precision resistors. On larger units the current through the shunt could be as much as 1000A and would result in a 50 mV drop - 50 microohms. Not knowing the rated current of this motor, it's hard to say which range is best but a 10 millohm shunt would drop 50 mV at 5A. If the motor full load current is 20A, the shunt should probably be smaller - like 0.05V / 20A or 2.5 milliohms.There's nothing magic about the 50 mV spec except that it is commonly used in industrial systems. The meter manufacturers just produce one meter range (0-50 mV) and various scales for the application. 1000A, 100A, 1A - doesn't matter. It's the same meter with a different scale. Only the shunt resistor is different.Using a DMM, there is truly no magic in 50 mV. It doesn't matter what the drop is as long as it is insignificant to the circuit. But the voltage drop must be high enough that it can be measured and the comment about the Brymen meter is that it will measure to 0.01 mV. Some meters have better resolution, some have worse. A bench meter might measure to 0.001 mV and my HP will measure to 0.0001 mV or 0.1 uV.When using high side current sense amplifiers, the maximum mV range will be in the datasheet and then it's simply a matter of picking the right resistance. The range should probably include stalled rotor such that the amplifier isn't overloaded during motor startup.The Brymen has MIN-MAX-AVG recording and AVG might be just the thing to smooth out the readings during an experiment. Many meters have this feature, it's not uncommon although it is often overlooked. MAX may also be useful.The BM237 is just a 6000 count meter. It has decent accuracy, great safety ratings, and Dave's required uA scale. Other than that, it's nothing particularly special. I bought a couple of them simply for the branding and to support Dave's efforts.It is currently unavailable at Amazon:https://www.amazon.com/EEVblog-BM235-Brymen-Multimeter/dp/B01JZ1ADCOShorter answer: You need a way to measure the current to a satisfactory resolution such that you can see the difference in current between having the gearbox installed and not having it. The measurements need to make sense.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 04, 2017, 06:14:59 pm
content: A local store has: 16 gauge and 14/2 with ground 300volts which have three wires inside. Which one should I buy?

author: rstofer
author_rank: Super Contributor
timestamp: May 04, 2017, 09:57:31 pm
content: First of all, you want stranded wire. It appears from your photo that the motor uses #14 AWG but that is just a guess. Knowing the full load current would be helpful.I sometimes cut up 2 wire extension cords. These zip cords are neatly identified. If you feel the edge, one will have a couple of ridges and the other will be just rounded. The wire with the ridges is the 'identified conductor' and usually connects to the neutral of the AC system. I use if for the ground wire on electronics related projects.Cutting up 3 wire extension cords gets pricey.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 05, 2017, 02:17:46 am
content: That 14/2 is difficult to bend. Each wire seems to be made of 1 solid wire.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 05, 2017, 12:36:51 pm
content: Quote from: Ian.M on May 04, 2017, 04:22:21 amDo I connect the Shunt in series with the motor of in parallel just like the voltmeter?At the recommended voltage, the no load current is 1.1A and lock current is 19A. However, I am driving the motor at a few voltages higher.

author: Zero999
author_rank: Super Contributor
timestamp: May 05, 2017, 12:57:14 pm
content: Quote from: fishandchips on May 05, 2017, 12:36:51 pmWhat do you think?Consider what would happen if:1) A low value resistor is connected in parallel with the motor? Work out the current, usingOhm's law. Would it achieve your goal? Will the current through the resistor be any different, if the motor wasn't there?2) The same low value resistor is connected in series with the motor. Hopefully you'll see that the resistor's value is much less than that of the motor. The voltage across the resistor will be small, compared to the motor and dependent on the current.

author: rstofer
author_rank: Super Contributor
timestamp: May 05, 2017, 01:53:49 pm
content: Quote from: fishandchips on May 05, 2017, 02:17:46 amProbably a better source is a 3 wire extension cord. Cut off the molded plug, cut back whatever length you want for test leads and put a new plug on the cord.There's a lot to know about wire and stranding. The stiff stuff tends to have fewer strands and the nice stuff has a lot of strands.#16 AWG has more resistance than #14 but it's a whole lot less than #22. For short leads, #16 is probably adequate. You would be looking at about 4 milliohms per foot of wire versus 3 milliohms for #14 or 18 milliohms for #22. If you knew the exact resistance, you could use a foot of wire as the shunt resistor.http://www.calmont.com/pdf/calmont-eng-wire-gauge.pdf

author: rstofer
author_rank: Super Contributor
timestamp: May 05, 2017, 03:50:28 pm
content: If you decided to use a foot of wire (more or less) as the shunt, all you need to do is put a voltmeter across the 1 foot of wire and compare the reading with an ammeter placed in series with the wire (temporarily).It would be better if you had a fixed load. Automotive tail lights draw about 1A at 12V. You would expect to see somewhere around 4 mV across 1 foot of #16 AWG. Whether that is useful depends on the resolution of the meter. If you can measure 0.01 mV then it probably works.The other way to do is would be to use a high side current sense amplifier across the wire and multiply the voltage.Linear Technology has a demo board using their LT1999 HSCSA but it's kind of pricey at $50.http://www.linear.com/solutions/3867For the LT1999-10, you would need a shunt resistor of around 0.008 Ohms or 2 feet of wire (#16 AWG). The maximum input sense voltage of the -10 device is 0.35V so 0.35 / 0.008 => 43.75 Amps - more than adequate but not so high that resolution is lost.For the LT1999-20 with a maximum sense of 0.2V, 1 foot of wire at 0.004 Ohms/foot would give a maximum range of 50A.http://cds.linear.com/docs/en/datasheet/1999fd.pdfThe point of all this is that DMMs measure voltage with pretty good accuracy. Current measurements are a little more iffy.

author: Zero999
author_rank: Super Contributor
timestamp: May 05, 2017, 06:27:14 pm
content: One thing to bear in mind with a wire shunt is the positive temperature coefficient: as the temperature rise, the resistance increases. If you can keep the power dissipation low, then you should be able to ignore it. Otherwise you'll need to measure the shunt, as it heats up and take it into consideration.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 05, 2017, 06:56:40 pm
content: I changed the wires from the power supply to the motor into a thicker ones. I also eliminated those "thin wires with clips". This time, I set the power supply voltage so that the voltage across the terminals across the DC motor was 4V. I stalled the motor to calculate the resistance. I did 10 runs.When I used the thin wires in the past, the averaged measured voltage was 0.55V while the averaged measured current was 5.83A. In this revised experiment, the averaged measured voltage was 1.95V while the averaged measured current was 21.23A. The calculated resistance is still around 0.09 Ohm.Given a 50% drop in measured voltage across the motor during stall, does that mean the setup is still not good enough to measure the Back EMF and resistance correctly?

author: Ian.M
author_rank: Super Contributor
timestamp: May 05, 2017, 07:07:07 pm
content: I'd say its good enough as long as the worst case RUNNING drop stays under a few percent. Its giving you the same calculated resistance which increases confidence in your previous measurements. Do check how much the voltage drops at the PSU output terminals when stalled as if the PSU's regulation isn't 'stiff' enough you could rewire with weldong cable and not see a significant improvement.You may want to re-run your stall tests at an initial voltage of 3V as you are overloading the cheap multimeter, so the current measurements may be even less accurate than its poor specification.As a matter of interest, what's the rated maximum output current of your PSU?

author: fishandchips
author_rank: Regular Contributor
timestamp: May 05, 2017, 07:16:57 pm
content: Thanks. This time, the motor did get hotter even at 4V. The most drop was 1.8V (23A) during stall.As I only have two multi-meter, I took away the MasterCraft meter (used to measure the current in serial to the motor) and used it to measure the voltage across the output of the power supply while the motor was free running. The measured value was 4.05V while the voltage across the motor (without stall, no load turning) was 4V. Is this what you called acceptable? Few percent voltage drop while the motor was RUNNING rather than being STALLED?Strangely speaking, when I tried to repeat the experiment at 3V, even I turned the dial to the minimum and without stalling the motor, the measured voltage across the motor was 3.8V. I could not make it lower. How come?I am using the following PSU:http://www.alinco.com/Products/ps/DM-330/In the past, somebody mentioned that the big drop in voltage might be due to impedance of the PSU. I asked the manufacturer but they also do not want to give out the value. They wrote:"The impedance may be calculated by the formula below just for your information in case you already have the device.Impedance (ohm) = unloaded output voltage V1 - 30A loaded output voltage V2 / 30 (max current of 330MV)"Know what that means? I forgot to mention that this time, I used the outputs at the back rather than at the front of the PS. The back outputs could provide 30A max.

author: rstofer
author_rank: Super Contributor
timestamp: May 05, 2017, 09:41:32 pm
content: Quote from: fishandchips on May 05, 2017, 06:56:40 pmYour supply setup has always been less than marginal. You're getting closer to something that can actually test the motor but you need to limit the run (or stall) time. In a perfect setup, there would be NO voltage drop from no load to stalled. That isn't practical unless you want to use a car battery and booster cables.Your voltage source has an internal resistance (Thevinin resistance), your wire has a resistance and the motor has a resistance. You would want the sum of the internal resistance plus the wire resistance to be much less than the internal resistance (factor of 10?) but that's just not going to happen.You have the motor internal resistance and that's all you need to compute the back emf versus RPM where RPM is dropping as a result of increasing load (some kind of Prony brake or whatever). We've already discussed using applied voltage and running current to calculate back emf. You have everything you need.Motors, in general, aren't intended to be stalled and they self-destruct pretty quick. The outlier being torque motors but that's not what we're talking about.So, take your internal resistance as 0.09 Ohms and go to work on the rest of the simulation. Just realize that if you didn't have that Thevinin resistance and wire resistance, at 12V applied, the current would be 133 Amps. Even at 4V, the current would be nearly 45 Amps.What the manufacturer was telling you was how to compute the Thevinin equivalent resistance which is a whole lot like this back emf thing.Consider a 10V battery with 0 Ohms internal resistance and put a 100 Ohm resistor in series with it. Now put the whole thing in a black box (yes, it has to be black, the instructions say so). You measure the open circuit voltage and, voila' you get 10V because your meter impedance is MUCH higher than the 100 Ohm series resistor.Now, put a 10 Ohm resistor across the terminals and measure the voltage again. You will get 0.9091V (almost every bit of the battery voltage is dropped internally). Since you know the voltage across the 10 Ohm resistor, you know the current through the 3 devices in series. 0.09091A. You also know that the internal resistor, at 0.09091A dropped 9.0909V and dividing 9.0909/0.09091 gives 100 Ohms for the internal (Thevinin equivalent) resistance.So, what they told you to do is set a voltage and measure it unloaded. Then apply a 30A load (however you want) and measure the voltage again. There will be a lower voltage and using the ideas above, you can calculate the Thevinin resistance.http://www.facstaff.bucknell.edu/mastascu/elessonshtml/source/source2.htmlMy numbers above were deliberately skewed. It would be more reasonable for the black box internal resistance to be 10 Ohms and the load to be 100 Ohms. But, in your motor case, your equivalent resistance (power supply plus lead resistance) IS 10 times higher than the load resistance.

author: rstofer
author_rank: Super Contributor
timestamp: May 05, 2017, 09:50:46 pm
content: There's also a Maximum Power Theorem that states, more or less, that maximum power is delivered from the source to the load when their impedances match.https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/maximum-power-transfer-theorem/So, your power supply plus lead resistance should be on the order of 0.09 Ohms to deliver maximum power to the motor. I doubt that this is going to happen.It's just another theorem they made us study.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 05, 2017, 09:58:41 pm
content: Thanks rstofer. I repeated the experiment to obtain the Back EMF. This time with the thick wires connected from the back of the PSU.Averaged voltage measured across the DC motor: 11.1 V(This time I made sure that the voltage across the motor was 11.1V by turning the knob of the PSU manually while looking at the value of the multi-meter)Averaged current in series with the motor: 1.4AVoltage drop across resistance (used 0.09 Ohm as resistance): 0.1VCalculated Back EMF: 11V.Since the averaged measured spinning speed at no load is about 26000 rpm,the Back EMF constant is:11V/26000 rpm = 11V/2723 rad/s = 0.004 V-s/radSince Kt = Ke, the Torque Constant is also 0.004 but in Nm/A.Does this sound good? The constants seem to be a bit small.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2017, 12:24:19 am
content: As I said earlier, I haven't thought about motor constants in 45 years or so. No point in starting now...My basic problem remains: The readings show such a tiny difference between applied voltage and back EMF that I can't tell if the measurement is real or just a bobble in the display. There are simply not enough digits. Perhaps you rounded off but, bobble wise, you are talking about 0.05V rounded up or down.Basically, I think you have the right answer because, unloaded, the back EMF will be quite high and the running current fairly low. Those are just intuitive kinds of things.Math wise, I think you have enough to model the motor. You certainly have confirmation on the internal resistance. What will be most interesting is the correlation between the model and whatever application you come up with.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 06, 2017, 01:46:27 pm
content: Thanks. I tried to simulate the motor using the equivalent circuit approach in Simulink. I applied a step input to the model and simulated it for 10 seconds. At the 5th second, the input jumped from 0 to 11.1 V. Although the maximum of the speed curve does not match the averaged measured rpm exactly, the curve jumped from 0 rpm to about the averaged rpm at the 5th second and stayed fairly constant. I guess the discrepancy might be due to measurement errors and modelling inaccuracy. To my limited knowledge, I guess it is fairly OK. Am I right?However, the current vs. time curve is another story. At the 5th second, I saw a spike to around 85 A (shouldn't it be about 1.4A?). The rest were zero or very very near zero. Anybody knows what is going on with such large spike? Shouldn't the current curve be like a step curve as well? As the current dropped to very very close to zero, shouldn't the motor stopped spinning. However, the rpm curve seems to look ok as it has a step response curve.Equivalent circuit approach:I got the same issue with the spike current using another model found on the internet:http://ctms.engin.umich.edu/CTMS/index.php?example=MotorSpeed&section=SimulinkModeling#3I measured the inductance using a DE-5000 LCR meter. I calculated the moment of inertia of the rotor by treating it as a solid cylinder of similar height and diameter.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2017, 04:16:03 pm
content: That's a terrific video series - worth every minute!Why don't you zip the .slx file and post it? You can not post a .slx file.What is happening in your model at t=5? Usually the step function occurs at t=1 (from the video).What values are you using for the motor constants: KT,KB,R,L,IL?That model from the video is independent of the actual constants, that is, the integrators, summer, gain blocks, are all the same regardless of the motor. Only the constants change.You can add more scopes or more signals to a single scope. You can also do XY plots if they are helpful.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2017, 04:41:43 pm
content: I am going to attach my Predator Prey Simulink model. Not that it is important but it shows how to use a .mat file to hold the constants for the .slx model.So, if I were modeling this motor using the video series, I would leave the named variables in the Simulink Model and have them linked to a table in the .mat file. Then I can play with the constants and not change any aspect of the model.Predator-Prey:There is a large field with rabbits. Were it not for the presence of foxes, the rabbit population would grow without bound. So rabbit-fox interactions (where the fox eats the rabbit) reduces the rabbit population.Foxes, OTOH, would go extinct were it not for a population of rabbits.So, we can write the differential equations:dR/dt = aR - bRF - the rate of change of the rabbit population is proportional to its population 'a' and inversely proportional to 'b', the number of rabbit-fox interactions.dF/dt = cF + dRF - the rate of change of the fox population is proportional to its population 'c' and the number 'd' of rabbit-fox interactions.Over time, the rabbit and fox populations rise and fall. More rabbits means more food means more foxes which eat more rabbits which means fewer foxes, etc.Varying the constants 'a'..'d' results in different graphs but there are only two stable solutions: R=F=0 (nothing alive) or R=c/d and F=a/b (or so the text says, I haven't looked).To run the model, put the files somewhere, find them in Matlab, double click on the .mat file first to load up the constants and then double click on the .slx file to load it. Run the simulation.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2017, 08:04:24 pm
content: Quote from: fishandchips on May 06, 2017, 01:46:27 pmSure, you are seeing the inrush current limited by the inductance but a function of the resistance. A 11V supply with a 0.09 Ohm resistor will result in a maximum locked rotor current of 122A. Inrush can be quite high until the motor starts turning and generating back EMF.For giggles, I built up the Matlab model from the video. Open the .mat file to fill up the workspace and then open and run the .slx file. In theory, the scopes will be open and displaying the results. If not, double click on the scopes.It might be fun to experiment with variations of IL. Make the moment of inertia a good deal larger and the motor won't accelerate so fast.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 06, 2017, 10:20:17 pm
content: Thanks rstofer.I also followed the video and got the same results as you did. Then, I changed the parameters based on the measurements and calculations that we have been discussing about. The parameters I used are:R = 0.09 (in Ohm)L = 80e-6 (as the DE-5000 LCR meter gave me 80 micro H, meter also displayed 1K Hz and Q = 0.579)IL = 5e-6 (in kgm^2)KB = 0.004 (in V-s/rad)KT = KB = 0.004 (Nm/A)Using the step input unmodified, the ang speed stayed at around 250 (rpm?) while a strange current spike showed up with a peak of about 10 (amp?).When I changed the Final value of the step input from 1 to 11.1V to simulate the measured voltage across the DC motor, the ang speed stayed at around 2800 (rpm?) while the current spike was peaked at about 115 (amp?).From Post 69, the averaged measured ang speed was: 26000 rpm while the averaged measured current was 1.4A. Somehow, these values are very different from the simulation results. Am I missing something?

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2017, 11:14:41 pm
content: Your resistance is quite low compared to the two models but it is what it is. It allows for very high current.The idea that KB and KT are equal may be questionable. The video has KB=0.22, KT=0.02 while the UMich model has both values as 0.01. UMich also has viscous damping - a real load. Their inertia seems low and the motor still takes a LONG time to accelerate. Try t -> 500.There is no rotational inertia so the motor can accelerate in zero time.There is no inductance so the motor current can spike quite high quite fast which allows the motor to accelerate in 0 time.There is no load so the back EMF will approach the supply voltage and current will fall to 0.The output is in radians/second so divide by 2*pi to get revolutions/second then multiply by 60 to get RPM.250 rads/sec = 2387 RPM.http://www.kylesconverter.com/frequency/radians-per-second-to-revolutions-per-minute

author: fishandchips
author_rank: Regular Contributor
timestamp: May 06, 2017, 11:41:09 pm
content: Quote from: rstofer on May 06, 2017, 11:14:41 pmThanks rstofer.Under the youtube model, the simulated speed is: 26738 rpm which seems to match the averaged measured speed of 26000 rpm.The current is the problem.Do you mean the inertia and inductance values are so small that it is like there is no inertia nor inductance?I got the inductance value by plugging the two probes from the DE-5000 to the terminals of the DC motor. Did I do it incorrectly?What suggestion do you have? We have done all the measurements already.I tried the UMich model using the same set of values again. This time, the spike disappeared. It was there when I tried it before. With a step input of 11.1 at Step time = 5, both speed and current curves have a step-like shape. However, for the current curve, it jumped to about 125 (amp?) and stayed with that value. As for the ang speed curve, it jumped to 5! and stayed around there (lots of small dots on the fairly flat curve after the 5th second.)If I set the b = 0, the ang speed reached about 2700 at the 5th second and stayed fairly constant at that value. As for the current output, the spike with a peak of about 115 happened at the 5th second.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2017, 11:44:13 pm
content: Quote from: rstofer on May 06, 2017, 11:14:41 pmWith b=0.1, the motor comes right up to speed and there doesn't seem to be a spike in current. Setting b to 0 causes the motor to take a long time to accelerate. This doesn't seem right so I'll have to look into it.

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 02:38:36 am
content: Quote from: fishandchips on May 06, 2017, 11:41:09 pmUsing your constants, given above, and the video model (lacks damping), I get a 10A current spike at t=1 and the tail current is 0. I get a rotation velocity of 250 rads/sec. This is at 1V, see a below where I increase the voltage to 11.1.The current spike makes sense. First, the current in my example is based on 1V and approximately 0.1 Ohms - of course the current is 10A. There's no inductance to cause it to be anything else.The tail current also makes sense. There is no load on the motor so the back EMF is very close to the terminal voltage so no current flows (or very very little).If I change the step voltage to 11.1V then my current spike is a little over 120A (makes sense, 11.1V / 0.09 Ohms) and w=2750 rads/sec so the RPM is about 26000.Change IL to 1x10-3and the motor will accelerate a lot slower and the current curve will make a lot more sense.QuoteExactly! The inductance should throttle the current spike or at least cause something of a slope on the rising edge. The inertia is so low that it can be instantaneously accelerated to full speed where the back EMF kicks in and shuts down the current by counteracting the applied voltage. Hence the spike!QuoteI'm not sure I can recommend anything. The rotational inertia is clearly wrong - by a lot. I don't know anything about that LCR meter but here's the thing: It is measuring the AC impedance at some frequency. Who cares about the frequency domain, I want the time domain and essentially DC. v(i) = L di/dt. Note that this is exactly the term used in the simulation.QuoteThat damping factor is a heck of a load. It keeps the motor running so slow that the back EMF never helps reduce the running current.QuoteYup! And it all follows from the above replies. Your spike is at t=5 by choice, mine is at t=1 but the numbers and description are the same.

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 03:08:52 am
content: We have discussed measuring the motor inductance in another thread. Apparently without getting what we need.Bruce Abbott's reply (with the scope traces) makes a lot of sense:https://electronics.stackexchange.com/questions/182116/whats-the-easy-way-to-measure-a-dc-hobby-motors-inductanceWe're only interested in the time constant and we know that 1 time constant is 63% of the upper value. So, if we hit the motor with a square wave, we will see a rising edge and a falling edge, both are curved. What we want to know is the time it takes to get to 63% of the applied voltage or drop to 37% of the max voltage. I would go for the rising edge...You will need to put a known resistor in series with the motor to limit the current from the square wave generator. Otherwise, a 1V square wave would need to drive 11A through the 0.09 Ohm resistor.Nevertheless, Tau = L / R. You have measured Tau (the 63% thing) and you know R which is primarily the limiting resistor because it will be much larger than 0.09 Ohm). Getting L is easy.That curve is 1-e-(t/Tau), When t=Tau (a period equal to one time constant), the result of the expression is 63%.None of the methods based on applying an AC frequency are going to do any good at all.The video example used 0.2 Henries - That's a lot higher inductance than your 80 microhenries. The UMich example used 0.5 Henries.Plug in 0.5 Henries and see what happens to current.

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 03:37:13 am
content: If you use a 1k resistor in series with a 0.5 Henry inductor,Tau = 0.5 / 1000 = 0.0005 seconds.It takes 6 Tau to be effectively at full voltage so 0.003 seconds.The square wave would be high for 0.003 seconds and the total period would be 0.006 seconds. Call it 0.01 seconds. So, 100 Hz ought to do it.If you have nothing else, an Arduino can do this pretty easy. It doesn't matter if the frequency isn't perfect as long as the signal is high for at least 6 Tau and low for at least 6 Tau. 60 Hz would work, square up an AC wall wart (I don't know how well that will work...). Better yet, use a 555 timer. Accuracy isn't a factor, we just need sufficient ON and OFF time for 6 Tau.At t=6 * Tau, you are at 99.75% of applied voltage which should make the inductor voltage trace look every bit as high as the applied square wave. Now it is just an exercise to find 63%. I'll bet the cursor capability will work nicely.I guess we never did discuss whether you had access to a scope. If we did, I have forgotten. That happens a lot lately...At t=6 * Tau, the equation is 1-e-6or 99.75%

author: fishandchips
author_rank: Regular Contributor
timestamp: May 07, 2017, 01:22:41 pm
content: Thanks rstofer for the analysis and suggestion. Let me check with DE-5000 users to see if I measured the inductance correctly using that device. I do not have an oscilloscope but there is a small possibility that I might be able to borrow one.I have a function generator but I no longer own a scope. I may need a multi-channel data logger later in another project. Do you think a logic analyzer like the saleae's could do the work?https://www.saleae.com/

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 04:31:44 pm
content: I don't see how a logic analyzer could help as we have no digital data.I was going to suggest using an Arduino but I don't think it will be fast enough. If my math is anywhere near right, Tau for your motor is 500 uS and you would probably want 10 samples in that interval so 50 uS per sample. That's 20 kHz. Maybe it works out, maybe not. There are examples of testing audio at 48 kHz:https://forum.arduino.cc/index.php?topic=205096.0The absolute perfect way to do this is with a Digilent Analog Discovery device.So, I tried it. I used a very small robotics motor and stuffed in a 1V 500 Hz square wave through a 220 Ohm resistor (blue trace) and got back the orange trace across the resistor. The rise to 63% is about 22 usec. This converts to about 5 millihenries. I expect your motor to have much more inductance. My motor has nearly 7 Ohms of internal resistance.I don't have an RLC meter so I can't check against some other method.

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 04:39:35 pm
content: I haven't quit looking for the specs on a small motor (particularly inductance) but I did run across the specs for a small motor that shows the stall current 77 Amps on a motor with a no-load current of 1.5 Amps.http://www.robotshop.com/media/files/pdf2/rb-wtc-03.pdfThis is a fairly small motor, it seems yours is somewhat larger, so the number you have for stall current (from the simulation) isn't completely unreasonable. Of course that is based on 0.09 Ohms and that has been cross-checked. Probably pretty close.Fortunately, your power supply won't deliver 12V at 120 Amps!

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 04:44:32 pm
content: Another interesting reference:http://www.gearseds.com/files/Mathematical%20Models%20of%20Motors.pdf

author: fishandchips
author_rank: Regular Contributor
timestamp: May 07, 2017, 08:56:34 pm
content: Thanks rstofer. I re-measured and re-calculated the moment of inertia of the rotor. The weight is 65g (0.065kg). The radius is 1.2cm = 0.012m. Inertia of a solid cylinder is (1/2)*mr^2= 4.7e-6 kgm^2. Is the calculation wrong?

author: rstofer
author_rank: Super Contributor
timestamp: May 07, 2017, 10:58:39 pm
content: Quote from: fishandchips on May 07, 2017, 08:56:34 pmI went back and looked at that armature photo you posted above - Reply 18. It doesn't have a scale so I guess I really don't know how big it is but, from the above, the armature is about 1" in diameter. If armature assembly weight is 2 oz then your calculation is correct. I would have thought it would be more.Given a low moment of inertia, acceleration would be quite high and that's what the model is showing. Just plug in a lot more inertia and see what happens. That's the cool thing about keeping the constants outside the model.From the photo, it looks like the winding is made from pretty large wire so I'm not surprised the resistance is so low. You have measured the resistance and it seems like 0.09 Ohms is correct even though my small robotics motor has 7 Ohms.That's an interesting motor!

author: fishandchips
author_rank: Regular Contributor
timestamp: May 08, 2017, 12:45:04 am
content: Yes, the armature is about 1" in diameter.Thanks rstofer for double checking.At IL = 5e-5, the spike is less sharp. The current peak is at around 120 while the speed is about 2700.At IL = 5e-4, the spike disappeared. The current gradually drops to about 5 amp at the 10th second while the speed increases from 0 to about 2600 and keeps increasing. Once I have increased the simulation to 20 sec. I see the current gradually drops to zero while the speed gradually stays at around 2700.At IL = 5e-3, current drops from about 120 to 60 in the 20th second while speed increases from 0 to about 1400. I need to increase the simulation time to 150 to see the current dropping to zero and the speed reaches about 2700.How come without current (i.e. voltage), the motor can still spin at constant speed?How come unlike the simulation, in the real life the current does not spike to about 120A?At the end of the youtube video, the author mentioned that he did not model the friction, etc. Perhaps with friction, the motor may gradually slows down when there is no current?The only thing that is left for checking is the inductance then.

author: rstofer
author_rank: Super Contributor
timestamp: May 08, 2017, 04:27:35 am
content: As long as you don't have friction, the motor will eventually get to zero current.ETA:The current is based on the difference between the applied voltage and the back EMF. In the absence of friction, the motor accelerates until the two voltage are equal (zero current) and perpetual motion takes over. This won't happen in real life because there is always friction.The current might very well try to spike to 120A (12V/0.09Ohms) but it won't get there because there actually is inductance sufficient to keep it from happening. By the time the inductance allows current to flow, back EMF is already being generated and this reduces the current. The fact that there is little inertia allows the armature to accelerate quickly and generate back EMF.The UMich model does include 'b' for friction. A value of 0.1 is a lot of friction. It limits the top RPM and holds the current high and steady. The higher the 'b' coefficient, the higher the friction and the slower the motor 'runs' and the more current it draws at steady state.

author: rstofer
author_rank: Super Contributor
timestamp: May 08, 2017, 04:03:51 pm
content: Thinking about that 'b' term in the UMich model, there is no reason there can't be several such terms added together.One term might be some friction loss in the motor itself to help account for the difference in no-load RPM between the modeled motor and a real motor. We could get fancy and make it a function of omega (w).Another term might account for the load. Perhaps it is gated by another step function or formed from a triangle wave that would linearly apply and remove load. We could see changes in RPM as the load varies with a constant applied voltage.We could also vary the applied voltage and watch the RPM vary. Including these other terms would make the model more realistic. It might be interesting to create the voltage as a PWM waveform.Matlab and Simulink are terrific tools.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 08, 2017, 05:06:51 pm
content: Quote from: rstofer on May 08, 2017, 04:27:35 amSince Torque = Kt*I, does it make sense that without friction and in the case of very small inductance similar to none, there would be a spike of large torque and then the motor produces zero torque while it keeps spinning at constant speed?I also tried to display the torque curve by multiplying the current by Kt. As I recall, the shape looked like the spike current curve due to the constant Kt.

author: rstofer
author_rank: Super Contributor
timestamp: May 08, 2017, 05:59:26 pm
content: Quote from: fishandchips on May 08, 2017, 05:06:51 pmAbsolutely! If current is zero, torque is zero (regardless of Kt) and, since we don't have any friction, keeping the armature rotating at a constant speed doesn't require any torque. Torque T, like force F, is only required if we want to accelerate something. F=ma, T=Iw' where w' is rotational acceleration and sometimes just called alpha and I is the moment of inertia.Maintaining a constant RPM in the absence of friction doesn't require torque. It wouldn't! There is no need to accelerate the moment of inertia so w' is zero (no change in velocity).This math thing works out pretty well!

author: rstofer
author_rank: Super Contributor
timestamp: May 08, 2017, 08:16:11 pm
content: My DE-5000 came in so I made some tests.First, I wanted Rsso I pushed LCR Auto until I got Rpand then pushed SER/PAL until I got Rsand measured 6.54 Ohms. This is very similar to what I got on my bench meters. Things are looking good!Next, I wanted L at 100 Hz so I pushed LCR AUTO until I got Lpand FREQ until I got 100 Ohms. I got a reading of 6 mH and a Q of 0.58. Lsdoesn't seem to be an option at this point.Let's cross-check using Q=2*pi*f*Ls/RsLs= (Q * Rs)/(2*pi*f) or 0.58*6.54/200*pi or 6x10-3= 6 mH. It checks.So, my little robotics motor has 6.5 Ohms series resistance and 6 mH of inductance. Remember, I measured 5 mH using the oscilloscope method. So, there is some basis for using the DE-5000 to measure the inductance of a motor.Try Rson your motor with the DE-5000. I still wonder about that 0.09 Ohm series resistance.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 08, 2017, 09:27:32 pm
content: Quote from: rstofer on May 08, 2017, 08:16:11 pmI followed your procedure but I had a different experience.First, I tried to push LCR Auto until I got Rp. However, Rpnever showed up. Only Rs. I have tried several times already.Rs1.035 Ohms. (The value is different from the one I got in my first trial.)Then, I pushed SER/PAL. It changed to Rp. Rp1.358 Ohms. Again, I got a different value compared with the previous trial.Next, since I wanted L at 100 Hz as well, I pushed LCR AUTO. I got DCR 0.97 Ohms. Pushing LCR AUTO and pressed FREQ until 100 Hz was shown. This time, I got: Rs 0.97 Ohm. Pressed LCR AUTO twice got: 100 Hz, 3.6 degrees, 0.96 Ohms. Pressed LCR AUTO again, got: APO Q 0.065, Ls = 101uH.I could not get to Lpby pressing LCR AUTO repeatedly. Each time I saw Lp, it quickly turned to Ls.To cross check like you did, 2*pi*100*101e-6/1.035 = 0.0613 ~ 0.065 shown under Q on the display.Ls = 0.065 * 1.035/(2*pi*100) = 1e-4 = 101uH. Checked.

author: rstofer
author_rank: Super Contributor
timestamp: May 08, 2017, 10:32:24 pm
content: But your resistance reading is 10 times more reasonable. One Ohm is a lot more comforting than 0.09 Ohms. For one thing, the stall current would be just 12A, not 120A. That's a lot more believable.Your motor has 1/6th the resistance but that's ok, the wire gauge seems much larger. But not 1/100th of the resistance...Your motor has much less inductance (1/50) but, again, there is a lot less wire around the armature.Looking at the constants presented in the two links, we really don't know what kind of motors they were testing. I can imagine that a 0.2H inductance might imply a fairly large motor.I just measured a 1 HP 24VDC motor:Rs=0.35 OhmL=180 uHQ=0.32f=100 HzIt all checks.The 0.35 Ohm versus your 1 Ohm is also a lot more reasonable. This is a very large motor so it is reasonable that your resistance measurement would be larger. I'm a little surprised at the inductance but that's what I got. I wonder what kind of motor has 0.2 Henries?

author: fishandchips
author_rank: Regular Contributor
timestamp: May 08, 2017, 11:22:56 pm
content: Thanks rstofer. I measured the diameter of the wire and it is about 1mm. I gathered that at a lower voltage, the dynamic resistance of the motor is 0.123 Ohms while the EMF/Torque constants are 3.55 V-s/rad and 3.55 Nm/Amp. So, even the resistance should be the same regardless of supplied voltage and current, my 0.09 Ohms seems OK? As far as I understand, the EMF/Torque constants should be the same regardless of the applied voltage. So, even I drive the motor at a higher voltage, the Back EMF should be 3.55 V-s/rad. However, using the method we have been discussing, I got only 0.0042 V-s/rad ?!

author: rstofer
author_rank: Super Contributor
timestamp: May 09, 2017, 12:56:17 am
content: I hope I get this right...I wonder about 3.55 V-s/rad. That implies that the back EMF would be 3.55V if the motor was turning 1rad/sec or 1/6 RPS or 10 RPM whereas I would think the value would be the applied voltage divided by the steady state RPM. Less friction, of course.It is the back EMF at some RPM that balances the applies voltage.Edited to account for seconds and minutes. Duh...

author: fishandchips
author_rank: Regular Contributor
timestamp: May 09, 2017, 12:26:06 pm
content: So, which inductance and at which frequency are correct? I got different inductance values each time I measured it. However, the values are always in uH (so small that it is like none) range.

author: rstofer
author_rank: Super Contributor
timestamp: May 09, 2017, 01:29:20 pm
content: I would tend to go with the scope method simply because the work is being done in the time domain. We are interested in the step response, not the frequency response. The thing is, these turn out to be the same at low frequency. Or, they did for my motor, give or take 15%. I would probably use the 6 mH number I got from the meter knowing that I got a slightly higher number from the scope approach. I don't think I want to use a higher frequency value simply because nothing is happening at high frequency. I didn't spend any time comparing inductance versus frequency, I picked 100 Hz from the beginning.I have two independent methods of determining the inductance and they both agree within a few percent. At least they're not different by orders of magnitude.As to KE, going back to your reply #20, I get a value of 0.004 V-s/rad. This makes sense because we can calculate the back EMF from the applied voltage. If we ignore friction (or any load), the applied voltage and the back EMF are identical because there is no current flowing. So, from the applied voltage and RPM, we can convert to V-s/rad.I wonder what you would get if you just spun the motor at some high speed and measured the terminal voltage.

author: rstofer
author_rank: Super Contributor
timestamp: May 09, 2017, 02:46:29 pm
content: Searching the Interweb thing, I found other references to the fact that KT=KEfor DC motors. That beats having to try to drag the info out of my 45 year old text book.https://electronics.stackexchange.com/questions/33315/understanding-motor-constants-kt-and-kemf-for-comparing-brushless-dc-motors

author: fishandchips
author_rank: Regular Contributor
timestamp: May 09, 2017, 07:18:14 pm
content: Thanks. The motor you have gave you 6mH but the one I have gave me Ls = 101uH. So, I use 101uH for Inductance?As for the resistance, should I use the calculated 0.09 Ohms from the stall test or 0.97 Ohms from the LCR meter?From previous discussion, we found that KE = 0.004 V-s/rad and KT = KE. How come the KE is much smaller than the one I was given (i.e. 3.55 V-s/rad at lower voltage)? Shouldn't the KE constant be the same regardless of the supplied voltage?

author: rstofer
author_rank: Super Contributor
timestamp: May 09, 2017, 07:48:58 pm
content: Quote from: fishandchips on May 09, 2017, 07:18:14 pmI don't see any reason to not use the 110 uH value. My 1 HP motor had a similar value. We either accept the readings or we don't. The fact that I have two different approaches yielding similar results makes me a little more comfortable for both resistance and inductance.I still have a problem with the 0.09 Ohm value. Yes, your calculations appear to justify it but I just can't get to 90 milli-ohms. That's a VERY low value. Were I you, I would use the 1 Ohm value from the LCR meter. My LCR meter agrees with my bench meters so I have no reason to doubt it.QuoteThink about 3.55 V-s/rad at, say, 20,000 RPM. That's about 2100 rad/s so the motor would be generating:3.55V/(rad/sec) * 2100 (rad/sec) = 7,400 Volts - that simply isn't possible!ETA: The back EMF and applied voltage should be about the same if the motor is running at a constant speed and unloaded. If you put 12V into the motor, the absolute maximum back EMF is 12V (with 0 friction). There's no way you could put 12V into the motor and get a back EMF of 7400V. Since the voltages need to be similar (just call them equal), you can see where 12V couldn't even get that 3.55V motor up to 4 rads/sec.You can get somewhat close to KEby just applying a voltage to the unloaded motor and measuring the RPM. You could just use these values. Or you could simultaneously measure current through the 1 Ohm resistance. Subtract that value from the applied voltage to get the back EMF.Take several readings and plot a graph. It should be a flat line but it probably won't be. Particularly at very low RPMs. If you are into the math, you could apply regression analysis to get an equation from the sampled points. The more points the better. You could try linear regression if the line is nearly flat or polynomial regression if necessary. I haven't tried it but it looks like Matlab can do this.https://www.mathworks.com/help/matlab/data_analysis/programmatic-fitting.html

author: fishandchips
author_rank: Regular Contributor
timestamp: May 21, 2017, 04:02:30 pm
content: Thanks. The simulink model seems to behave similar to the real one but there is a problem when I considered adding a gearbox. In the UMich model, I added a gain factor (of 1/G where G is the gear ratio) at the speed output so that the virtual scope displays a speed similar to the real one. At the torque output, I also added a gain of G so that the torque is amplified by a factor of G. The virtual scope also displays a torque similar to the real one. However, the actual turning does not reduce by a factor of G. I am sure that the current output also does not take the gear ratio and the actual load at the motor shaft into consideration.For example, assuming that I have a test square input voltage signal of 1Hz. Before half a cycle (positive voltage) is completed, the motor have already completed way too many turnings already. This is probably due to the fact that although I amplified the output of the torque, the speed of the shaft is not geared down. What can we do to modify the model so that the gear ratio is taken into account?

author: rstofer
author_rank: Super Contributor
timestamp: May 21, 2017, 04:24:52 pm
content: Matlab has a gearbox model that works for mechanical modeling:https://www.mathworks.com/help/physmod/simscape/ref/gearbox.html?requestedDomain=www.mathworks.comThere is a UMich tutorial for creating the model in Simscape using parameters identical to what is used in Simulink (part of the original page we used earlier):http://ctms.engin.umich.edu/CTMS/index.php?example=MotorSpeed&section=SimulinkModelingWhy not .zip up your model and post it?

author: jmelson
author_rank: Super Contributor
timestamp: May 21, 2017, 04:51:32 pm
content: Yes. On Motor B, all you want to know is RPM and voltage. You can then compute Volts per thousand RPM, which is one of the important motor characteristics.The lower voltage on motor A indicates your power supply was sagging under load.The reason motor A had a higher voltage is it was under load (driving both itself and motor B.) So, the voltage on motor A was back EMF PLUS IR drop in the windings.Jon

author: jmelson
author_rank: Super Contributor
timestamp: May 21, 2017, 04:53:48 pm
content: Quote from: neko efecktz on May 01, 2017, 01:53:12 amBack EMF is defined EXACTLY as the voltage produced when a motor is spun with no load. (It gets way more complicated on series-wound motors, but I assume we are talking about permanent magnet field motors, here.)Jon

author: rstofer
author_rank: Super Contributor
timestamp: May 21, 2017, 06:02:48 pm
content: Quote from: jmelson on May 21, 2017, 04:53:48 pmBut, with no load. If you know the 'R' and measure the running current 'I' along with the terminal voltage, you can get back EMF under the operating condition.Nevertheless, this part of the problem has been solved and an analog model created. Now the issue is adding a gearbox to the analog model. This should involve multiplying the torque output and dividing the RPM. The question is: How to model this in an analog model like Simulink.A side issue: Less losses in the gearbox, which we can model as some scaling factor (related to torque?) and viscous damping (which we already know how to model) the horsepower should remain constant. The product of torque and RPM (divided by 5252) gives horsepower. So, torque goes up, RPM goes down and vice versa.We have a pretty decent analog model:

author: fishandchips
author_rank: Regular Contributor
timestamp: May 21, 2017, 06:36:16 pm
content: Thanks. I just added the scales and scopes. For your convenience, I highlight the parts that I have added.https://flic.kr/p/TMWmoTThe scope outputs look correct but it is kind of cheating because even the speed is decreased as shown in the scopes, the actual speed output is the one without gearbox. Similarly, although the torque have been amplified by the Gear Ratio G, when driving by a periodic signal, it does not oscillate with the signal but oscillate at much higher speed. I guess the output of the current also does not match the real one that has both gearbox and load mounted to the unloaded motor that we simulate.I know about the matlab gearbox block. I think the effect is similar to what I did but adding scaling factors at the speed and torque output. Am I correct?

author: rstofer
author_rank: Super Contributor
timestamp: May 21, 2017, 07:33:53 pm
content: The unloaded motor ('b'=0) generates no torque so there is nothing to multiply. It also draws no current because the applied voltage and back EMF are equal.I put b=1e-3 and things get a little more realistic but we still have no load on the gearbox. I might add a bit of viscous damping to the torque output of the gearbox and feed it back to the same adder than handles 'b'. I don't know if I would leave the existing damping or just damp the output.As it is, 'b' is the only load once we get up to speed. Of course, the inertia thing comes into play at startup but it doesn't last long.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 21, 2017, 08:45:24 pm
content: Quote from: rstofer on May 21, 2017, 07:33:53 pmDo you mean if I connect the DC motor circuit so that the torque from the motor is used to drive an object, some kind of feedback loop is created which affect the current, speed and torque of the motor?I recall testing it by using the torque output of the DC motor circuit to make an object to turn. The actual turning of the object happened much faster. For example, the object is supposed to make a full rotation in one second. However, in simulink, it completed the rotation in 0.004 seconds. Anybody knows what is wrong?

author: rstofer
author_rank: Super Contributor
timestamp: May 21, 2017, 10:01:13 pm
content: Quote from: fishandchips on May 21, 2017, 08:45:24 pmI would think so... If I load the motor, it slows down. This reduces back EMF and increases current. The increase in current also operates on the internal resistance causing heating (about which we aren't concerned at the moment, it has to do with efficiency). About the only parameter that isn't affected, long term, is the inductance and its di/dt. This makes sense because we are talking about a constant speed.Here is some more math:https://www.precisionmicrodrives.com/tech-blog/2015/08/03/dc-motor-speed-voltage-and-torque-relationshipsOr a graph where RPM drops linearly with torque and current increases linearly with torque. Notice the equation a little farther down:M0=Mf+MlTotal torque = Motor friction torque ('b') + load torque.http://www.micromo.com/technical-library/dc-motor-tutorials/motor-calculations#numericalcalculationI can see another input to the adder where 'b' is applied.I like your idea of doing units conversion. That's the one thing I don't have a good feel for in the simulation, units.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 21, 2017, 10:41:58 pm
content: Do you have an idea why an object driven by the torque of the DC motor model completed the rotation in 0.004 seconds in Simulink rather than 1 second in the real world?

author: rstofer
author_rank: Super Contributor
timestamp: May 22, 2017, 12:47:05 am
content: I have always had a problem with the value for J (we're calling it IL), the rotational inertia. The motor accelerates in near zero time because there is no inertia to overcome.You calculated the inertia much earlier in the thread and the numbers seemed right but somehow they don't present enough of a load to slow the acceleration.

author: fishandchips
author_rank: Regular Contributor
timestamp: May 22, 2017, 01:26:17 am
content: Is there another good way to measure the moment of inertia of the rotor?

author: rstofer
author_rank: Super Contributor
timestamp: May 22, 2017, 04:24:49 am
content: I found a recommendation to just measure RPM versus time for a step input. The problem is, how to measure the RPM as it is changing so fast.Maybe it can be done by connecting another motor as a generator but that will add even more inertia. So will any other physical method I can think of. But if the motor and generator have similar moments of inertia, maybe the math is pretty easy.You still need a way to capture the voltage waveform from the generator. A scope is the way to do this. I know you're considering one so that may turn out to be good timing.Are you confident in your measurements, units and calculation of 'J'?BTW, your gearbox and ultimate load also contribute to rotational inertia. Not just viscous damping (drag) but another source of inertia. Probably modify the existing constant being used for IL

author: fishandchips
author_rank: Regular Contributor
timestamp: May 22, 2017, 12:41:35 pm
content: Thanks. I increased the inertia in steps of 100 and tested. Even it is amplified to a 7 digit number, it is still spinning very quickly.I have a tachometer to measure the rpm of the real motor.

author: rstofer
author_rank: Super Contributor
timestamp: May 22, 2017, 01:41:41 pm
content: The thing about the tach is the issue of capturing the change or ramp in RPM in fine enough steps to be useful in determining inertia. If the motor accelerates in near zero time, the tach will never see anything other than the final value.Ramping of the terminal voltage in a coupled unloaded motor seems ideal. Adding resistors across the terminal will add even more load to the driving motor.I don't have any problem with inertia causing a huge curve in the various parameters. I have attached a .zip of my model and the .mat files with the variable definitions. I'm sure you know but 'load' DCmotor2.mat and 'open' DCmotor2.slx then run.

author: rstofer
author_rank: Super Contributor
timestamp: May 23, 2017, 01:46:18 am
content: About that 'b' value, I wonder if we should add just enough damping to account for the no load current perhaps including the gearbox.

author: fishandchips
author_rank: Regular Contributor
timestamp: July 22, 2017, 12:00:29 pm
content: I am still stuck with this problem so I bought an oscilloscope and used it to observe the input signal to the DC motor. It is a train of square pulses with amplitude 11.1V and width 300 ms. I also observed that with each square pulse to the motor, the shaft of the DC motor makes about 1.25 turns.I used a signal generator block in simulink to generate one square pulse with the amplitude and width mentioned above. Then, I used it as a input signal to the DC motor model to turn a rotational joint. As soon as the signal generator generates the rising edge of the square pulse, the joint turns many times very quickly even I attached a very heavy load to it. I am unable to make the DC motor to turn slowly like the real one (i.e. shaft makes 1.25 turns in 300 ms).

author: rstofer
author_rank: Super Contributor
timestamp: July 28, 2017, 06:15:02 pm
content: I don't understand the pulse. What is the frequency and what is the duty cycle? 300 ms doesn't seem to be enough information (to me). Unless it is a one shot pulse of 300 ms width.If you pulse the motor and it turns 1.25 revs, does that mean it moves from stop to some rpm and back to stop in 300 ms?If the model turns more than the real motor, there is something wrong with either the friction or inertia in the model. I don't have a similar motor so I really can't do much in determining the constants.It looks like 'b' in my Motor2 Matlab project is viscous friction.I revised the project to display 'w' the total rotation and I get around 9 revolutions in 300 ms. Reduced by 10x for the gearbox, this would be about 0.9 revs. The thing is, inertia keeps it rotating. The RPM very slowly drops off. You can see this by changing the simulation run time from 0.5 sec to, say, 100 sec. You will see that it takes nearly 1 minute to stop rotating. So, we have a boatload of inertia and no friction. If 'b' in my .mat file is truly friction, and I think it is but I forget, it's set to a very small number 5x10-6.Since I didn't know how to define a 300 ms one-shot pulse, the pulse generator sends a 300 ms pulse every 100 sec (I guess I could have chosen a bigger number) so don't simulate beyond 100 seconds without changing the generator.Attached .zip file contains the .mat and .slx files.

author: rstofer
author_rank: Super Contributor
timestamp: July 28, 2017, 08:17:53 pm
content: Set b=0.012 and simulation time to 2Look at the 'w' graph, it maxes out at 12 which when reduced by the gearbox is 1.2 revs.Originally b=5.0e-6which is a very small amount of friction.I still don't have any confidence that after changing 'b' that the model represents anything like the real motor. I'm still not convinced about inertia. I suspect that the parameters still need a little tweaking. I didn't go back and put in an 11.1V step function instead of the pulse.ETA: Nuts! 'w' is in radians, not revs. w(t) is in rads/sec and so on. So a little more tweaking of the constants will be required. But, it seems like some adjustment to 'b' is the way to go.

author: rstofer
author_rank: Super Contributor
timestamp: July 28, 2017, 10:41:58 pm
content: So, 1.25 revs is about 8.2 rads. Multiply backwards through the gear box and w=82b=1.67e-3 gets very close.