1
\$\begingroup\$

Inspired by the video Fake Op-amps from China, I tried to measure the Offset Voltages of my Op-amps using the voltage follower scheme

enter image description here

Note that I used a voltage divider (R3 and R4) to create a "fake" ground, because I do not have a symmetric voltage source.

  1. When I measure the voltage between A and B with a multimeter, I get quite good results, the voltages are around 100uV. Interestingly, the voltage changes to almost 0uV when I hoover my hand over the assembly (without touching anything). How is this possible?

  2. If I use A' instead of A and B' instead of B, I get something on the order of mV! This is much worse and I wonder why this is possible and which result is the correct one?

  3. Finally, if I connect Vin- directly to Vout, I also get voltages of the order of mV. Does this mean that you should always insert a resistor between Vin- and Vout even in a voltage follower scheme?

EDIT: VCC is 12V, so this would represent +-6V voltage source.

\$\endgroup\$
2
  • \$\begingroup\$ You slap have to consider bias current to the opamp \$\endgroup\$
    – Tyassin
    Commented Jul 28 at 12:34
  • \$\begingroup\$ There are other non-ideal characteristics which are modeled as either voltage sources in series with the inputs or current sources which create voltages at the input pins. You can't measure just one of them pretending the other don't exist, so you'll need circuits and procedures which minimize the effect of other non-idealities. For this reason, it is hard to answer your questions as posted. Also, a stable symmetric power supply will help a lot. \$\endgroup\$
    – devnull
    Commented Jul 28 at 13:22

2 Answers 2

Reset to default
2
\$\begingroup\$

The OP-07 - even the 'D' version- is a pretty good (but very old) op-amp, but you have to give it high enough supply voltage and account for input and output range and bias/offset currents.

Here is AD's recommendations and basic circuit for measuring Vos:

enter image description here

As the app note describes, thermoelectric voltages will come into play with precision op-amps. Note the use of low (10Ω) resistors to minimize the effects of bias and bias offset currents.

Rather than a matched 10Ω||10kΩ source impedance yours has mismatched and much higher (1.5kΩ and 1kΩ) source impedances. You should make sure you have enough power supply voltage for the op-amp you are testing (OP-07 needs 6V but is happier with more like +/-15V) and that you are well within the input and output ranges. For a virtual ground you should bypass it and the supply to the op-amp, since even these relatively poky precision op-amps can oscillate.

\$\endgroup\$
7
  • \$\begingroup\$ Thank you for your answer. Yes, I already came across the thermoelectric effect, but there is no reason why components/parts of components should have different temperatures. My power supply is 12V or +-6V for symmetric supply and that should be enough. Can you please explain to me what you mean by bypassing the virtual ground and the supply to the op-amp? \$\endgroup\$
    – Pygmalion
    Commented Jul 28 at 16:08
  • 1
    \$\begingroup\$ Well, with thermoelectric effects there's no reason for everything to be the same temperature. There is significant heat flowing down the op-amp leads even if you can't perceive it. Shadow a part of the circuit from room lighting and it will be slightly cooler. When you are measuring microvolts, it can matter. For +/- 50uV measurements, not so much. Bypassing- a capacitor across R4 and another from Vcc to GND near the op-amp. Something like 100nF ceramic is appropriate. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 28 at 16:26
  • \$\begingroup\$ OK, I will try the above circuit with two bypass capacitors. What bothers me about the circuit is what is the point of connecting 10 and 10k in parallel? \$\endgroup\$
    – Pygmalion
    Commented Jul 28 at 16:56
  • \$\begingroup\$ They're not connected in parallel in the AD circuit, that's the Thevenin equivalent resistance for the bias current and it's essentially 10Ω, I was just being 100% accurate (or pedantic if you prefer), partly because that idea seems to have been missed in your circuit- the equivalent DC resistance your non-inverting input sees is 10kΩ+(10kΩ||10kΩ) or 1.5kΩ. The mismatch of 500Ω, even if the bias currents are perfectly matched, yields 1uV offset per 2nA of bias current. With AD using 10Ω rather than 1KΩ-ish, mismatch is ~2 orders of magnitude less significant (& they've matched the resistances) \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 28 at 17:05
  • \$\begingroup\$ Thank you. Your circuit gives -30mV without bypass capacitors, -3mV with bypass capacitor in parallel to R4 and 15mV with bypass capacitor between VCC and GND. It is amazing how capacitors affect what is essentially a DC driven circuit. Do these data show that the offset voltages are really in the 10 uV range and that this is a real OP07 chip? \$\endgroup\$
    – Pygmalion
    Commented Jul 28 at 17:42
2
\$\begingroup\$

The voltage between A and B includes the voltage from input bias offset current through R1 and R2, and can add or subtract from the input offset voltage (apparently it subtracts in your measurement). The voltage between A' and B' is the offset voltage.

So an equal equivalent-value for the resistances into the two inputs will cancel the effects from the input bias current but not from the input bias offset current.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you for your answer. Since the voltage between A' and B' on my chips is 1mV and 1.2mV, I assume this means that these op-amps are not proper OP07s for which the expected input offset voltage is only 75uV? \$\endgroup\$
    – Pygmalion
    Commented Jul 28 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.