Question on zero-center ammeter - Page 1
Forum section: electronics / beginners

author: RLSprouse
author_rank: Contributor
timestamp: April 20, 2016, 11:12:54 pm
content: I'm a beginner, working through several resources including Owen Bishop's "Electronics: A First Course, Third Edition." In that book, in Topic 17 on p.59, he shows a simple circuit to demonstrate a capacitor charging and discharging. The circuit includes an analog ammeter rated at -5 mA to +5 mA. He's using analog meters so we can see the rate of change, i.e. the time it takes for the cap to charge and discharge.I've only been able to locate a few little zero-center analog ammeters, and nothing close to that rating. Most are 1A and up, I guess for automotive use, and then 100 uA or 250 uA at the other extreme. I was thinking about how to rig up a substitute, and I am wondering if the following would work. Suppose I have two identical bare movement ammeters of, say 10mA capacity. Could I hook them up in parallel, but with the polarity reversed on one, and with diodes on each to prevent reverse polarity through the movements? I figure that way one of the meters would show the positive current and the other would show the negative current.Is my thinking reasonable?

author: Delta
author_rank: Super Contributor
timestamp: April 20, 2016, 11:35:53 pm
content: Don't bother with diodes - they'll mess up what you're trying to do. Just use the back-to-back ammeters, these things are tough, and will happily take reverse current - worst case: a bent needle!You might even be able to adjust the needle on one so it sits mid scale at zero current...

author: CatalinaWOW
author_rank: Super Contributor
timestamp: April 20, 2016, 11:46:08 pm
content: The other approach would be to get one of the low current ones and make or buy an appropriate shunt to bring it to the desired range.

author: Ian.M
author_rank: Super Contributor
timestamp: April 20, 2016, 11:47:19 pm
content: Don't use diodes, because you cant get an ideal diode with a simple circuit, so the series diodes will cause a non-linear voltage drop across the combo. Also, it will be very clumsy to use.What you should do is take a more sensitive center zero meter movement, and add a shunt across it. The shunt is a precision low value resistor that takes the bulk of the current only leaving the tiny fraction the meter movement can handle to pass through the meter coil. The details of how to select, connect and calibrate a suitable shunt are rather esoteric, and there are several pit-falls that can destroy the meter movement by overloading it (even something as simple as trying to measure its resistance with an ordinary DMM can be disastrous) so if you can get hands-on assistance from someone who has experience as an analog instrument tech, do so, otherwise you've got a fair bit of research and work ahead of you learning how to shunt a meter movement.Another option would simply be to use a simple cheap analog multimeter and reverse the meter's connections as needed to keep the meter deflectioin on-scale. You'll only have to do so when swapping between charging and discharging.You'll need a *VERY* large capacitor to get a reasonable time constant if initially charging/discharging at 10mA, something like 100,000uF, but large capacitors generally have a very inaccurate value, and can be as bad as +100%,-50% of the nominal value which will confuse the issue, so are you sure this circuit is a practical experiment not just a thought experiment?

author: barry14
author_rank: Regular Contributor
timestamp: April 20, 2016, 11:50:00 pm
content: You can use a more sensitive meter by placing a resistor in parallel to the meter to shunt most of the current (which is why it is usually called a shunt resistor). Its value depends on the internal resistance of your meter and the desired current scale (the meter resistance and the shunt resistance form a current divider). If you don't know the meter's internal resistance, you can experiment with different resistors until you get the desired full scale current.

author: Richard Crowley
author_rank: Super Contributor
timestamp: April 20, 2016, 11:55:19 pm
content: Is there some particular reason you need an antique analog meter? Most (all?) DMMs will be quite happy to read - and + milliAmps.What is your budget? How much money is it worth to have a low-current zero-center analog meter?I see many zero-center low-mA meters available online. For example:https://www.surplussales.com/Meters/MtrDCMA.htmlhttp://www.ebay.com/sch/i.html?_from=R40&_trksid=p2050601.m570.l1313.TR0.TRC0.H0.Xpanel+meter+zero+cent.TRS2&_nkw=panel+meter+zero+center&_sacat=0

author: RLSprouse
author_rank: Contributor
timestamp: April 21, 2016, 12:00:47 am
content: Thanks, guys, for the good input.I thought of another solution, which is to set up the switch so that when you throw it to discharge the cap, it also reverses the polarity of the meter connections.@Ian.M: "...are you sure this circuit is a practical experiment not just a thought experiment?" Well, in most of the Topics (chapters) of the book, he has sections labelled "Things to do" and they are written so as to appear to be hands-on experiments, which is the sort of thing I love doing. I seem to learn much better that way.For example, in this particular one, he has steps like this:1 Set the switch to A to discharge the capacitor if it is already charged.2 Set the switch to B and watch the meters as the capacitor is charged. You and a partner could watch one meter each.3 Set the switch to A and watch the meters as the capacitor is discharged.4 Repeat 2 and 3 until you can answer these questions... (and so on...)This doesn't sound like a thought experiment to me, it sounds very hands-on.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 12:28:28 am
content: Quote from: Richard Crowley on April 20, 2016, 11:55:19 pmAt SurplusSales they have a 1-0-1 mA meter for $25 that would be handy to have around. They have a nicer one for $39 that has a mirrored scale. Although it doesn't look like it, the $39 meter is somewhat larger and has a more useful scale.I would figure out a shunt resistor equivalent to 1/4 the meter resistance to cover the 5-0-5 range. Or I would scale down the experiment.I could really use one of those meters...

author: RLSprouse
author_rank: Contributor
timestamp: April 21, 2016, 12:35:21 am
content: Thanks again, guys.The reason for the "antique analog meter" is so that you can observe the rate of charging and discharging. Analog meters give you a nice "feel" for what's going on.I think I will order one of the -1 - 0 - +1 mA meters from Surplus Sales and see if I can figure out the right shunt resistances to make it measure a -5 - 0 - +5 mA range and -10 - 0 - +10 mA range.For me, it's all about the learning... and the fun I have while doing it.

author: Ian.M
author_rank: Super Contributor
timestamp: April 21, 2016, 12:46:48 am
content: Ok. that does sound like it is meant to be a real practical circuit. What value capacitor does it use? Unfortunately page 59 of the book isn't available on Google Books, but you could post a scan of just that circuit diagram and the instructions under fair use, and we could then suggest where to wire in two ordinary milliAmmeters, one in the charging circuit and one in the discharging circuit, so they each only have to deflect one way, to replace the center zero meter you don't have.OTOH a 1ma-0-1mA meter + a selection of shunts is a very useful item to have handy for a serious analog electronics student.If you cant easily get the 1mA meter, assuming you have a good DMM, an adjustable PSU and a reasonable selection of resistors, we could also talk you through measuring the coil resistance of a 100uA or 250uA meter movement without blowing it and discuss how to add switchable shunts and series resistors to make a low-spec center-zero analog multimeter (DC mA and V only)

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 01:14:04 am
content: Or draw a hand sketch of the circuit using whatever method you prefer and post it.I ordered one of the 1-0-1 meters ($39) just to have it on my bench. The meter internal resistance should be marked on the unit, somewhere.I built an analog computer project a year or so back and I could have really used a center zero meter. Now I can get rid of the complex precision rectifier and LEDs to indicate polarity. Outstanding!Assuming you have a way to create a .pdf, almost any program can be used to draw the schematic. LTSpice, ExpressPCB, Eagle, whatever.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 01:40:25 am
content: Quote from: RLSprouse on April 21, 2016, 12:35:21 amElectronics is a subject that is best learned by doing. There's nothing quite equal to actually building a circuit and making real measurements.

author: RLSprouse
author_rank: Contributor
timestamp: April 21, 2016, 03:11:27 am
content: I am attaching a scan of the specific "Things to do" section I am inquiring about. It would be interesting to see how any of you knowledgeable folks would approach the question, lacking the specified zero-center ammeter.I definitely enjoy learning by doing. It is the only way it all seems to "stick" in my slightly ADHD afflicted brain.

author: Ian.M
author_rank: Super Contributor
timestamp: April 21, 2016, 03:38:09 am
content: Thanks for that.The circuit has a few issues. As the supply voltage is 6V and R1 is 10K, the maximum possible current is 0.6mA. A 1mA-0-1mA meter would be ideal - you certainly don't need a 5mA meter.You will be lucky if the analog voltmeter has an impedance of more than 500K, and a cheap one might be as little as 100K, so the capacitor will discharge through the voltmeter in a couple of minutes if the switch was left in the charge position but the supply disconnected.To observe the circuit's behaviour without a center zero milliAmmeter, remove the milliameter (shorting it out) and the switch from the schematic. In place of the switch use an analog multimeter on a low mA DC range or a panel meter. It needs to have 1mA or greater FSD, but as low as possible. To observe charging, connect its + lead to the +6V supply and its - lead to the left end of R1. To observe discharging, connect its + lead to the left end of R1 and its - lead to 0V.*DO* *NOT* connect it between +6V and 0V or you'll damage it.Alternatively, use a DPDT switch and wire the circuit as below.which will automatically connect the milliAmmeter in the correct direction for charging and discharging. The DPDT switch is shown in the discharge position.Sorry about the horrible digital meter symbols - they were the only meters the schematics package I drew it in had. You should use analog meters or at least DMMs with a bar graph 'trend' display.N.B. when the capacitor is fully charged the milliAmmeter wont read exactly zero due to the load current of the analog voltmeter.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 02:08:35 pm
content: A word of caution about the 1-0-1 mA meter: When you receive it, there will be a wire shorting the terminals. This keeps a short circuit on the winding and this prevents damage to the meter during transport. Some multimeters, like the popular Simpson 260 have a specific 'transit' setting on the selector switch. In effect, it places a resistor across the meter.When you are done with your experiment, replace the shorting wire.t = RC or 10^4 ohms * 10^-3 farads = 10 seconds. That is ONE time constant, 10 seconds, and you will want to observe out to 6 time constants.This is an application where there is nothing quite as useful as the Digilent Analog Discovery tool in Scope mode with a setting of 6 seconds per division. The entire 60 seconds will be on one graph which can be measured using the mouse or cursors and annotated prior to printing. Very cool!There's a lot of utility in that little box:http://store.digilentinc.com/all-products/scopes-instruments/I have the older model.

author: RLSprouse
author_rank: Contributor
timestamp: April 21, 2016, 07:34:52 pm
content: @Ian.M: Wow, it is fantastic that you have redone the circuit for me, thank you so much! The DPDT switch is exactly what I had in mind, but I had not figured out how to hook it all together. This is perfect! I was planning to use an inexpensive little Radio Shack analog multimeter for the voltage measurement. I don't know the impedance... The User's Guide pamphlet says, Sensitivity - DC Voltage 20000 Ohm/Volt. Does that tell us anything? I also have an old RCA WV-77E VoltOhmyst VTVM that miraculously still works. I have no idea what the impedance of that would be, nor how to determine that.@rstofer: Thanks for the warning and additional info. The Digilent Analog Discovery tool looks VERY interesting; I am a sucker for such gadgets. I already have a National Instruments myDAQ and the ELVIS software. I wonder if that is similar to the Digilant gizmo, and if I could use it for the purposes of this exercise.So much to learn!

author: Ian.M
author_rank: Super Contributor
timestamp: April 21, 2016, 08:09:31 pm
content: There's a RCA WV-77E manual here:http://bama.edebris.com/manuals/rca/wv77e/It says 11Meg when using the original DC probe. The DC probe has a 1M resistor in series with the tip, so if you've lost it you'll need to make up a test lead with a 1Meg 1% resistor (preferably a high voltage one) in series with the probe tip, as if you use it without it will be 10% out of calibration.Its an ideal analog voltmeter for this sort of stuff, and if you connected it accross the 10K resistor in the original circuit you could use it in center zero mode on its 15V range to read the voltage across the resistor which would directly give you the current by ohms law.The multimeter's 20K/V sensitivity is simply multiplied by the FSD of the range you are on to get the input impedance of all its voltage ranges. On a 10V range it would be 200K.I'd use the VTVM for voltage as it will not discharge the cap significantly for a considerable time, and the cheap multimeter for current. If you use a DPDT center off switch, you can pause the charge/discharge at any point by putting the switch in the off position and observe the change in capacitor voltage due to the load of the voltmeter and the capacitor's internal leakage.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 09:01:49 pm
content: Quote from: RLSprouse on April 21, 2016, 07:34:52 pmThe switch change is only used if you have a 0-5 mA type of meter. You don't need to do anything other than what the original schematic shows if you are using the 1-0-1 meter.QuoteThat 20k ohm/volt gives you the impedance for the range you are using. If your range is 5V full scale, the impedance is 100k ohms. 5V * 20k Ohms/volt gives 100k ohms. If it is 100V, the impedance is 2M ohm. Ohm/per volt times full scale, simple as that.That VTVM will have very high impedance as will most digital voltmeters.QuoteInteresting question considering that National Instruments just bought Digilent! That's why you see even more education related boards at Digilent as well as training courses and a really well thought out wiki.The concepts are similar and it would take a detailed analysis of the capabilities to tell them apart. Clearly the myDAQ runs with LabVIEW (including scripting) whereas the Analog Discovery comes with an SDK which allows for a great deal of automation. Toss-up!The myDAQ has a better multi-meter but fewer digital channels. I need digital channels. Lots and lots of digital channels. The AD has 16, I want 32. I may have a way around this requirement with the new Xilinx Vivado software and the licensed Internal Logic Analyzer. Sixteen channels may be sufficient.At $400 for the myDAQ versus $279 for the Analog Discovery, I know where my money is going. OTOH, if a college class REQUIRED the myDAQ, it wouldn't be a bad investment simply because it ties into LabVIEW.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 09:15:41 pm
content: In looking at the book schematic, there are a couple of points. First of all, the resistance of the meter is in series with the 10k timing resistor. The time constant will not be exactly 10 seconds but then it won't be anyway simply due to component tolerances. Ammeters usually have very low resistance but I have no idea what these specific meters have.Since we know the meter has some resistance and therefore has a voltage drop, the voltmeter is seeing the middle of a voltage divider, not the voltage on the capacitor. The meter needs to be moved so that it is in parallel with the capacitor only, not at the junction of two resistors.For a Simpson 0-1 mA meter, the internal resistance is 43 Ohms. Compared to 10k, this is less than 1/2% change in charging current, maybe it isn't worth thinking about.The capacitor is likely 10% gadget, the resistor might be 5% and the voltage is bound to sag over time so maybe the 1/2% caused by the meter just doesn't matter. But it's worth knowing about!

author: rstofer
author_rank: Super Contributor
timestamp: April 22, 2016, 02:45:22 pm
content: Quote from: RLSprouse on April 21, 2016, 07:34:52 pmI was napping yesterday when you mentioned you had the myDAQ. Not paying attention, I guess...You don't need an external meter all you need to do is decide that the junction of the resistor and capacitor is analog ground. This works as long as you are using a battery and the circuit doesn't have any other ground reference.Connect a jumper to AGND from that analog ground point. Connect channel 1 (-) and channel 2 (+) to this common point as well. Then connect channel 1 (+) to the junction between the switch and the resistor. Channel 1 now measures the voltage across the resistor, hence the charging current. Connect the (-) side of channel 2 to the junction between the capacitor and the (-) side of the battery. Channel 2 measures the voltage across the capacitor.You should be able to see current and voltage simultaneously on the scope instrument. I'm assuming that after the graphs are displayed, you can print the screen (I use Alt-PrtScn to copy the screen to the clipboard and then I paste the image in Paint to actually get it printed. Maybe your software does a better job of this than the AD Waveforms version. It has no print capability, I have to do the print-screen/Paint process.Pages 36 & 38 of the User Manual show the voltage limitations for the analog inputs. Basically +- 10V and you're working with 6V - good enough. With the Analog Discovery, I have to watch what I am doing because it isn't a real scope, overvoltages will probably let the magic smoke escape.Pages 11..14 discuss the need for an AGND reference. The diagrams all show the (-) lead connected to this point but that is just by convention. The analog channels are differential so connecting the (+) lead should work just as wellNotice also that the input impedance is in the gigaohm range. The instrument will not affect the accuracy of the experiment nor will it alter the timing.http://www.ni.com/pdf/manuals/373060e.pdf

author: RLSprouse
author_rank: Contributor
timestamp: April 23, 2016, 01:08:10 am
content: @Ian.M: "Alternatively, use a DPDT switch and wire the circuit as below, which will automatically connect the milliAmmeter in the correct direction for charging and discharging. The DPDT switch is shown in the discharge position."I just wired the circuit you provided, using my cheapo Radio Shack analog multimeter for the voltmeter (in its 5V range - I figured 6V wouldn't kill it) and a 1mA panel meter I have in my collection. It worked brilliantly. When I throw the switch to the "charge" position, the ammeter jumps up to something in the 0.7mA range and then begins falling as the cap charges, until it comes to rest after about 20 seconds, at a reading of a little less than 0.1mA. At the same time, the voltmeter climbs steadily until it reads beyond the range of the 5V scale, so I am presuming it is approaching 6V. If I throw the switch into the 0 position, the ammeter drops to 0 and the voltmeter begins to drop very slowly. I attribute this to the drain through the voltmeter. When I switch to the "discharge" position, the ammeter again jumps up, then begins falling fairly rapidly, as does the voltmeter. In this case, the ammeter and voltmeter drop all the way to zero, as I would expect.I think this has become a nice little experiment, and it sure helps to see how the cap is behaving. Thanks so much for your help!I'll try it again when I get my zero-center meters (I ordered a few ranges).

author: RLSprouse
author_rank: Contributor
timestamp: April 27, 2016, 10:10:34 pm
content: I just received the 50-0-50 uA meter I ordered. Since 50 uA is just 1% of the 5 mA rating of the meter called for in the original schematic, I will need to put a shunt resistor across it so that it will measure the correct current range, right? And I imagine I can figure out what size resistor to use for the shunt by measuring the resistance of the meter, then dividing that by 99? My logic is that would allow 99% of the current to flow through the shunt resistor, and 1% to flow through the meter. Is that right?I assume I can divide by 100 and that would be close enough.

author: Ian.M
author_rank: Super Contributor
timestamp: April 27, 2016, 10:50:10 pm
content: The problem is to measure the meter resistance without damaging it. Most ohmmeters/multimeters will put more than 50uA through the movement, slamming the needle into the end stop and probably bending it. Take a 120K resistor and a 5V supply. Connect the resistor and the meter in series, note the indicated current, and measure the voltage across the meter. Calculate the meter resistance from Ohms law.You then add a preset in series with the meter coil to make up the resistance to the next preferred value, and to allow you to trim the FSD, and use a 1% resistor of 1/100 the total as the shunt.N.B. You cannot simply switch in different shunts, leaving the meter across the terminals - at least one side of the meter must be switched to the selected shunt, so the contact resistance of the range switch doesn't add to the effective shunt resistance. This also lets you switch in different presets so each range can be calibrated individually

author: RLSprouse
author_rank: Contributor
timestamp: April 27, 2016, 11:07:36 pm
content: I anticipated the problem of the multimeter putting too much current through the analog meter movement to measure the resistance directly, so I approached it this way. I took a conservative first step, and put a 100kOhm resistor in series with the meter movement. The resistor actually measured 100.1kOhms. In series with the meter movement, I read a resistance of 101.7kOhms, and the meter movement deflected just slightly. So I dropped down by an order of magnitude, and swapped in a resistor that measured 10.04kOhms. This time, in series, the total resistance was 11.63kOhms, so I figure the meter movement must have a resistance of about 1.59kOhms. Is my methodology and logic sound?I will try the method you have suggested, and compare the results.Thanks!

author: Ian.M
author_rank: Super Contributor
timestamp: April 27, 2016, 11:41:50 pm
content: Your method gets you a good enough result if you are adding a series trimmer resistor to pad it up to the next preferred value, but it looses precision if you need the exact coil resistance.

author: RLSprouse
author_rank: Contributor
timestamp: April 28, 2016, 12:00:39 am
content: I just did the measurements and calculations using your method. I used a 120kOhm resistor, which just happened to measure exactly 120kOhms, and I measured my test bench supply at +4.987V. Hooking these up in series with the meter movement, I got a reading of 42.5uA on the meter. I measured the voltage across the meter at 0.065V. Using Ohm's law, R = V/I = 0.065/0.0000425 = 1,529.41 or about 1.529kOhms. So my estimate wasn't too far off, but I assume this is a more accurate figure.

author: RLSprouse
author_rank: Contributor
timestamp: April 28, 2016, 12:11:10 am
content: Coincidentally, just five minutes ago UPS delivered the 1-0-1 mA meter I ordered from Surplus Sales of Nebraska. It is a beautiful thing, a Simpson panel meter measuring about 4.5" square and *way* more robust than the cheap plastic meters from China. I will go through the resistance measurement process with it now.

author: rstofer
author_rank: Super Contributor
timestamp: April 28, 2016, 12:27:18 am
content: Quote from: RLSprouse on April 28, 2016, 12:11:10 amI bought the same meter. That thing would retail for more than $200 based on meters currently in production but, alas, it isn't listed in the catalog. It is a superb meter.The current offering of a 1-0-1 meter shows an internal resistance of 50 Ohms. For a 5V power supply, we need 5k Ohms to get 1 mA. We can ignore the 50 Ohms because that is about 1%. We can use (2) 10K Ohm resistors in parallel. Or, we could use 4.7K + 270 in series and get a little closer.I would start with a single 10k just to be certain that I was in the ballpark. This would give about 0.5 mA.

author: Ian.M
author_rank: Super Contributor
timestamp: April 28, 2016, 12:48:01 am
content: You really need a three digit reading for the voltage across the meter coil to get good results. Also you should ideally compare the current through the meter with another microammeter (or by measuring the series resistor and the voltage drop across it). However with a series preset to pad the 50-0-50uA meter it up to 1.80K or 2.20K it will be fine as any errors will be compensated for during calibration.

author: RLSprouse
author_rank: Contributor
timestamp: April 28, 2016, 12:51:41 am
content: I put the new 1-0-1 mA meter in series with my 5V supply (which measures 4.987V) and a 1kOhm resistor (measuring 0.996kOhms) and the meter indicate about 0.49mA current. The voltage across the meter movement measured 0.050V. So, if I have things right, R(meter) = V/I = 0.05/0.00049 = ~102 Ohms. I am assuming there is a shunt resistor internal to this meter that results in this low resistance.

author: RLSprouse
author_rank: Contributor
timestamp: April 28, 2016, 12:58:23 am
content: Given my previous results, I went ahead and measured the resistance of the meter directly with my Fluke 87V and it indicates an internal resistance of 101.3 Ohms for this meter. The meter is indicating just a hair beyond 1mA current, so I should be able to determine what voltage my Fluke DMM is pumping out to measure the resistance.

author: RLSprouse
author_rank: Contributor
timestamp: April 28, 2016, 07:41:03 pm
content: I built the circuit from the book (which I posted earlier in this thread) but substituting my new 1-0-1 mA meter movement, and it worked perfectly.Thanks to you guys for helping me understand all this stuff.