"DIY Constant Current Dummy Load" [EEVblog #102] - Page 1
Forum section: electronics / general

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 16, 2010, 01:54:10 pm
content: Hi...I've two DC generators, namely Generator-1 and Generator-2.When the Generator-1 supplies 1.5A, the output current waveforms (see attached figure, measured using Tektronix TCP312 current probe with AC coupling) shows peak-to-peak current ripples of 40mA.When the Generator-2 supplies 37mA, the output current waveforms (see attached figure, measured using Tektronix TCP312 current probe with AC coupling) shows peak-to-peak current ripples of 8mA.The frequency of the current ripples is about 200HzKindly refer to the circuit as shown in the following link:EEVblog #102 – DIY Constant Current Dummy LoadWill the current ripples be eliminated or reduced if the positive output terminal of the generator is connected to the "constant current dummy load" presented in the EEVblog #102 (see attached figure on setup).Well, the circuit may need little modification. For example, the "1R" need to be replaced with a proper value to optimise the current measurement range.Kindly advise.Thanks

author: Simon
author_rank: Unknown
timestamp: September 16, 2010, 04:22:33 pm
content: why do you need to reduce the ripple so much, is it causing you problems ?You could use another LC cell to further filter the ripple. I'm not sure a constant current source (which is what your asking for) will be much help as that too will have it's own ripple. All ripple is generated by the fact that the regulator needs some time to readjust to new conditions and that it will over and undershoot slightly and then correct itself which is what the ripple is

author: Time
author_rank: Frequent Contributor
timestamp: September 16, 2010, 04:36:30 pm
content: Even if it does solve your problem, I think that would be a very inefficient way to drive your motor.

author: scrat
author_rank: Frequent Contributor
timestamp: September 16, 2010, 05:01:27 pm
content: If I understant it right, you want to do an active power filter, something that "absorbes" the spurious (ripple).Usually it is done by controlling the voltage, since the quality of your supply is measured in terms of voltage precision, not current precision (this last indeed depends on how your load varies). You could also put a simple L-C filter, since power is not so high.However, typical active power filters are made by controlling a switching converter which input is the line to "clean" and which output is connected to a capacitor. A controller forces the bus-capacitor voltage to be constant by imposing to the "filter" a proper input current._________------o-------------| |___ | |LOAD |bus------|-----o-------|_____|cap _|____|_| |____| active | ---| filter | --- cap (tank)|_______|-----

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 16, 2010, 07:04:09 pm
content: Quote from: Time on September 16, 2010, 04:36:30 pmThanks. I'm aware that it's not an efficient way to drive the motor. I need a constant load torque for some tests. The current ripples in the armature and field current of the generator may result in unwanted torque ripples.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 16, 2010, 07:08:12 pm
content: Quote from: Simon on September 16, 2010, 04:22:33 pmI suspect the current ripples introduce some error in my experimental measurement. If the bandwidth of the "constant current source" control circuit is high, it should be able to reduce the current ripples. Please correct me if I' wrong.

author: jahonen
author_rank: Super Contributor
timestamp: September 16, 2010, 07:29:25 pm
content: Have you tried measuring the current noise level when the constant current load is turned off, to see how clean your current measurement setup is? Or in other words, if the ripple you see is just a measurement artifact.It is also possible that the current source has low level oscillation. In that case, frequency compensation should be added. This will require some experimenting. For example, add a resistor between negative input and sense resistor (something like 1-10k), and then capacitor (something like 10 nF) from opamp output to negative opamp input.Regards,Janne

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 16, 2010, 08:06:31 pm
content: Quote from: jahonen on September 16, 2010, 07:29:25 pmHi JanneI've not built the "constant current source" yet. The measured current waveforms shown in the first post of this thread were captured using Tektronix TCP312 current probe without the "constant current source". If "constant current source" may help to reduce or eliminate the current ripples then I will build one.Thanks.

author: scrat
author_rank: Frequent Contributor
timestamp: September 16, 2010, 11:55:35 pm
content: Quote from: onemilimeter on September 16, 2010, 08:06:31 pmSorry for my previous post, I didn't read too carefully your question. However, the thing you want to do can be called an active power filter.You're right, if the bandwidth of your current regulator is much higher than the disturbance oscillation frequency, then it wil work.Current sensing has to be done on the current going to motor (I'd put it from armature return and gnd), while the transistor controlled by the opamp has to be in parallel with the motor, then shunting to gnd all non-DC current.The only problem I see is the fact that the "dummy load" can't source current, so you have to set generator's DC current output to be higher than the needed motor load current, and then make the "dummy load" draw a little DC current plus all the ripple (so subtracting to Igen an always positive current Idummy). This leads to the fact that Ref has to be carefully set, and that an amount of continuous power has to be dissipated by the MOSFET.--> Igengenerator o---------------o-----------------------| | Idummy | | Imotor| \ __| \/ | \/Ref o--|+ \______o_____|| | |__|- / | ||<-| (M)| | / | _|_ || | /// || | |------||-----o------/\/\/----------------------oC R |<> Rsense<>|_|_///

author: Simon
author_rank: Unknown
timestamp: September 17, 2010, 06:33:59 am
content: is that ScratCAD ?onemillimeter: is it possible that your over thinking this ? The ripple is more or less repetitive and predictable, if you take the voltage as the mean/average of the ripple then you will be ok, a motor will not react instantly to such a small change in voltage and will in fact smooth it out acting as a physical/mechanical low pass filterIf you want to supply a constant current to the motor then that's another matter, perhaps you should explain your project further

author: scrat
author_rank: Frequent Contributor
timestamp: September 17, 2010, 08:35:54 am
content: I saw this way of drawing in a post of someone at this blog, and have fun using this "CAD". I can't call it ScratCAD!Perhaps I like this "technique" because it makes me remember the days when I started programming in Pascal, and made some graphical things in char mode...About the schematic, "the night brought an advice" (as we say in Italy): to make the controller work you surely have to invert polarity of the opamp, since it has to increase MOS current when voltage on Rsense rises. So Rsense has to be connected to + input, while Ref has to go on the inverting one. There is a possibility to try estimating the bandwitdth of your control, but it depends heavily on real MOSFET transconductance, so I think the effort is not that worth.Another thought I had this morning is that in the case you disconnect the motor you would probably have the MOSFET shorted to gnd or nearly, which means it could go into overcurrent. This can be prevented by limiting gate voltage in some way (below you just have to tune the potentiometer to set max current to the needed value).o +Vcc||opamp \output diode /o-/\/\/\-----o---|>|---------> \ pot.| /| \| || || _|_| ///||___________\ to MOS gate/

author: DJPhil
author_rank: Frequent Contributor
timestamp: September 17, 2010, 10:17:28 am
content: If ScratCAD is open source, I have a suggestion.Consider using a code block for schematics. It uses a fixed width font, so it's easier to copy and paste fixed length segments.Code:+-------+-------+-------+-------+-------+-------+-------+-------+|One |Two | Three | Four | Five | Six | Seven | Eight |+-------+-------+-------+-------+-------+-------+-------+-------+Just an off base idea to add to the pile.Edit: Server errors are bad this . . morning?

author: jahonen
author_rank: Super Contributor
timestamp: September 17, 2010, 01:37:12 pm
content: Quote from: onemilimeter on September 16, 2010, 08:06:31 pmYou are using the other motor as mechanical constant torque load? Yes, it might work then, if voltage generated is DC (no polarity change).Regards,Janne

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 17, 2010, 07:13:37 pm
content: Quote from: scrat on September 16, 2010, 11:55:35 pmHi scratThanks for your advices.I guess many of you have not seen my application correctly yet. I'm very sorry for my poor explanation.I've three electrical machines. The first one (the one with red color as shown in the experimental setup diagram) is used as prime mover (i.e. as a motor). The other twos are used as generators and labelled as Generator-1 and Generator-2, respectively. For each test, only one generator (the one with blue color, which can be either Generator-1 or Generator-2) is coupled to the prime mover. Now, let's say Generator-1 is coupled to the prime mover. The armature terminals of the Generator-1 are connected to a resistor (the one with black color) as show in attached figure in this post. The current waveform captured using Tektronix TCP312 current probe is given the first post of this thread. Clearly, the armature current has ripples.Now, I wish to construct a "constant current load" to replace the resistor. I wonder if the "constant current load", if the regulation bandwidth is high enough, can eliminate or reduce the current ripples.

author: Simon
author_rank: Unknown
timestamp: September 17, 2010, 09:18:42 pm
content: so your trying to put a constant load on your generator ? why ? yes a generator will have unstable output but really you need to look at the generator's control circuitry and make that smooth before fixing the output

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 17, 2010, 10:15:16 pm
content: Quote from: Simon on September 17, 2010, 09:18:42 pmThanks. I use the generator as a load for the motor (prime mover). I wish to study the torque characteristics of the motor. If the armature current of the generator has ripples the generator may induce torque ripples which in turn may affect the the torque produced by the motor. However, recalling the Newton's "action and reaction" law, I'm not sure the way I think is correct or not. Please comment.

author: DJPhil
author_rank: Frequent Contributor
timestamp: September 17, 2010, 11:22:41 pm
content: I think I might have some insight, but this is mostly guesswork, fair warning.I've deleted a few paragraphs already but I think I understand now. If you're getting armature ripple with just a resistor then it's likely caused by the non-ideal nature of the generator. I don't think there's anything you can do electrically to smooth the generator's load behavior. The best solution I can think of for a smooth, steady motor load would be mechanical. There's a heap of purely mechanical ideas based on friction, gearing, and inertia. These are much more direct and linear loads than a generator, as they'll go straight from motion to heat or mechanical resistance. Imagine the equivalent of a vehicle torque converter from an automatic transmission with one side on the motor and the other side under heavy friction or stopped altogether (grounded emitter? hehe). Calling all mechanical engineers, we've got a live one for you!If you must use the generator for a load I think you're stuck with the ripple, but I had a useless idea/brain vacation about that as well. I wonder what would happen if you took an LM317 and wired up a variable supply with suitable protected pass transistors for high current operation and powered it with the generator. The regulator and pass transistor set will look like a current source. Then, error amplify the ripple and use it to modulate the vol . . . ok, yeah. That's useless, but I'm going to think more about it anyway.I hope that helps some, kinda goofy today.

author: Simon
author_rank: Unknown
timestamp: September 18, 2010, 07:07:32 am
content: oh dear,Look your electrical ripple is not going to make the load have "mechanical ripple" because your generator will not react that fast, the whole mechanical side will act as a low pass filter and the actual mechanical load will be quite smooth, your worrying over a non existent problem.it is also likely that the ripple is a produce of the generators electrical properties and therefore that is the result you get with a constant mechanical load

author: scrat
author_rank: Frequent Contributor
timestamp: September 20, 2010, 03:51:17 pm
content: Yes, since your load is only made by a resistor, ripple is only caused by motor non-idealities and/or measurement noise.The only way to be sure to apply a very stable torque only if you measure it (torque meters are quite expensive, unfortunately) and put its signal into a feedback loop.I partially agree with Simon, Generator-2 ripple is quite fast and, depending on your motor's inertia, could be negligible, while Generator-1 has a component at about 100Hz, which could easily end in some oscillation on speed.However, current regulator could be helpful in substitution to the resistor, and will make load easily variable. Don't know if you need it.

author: scrat
author_rank: Frequent Contributor
timestamp: September 20, 2010, 04:01:46 pm
content: Quote from: DJPhil on September 17, 2010, 10:17:28 amGood idea, I never considered that... But perhaps SomeOneCAD (as I said, it isn't an invention of mine) won't have a very wide usage.Is there any simple program to make schematics, with the possibility of making a symbol taken from an image or something like that?I mean something completely different from Orcad, WEWB, ... a program where you can make a schematic, add schetches and writings, and export an image (so anyone can open it), very quickly. And where if you need a symbol, you get an image and create it in 5 seconds. If there isn't, we could make one! If the saving format will be very simple to understand, there could even be a "command line" mode, to add components.Sorry for the off-topic, but a thread only for this would have been really ridiculous.

author: Simon
author_rank: Unknown
timestamp: September 20, 2010, 04:57:26 pm
content: express pcb is very sinmple and easy to use and create custom parts for, as for adding sketches that's down to your image editing skills, expresspcb can also save schematics as any number of image formats

author: george graves
author_rank: Super Contributor
timestamp: September 21, 2010, 12:51:26 am
content: Has anyone built this per Dave's blog?I was actually looking to make something like this, but I was going to use a series of automotive 12v incandescent bulbs, driven off a mosfet by a 555 controlled by a pot. I was just going to throw come large caps on the input and output to smooth the pwm ripple. Is there something wrong with that idea?Anyways, I though I would give Dave's plan design a try first. I was surprised that I did indeed have most of the parts on hand. That 50k 10 turn pot is *expensive* ($15-30) so I went with a trimmer instead. Not as elegant - but I might follow his advice and add a coarse/fine adjustment.The only question I had is this. The resistor(s) isn't there to dissipate the energy? That's weird to me, that - well that and the I'm not use to seeing anything between the "source" pin and ground. I'm use to seeing the "load" on the drain pin. What's up with that? Is there an easy explanation?

author: Zad
author_rank: Super Contributor
timestamp: September 21, 2010, 01:46:03 am
content: If you are driving it directly from a 555 without a feedback loop, then the big problem you are going to find is that 12v car lamps (and for that matter any incandescent lamp) are far from linear. For example, you could set the sink to 100mA. Initially the filament would be cold and the load current would rise quickly, but then the filament would gradually heat up, the resistance rises, and the current drops off again. It would be possible that this might make the load oscillate.I wouldn't worry about it being specifically a 50k 10 turn pot, 5k, 10k, 20k, 3 turn, 5 turn, 20 turn would all do a similar job. Separate coarse/fine is a perfectly good solution too.

author: george graves
author_rank: Super Contributor
timestamp: September 21, 2010, 02:46:29 am
content: Ahhhh. I see said the blind man. Hadn't planned on them being so non-linear. Thanks for the tip.

author: Simon
author_rank: Unknown
timestamp: September 21, 2010, 06:49:03 am
content: Quote from: george graves on September 21, 2010, 12:51:26 amthe mosfet is controlled by the voltage across the source and gate, the 1 ohm resistors will never see more than 1 V if I remember rightly ? so the gate will always see 1V less than is coming in as the source is held 1V up from GND, i suppose this also helps stability because if the current drifts to increase the voltage on them increases and so the voltage on the gate decreases. There is of course plenty of head room to so supply the gate enough voltage,

author: scrat
author_rank: Frequent Contributor
timestamp: September 21, 2010, 09:37:50 am
content: Quote from: george graves on September 21, 2010, 12:51:26 amThat resistor acts as a sensing resistor. Dave put it between source and ground because it is far easier to pick up a voltage referred to ground, to use it for feedback (it goes to the opamp). It has to be choosen so that its voltage is small (to avoid resistance variation due to heating), but sufficiently large to be measured. If you use an opamp in a differential configuration, you could pick up current proportional voltage from any other place in the circuit, but in this case it isn't much worth.I think that a good tool could be an MCU-controlled PWM load, that you could control from PC or from a simple LCD+buttons. Feedback could pass (after a little amplification, if needed) through MCU's ADC, since a large bandwidth is not required. Then you directly generate PWM signal for the MOSFET. To verify control precision you still can measure the resistor's voltage with a DMM.

author: Simon
author_rank: Unknown
timestamp: September 21, 2010, 11:39:11 am
content: sounds like a good idea

author: scrat
author_rank: Frequent Contributor
timestamp: September 22, 2010, 10:47:05 am
content: Quote from: Simon on September 20, 2010, 04:57:26 pmThanks! I tried the schematic tool inside ExpressPCB. It's quite good, but I will take another look around, since there are some annoying features (grid and wire connection behaviour, for example). Of course, there is no perfect tool...

author: Simon
author_rank: Unknown
timestamp: September 22, 2010, 11:31:57 am
content: its chaep (free) and nasty but can get a job done quick i'm trying to migrate to KiCAD

author: george graves
author_rank: Super Contributor
timestamp: November 13, 2010, 03:15:05 am
content: What do you think would be a good way to expand dave's dummy load to handle more current?I was thinking of just adding more 3055's and run them in parallel with more sense resistors not connected to the opmap? Or is there a better way?

author: scrat
author_rank: Frequent Contributor
timestamp: November 15, 2010, 09:55:13 am
content: Sense resistors, by definition, must sense current for feedback. If you don't connect them to the opamp, they just become dissipating loads, which could be good, but you have to sense ALL of the current and send it as a feedback to the opamp.Putting BJTs in parallel can be a problem if they do not have each a resistor in series. Otherwise, the one which conducts slightly more current will heat the more and so increase its conductivity. Then it will burn, and the other transistors will do the same one at a time.MOSFETs are good for paralleling.

author: george graves
author_rank: Super Contributor
timestamp: November 19, 2010, 01:18:38 am
content: Interesting. So how would you expand upon Dave's design handle more current?

author: scrat
author_rank: Frequent Contributor
timestamp: November 19, 2010, 10:21:18 am
content: I'll use many parallel resistors, with a BJT or MOSFET in each leg. Then all of these must be in series with a single sense resistor, which voltage will be the feedback for the opamp. So you can split power dissipation among many resistors and transistors.If you need a very high max.current/min.voltage ratio (= a very low min. resistance) then paralleled MOSFETs are the only solution, IMHO.There is at least another recent (the last posts were about a month ago, I think) thread on this topic, a good search should point it out. There was a big dicussion, and some schematics. If I find it, I will post the link..

author: scrat
author_rank: Frequent Contributor
timestamp: November 19, 2010, 10:31:22 am
content: I found the thread I mentioned above:https://www.eevblog.com/forum/index.php?topic=1049.0

author: scrat
author_rank: Frequent Contributor
timestamp: November 19, 2010, 12:54:20 pm
content: I'd try with paralleled MOSFET+resistor followed by sense, with a single controller.Maybe power resistors can help lowering the cost (smaller MOSFETs and wider power distribution).I think multiple controller won't be necessary, unless you need a very low resistance and thus you have to eliminate the power resistors.

author: scrat
author_rank: Frequent Contributor
timestamp: November 19, 2010, 02:14:11 pm
content: @AcHmed99: Pleased to learn from who has much more experience than me...I didn't consider a very wide output range.@george: The point is (as usual) that specs are the base for almost any design option, it is not possible to give generic answers.