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Help with high-impedance panel voltmeter

Discussion in 'General Electronics Discussion' started by Sigmoid, Jul 21, 2013.

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  1. Sigmoid

    Sigmoid

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    Jul 21, 2013
    Hello,

    I need to drive a voltmeter in a project I'm doing without affecting current flow. We are talking about very low currents in the milliamper range, so the only choice I have is to use an op-amp to connect the meter in order to prevent major impact from measurement.

    IMG_0546.png

    The leftmost three resistors "simulate" the actual device in this mockup. I need to measure the voltage drop over the center resistor, R2, the ohmic value of which is unknown. (Ie. it can be anything from a short circuit to a break.) This essentially means that the measured V2 can be anything between 0V and the supply voltage V+.

    My design uses a voltage divider (with appropriately large, megohm resistors) to keep the maximum voltage lower than the top rail.
    V+ is rather high (24V), so I may have to use a voltage divider to feed the positive supply leg of the amp as well, since most op amps seem to operate in a much lower voltage range - though this is not noted in the mockup.

    My main doubt is about the bottom rail. In order to get a proper measurement of 0 Volts, I was thinking I need to establish a "virtual ground" which is above the actual ground used for the bottom rail in the op-amp, using R3 which is part of the measured circuit anyway...
    (As for Rm, it's the resistor to adjust the measured voltage range. This is also where I will have to compensate for the voltage divider on the input.) Does this look like it will work?

    Also, which op-amp do you suggest I should use? (Should be cheap, ubiquitous and good enough for the job.)
     
    Last edited by a moderator: Jul 22, 2013
  2. Sigmoid

    Sigmoid

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    Jul 21, 2013
    Alrighty, it would seem I did miss something important. :) If I divide V12, I need to divide V23 as well.

    I think this looks better now, please do tell me if you think it's okay - or seem something wrong with it:
    IMG_0547.jpg
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Hi Sigmoid and welcome to the Electronics Point forums :)

    I'm having trouble understanding your drawings. What is the meaning of the V in a circle? Is this a component, or just a test point?

    I don't know why you have separated the centre resistor into two components, R2 and R3A. Can you explain?

    Your writing is messy. Could you redraw the diagram with tidy annotations and formulas please.

    You may find it useful to label absolute (0V-referenced) voltages Vn, and voltages between two points as Un. This is a European convention; V is an absolute voltage and U is a difference in voltage between two points.
    I would number the nodes in the circuit, then you can refer to V1, V2 etc for the voltages on those nodes, and U12 etc for the voltage differences between nodes.

    It looks like you're using a standard differential amplifier to convert the voltage drop across the central resistance into a 0V-referenced voltage to drive a meter.

    Many op-amps can operate with their inputs down to the 0V rail but few can operate with their inputs near the positive supply rail. Can you either drop the voltage at the top of RA (e.g. with a zener diode in series) or boost the voltage on the positive supply pin of the op-amp. You can then use any "single-supply" op-amp.

    Your op-amp should have a high input resistance, and low input voltage offset. I have had good results with the Texas Instruments TLE2022 (the single version is TLE2021) but you should look at other products from TI as well as offerings from Linear Technology (linear.com) and Analog Devices (analog.com).

    It would be helpful, and common courtesy, to explain what you're trying to do from the ground up, rather than giving an abstract definition.
     
  4. Sigmoid

    Sigmoid

    7
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    Jul 21, 2013
    Hey,

    Thanks for the reply. From the ground up, I'd like to build a current regulated lab supply, with the ability to see the voltage drop between the output terminals.
    The V in a circle is a mechanical panel meter, the one that will give the readouts.
    The reason I was abstract is that the part I'd like help with is building a simple amplified voltmeter circuit that runs from the same power source as the device itself.

    As for the signal approaching the top rail, I grounded the point of measurement (V12) through resistors Ra and Rb to act as a voltage divider, and keep the positive input of the op-amp well below supply voltage. Does this seem adequate?

    I split R3 into two resistors because I was thinking that this way I can divide the absolute voltage at both measurement points by the same factor (thus getting V12*f and V23*f).

    However, I think I overlooked that the measurement circuit has a significant current to ground, which would mess with the voltage division at the V23 point.
     
    Last edited: Jul 21, 2013
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    A power supply will have its positive output grounded, won't it? So your three-resistor series should not have the bottom resistor, and you can measure the output voltage directly, right? I don't see why the negative terminal of the supply output would be above ground.

    The circuit you have there is a differential amplifier. It converts a voltage difference (across the centre resistor) into a voltage that is referenced to the 0V rail. You need the four resistors labelled RA, RB, Rf and Ri in your recent diagram, but the voltmeter does not connect in series with Rf. It connects from the op-amp output to the 0V rail.

    You should also specify those resistances. I guess they will all be around 1 megohm? It's common to have them all the same value.

    You're right that the op-amp inputs will not see the full voltage. Assuming RA and RB are equal, they will never see more than half the supply voltage. So you're right, you don't need to worry about modifying the circuit to drop the voltage down, or boost the op-amp's positive supply.

    The input voltages could go all the way to zero, but that's no problem with a single-supply op-amp.

    I still don't understand the reason for splitting the centre resistor into two. If you really need to measure the voltage difference across the load on your current-regulated power supply, and neither of the output pins are grounded, then I don't see why you would simulate the load resistance as two resistors.
     
  6. Sigmoid

    Sigmoid

    7
    0
    Jul 21, 2013
    Thanks. :) Yes I overcomplicated it.
    I was trying to avoid having the same potential on the non-inverting output and the negative supply terminal. (And needing 0 Volts to ground at the amp output, which even most single-supply amps seem to be incapable of.)

    So here's the cleaned-up design:
    • Rc is a variable resistor modeling my current regulator circuit (using a regulator IC). As I want the output voltage to be variable between 1 and 30 mA, its value will likely be between a few hundred Ohms and 24 kOhms.
    • Ra and Rb are my voltage divider resistors. In order to keep their current below 1% of the minimum output current, they are both 1.2 MOhm.
    • Ri is for setting the voltage range of the meter. If I calculate with a meter which goes full scale at 1mA, and want to cover the entire theoretical 24V range of the supply from short circuit to break, divided by two at the point of measurement, it should be a 12 kOhm resistor.

    IMG_0548.png

    Does this look good? Also, will I get a "real" zero reading if the output terminals are short circuited with the TI amps you suggested?
     
    Last edited: Jul 21, 2013
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK, that looks better.

    You're still showing the meter inside the feedback loop. I understand why you're doing this, but you don't need to - you can convert a current meter into a voltage meter by adding a series resistor, then you can connect it from the output to the 0V rail.

    There are two reasons why it's normally done that way. First, if it's an electromechanical meter movement, it will have some inductance, and putting this in the feedback loop will affect the op-amp's behaviour when a change occurs. (It won't affect the steady state behaviour.) I don't know whether this could cause instability, but you should consider that possibility.

    Second, as shown, the meter is connected directly to the inverting input of the op-amp. This is a sensitive point in the circuit; for proper operation of the op-amp the inverting input should see the feedback signal as cleanly as possible. The wires to the meter, and the meter itself, will add capacitance, and some small amount of signal pick-up, which should be avoided at the feedback node.

    Regarding getting the op-amp output to reach 0V, a simple way is to add something in series with the output to give a voltage drop. This could be, say, two 1N914 diodes in series (anode to the output pin), or a zener diode (cathode to the output pin), or a transistor or Darlington transistor connected as an emitter follower. You need a resistor to ground after the voltage dropper, to cause current to flow through it.

    (Ideally this resistor should be returned to a negative rail, so there is current flow through the voltage dropping circuit even when the output is zero, so that the dropped output voltage can go all the way to zero; if the resistor is returned to the 0V rail, the dropped output will go very close to 0V but never completely to 0V. That shouldn't be noticeable though.)

    You can also add a load resistor, e.g. 1k, directly from the op-amp output pin to the 0V rail, to help the op-amp output to go as low as possible.

    When you add this voltage dropping device, remember that the meter and the feedback circuits must be connected after the dropper. Nothing connects to the op-amp output pin apart from the top of the dropper circuit, and optionally, the load resistor I just mentioned.
     
  8. Sigmoid

    Sigmoid

    7
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    Jul 21, 2013
    Thanks for the help. :)

    Actually, what I'm most concerned about regarding this design is that with most amps, the output is usually nonlinear (ie. doesn't react) for at least 0.7 V above the bottom rail. Since I likely need to keep the gain at 1 to stay within operating parameters, and the signal is already cut at least in half, this could mean a good 10% of the meter dial.

    I'm wondering if it would help to use a noninverting summing / averaging circuit, adding a few volts from the supply rail to the signal, and then dropping the same voltage across a Zener diode at the output... That would hopefully get me a linear characteristic across the entire range...

    EDIT: Actually, correct me if I'm wrong, but I've just gotten the impression that if I take a reference voltage of 1.4 Volts, and average it up with the DC signal in a passive averaging circuit, I get a flattened signal that is raised by exactly 0.7V... So if I put it through a simple common collector amplifier, I'd get a linear, low impedance signal on the emitter to drive the meter movement that starts at or extremely close to 0V... And as an additional upside, no op-amp needed at all, just a single transistor.
     
    Last edited: Jul 22, 2013
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    OK Sigmoid, let's take a step back

    That's what you want to do, right?

    It is an issue if the impedance of the voltmeter is not high with respect to the voltage source.

    So what can you tell us about it?

    Hang on there. Milliamps are not very low currents! A typical multimeter will have a 10M input impedance, so on a 20V range will only draw 2uA at full scale.

    So if there's about 20V, you're talking about a 0.1% error AT THE VERY WORST.

    No, I think you misunderstand. Or perhaps you've stated your problem poorly.

    Did you mean the current would be only several microamps?

    And unless the values of these resistors exceeds 10M in total, your cheap multimeter will perform better.

    No, you would use the real ground of a double ended supply. I presume the power rails for this would float (i.. none of them are common to the circuit under test).
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That won't be a problem with the changes I suggested. A voltage dropper from the op-amp output will give you an output that goes right down to 0V and there will be no deadband.
    You will be taking the feedback from after the dropper. As long as the feedback is taken from the same point as the output (to the meter), the dropper will not cause any kind of offset or deadband.
    No. There's no need to do anything like that.
    You could do something like that, but it won't be as accurate. For example, the voltage divider across the output you're measuring can feed a PNP Darlington emitter follower (common collector) with its collector connected to 0V and its emitter pulled up to VCC, and this emitter could feed an NPN Darlington emitter follower with its collector connected to VCC and its emitter driving the meter. The idea is that the PNP adds two base-emitter voltages, then the NPN subtracts them again.
    One problem is that the base-emitter voltage will vary somewhat with current, and the currents in the PNP and NPN will vary in opposite directions depending on the voltage. When the voltage is low, the PNP current will be highest, but the NPN current will be lowest, and vice versa.
    You could try it and see whether it's accurate enough for your purposes.
     
  11. Sigmoid

    Sigmoid

    7
    0
    Jul 21, 2013
    Okay I understand it now. :) Thanks a lot. Does this look correct?

    meter.png

    We aren't talking about a cheap multimeter, but about an approximately $3 Chinese panel meter movement, which eats 1mA for full dial deflection.
    And I think I'll try and see how far I can go with those resistors. :D The point isn't that much the accuracy of measurement, the movement itself is rated at 2.5%, but rather to have as little impact to the regulated current source as possible.
     
    Last edited: Jul 22, 2013
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes, that's right. You haven't shown the load resistor from the op-amp output to ground, but I don't think you need it in this application.

    Your voltage dropping zener needs to have low leakage current. Use a device rated for 500 mW or less; larger zeners have significant leakage current when the reverse voltage across them is less than the zener voltage.

    The 4.7V part from the BZX79 series from NXP Semiconductors has a quoted maximum leakage current of 3 uA at a reverse voltage of 2V. This would contribute only 0.3% error. http://www.digikey.com/product-detail/en/NZX4V7C,133/568-7964-1-ND/2762654
     
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