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Update (21.Aug.2024): from the answers I learned that the bias current should be only some pico ampere. But in my case, this is micro ampere. It turns out that either the components are broken, or they are counterfeit.

Please comment below if you think there could be other reasons.

schematic

simulate this circuit – Schematic created using CircuitLab

I'm learning to use a voltage follower as a buffer for my DC voltage measurement. The simulation figure below only shows the schematic. But I'm doing tests with a real system.

The component is an OP Amp (TLVx365 / OPA365) and the output pin is directly connected to the negative input pin. The power supply is a positive 3.3V single-supply.

In theory, the input impedance is very large / infinite. However, when I measured it with a multimeter, it shows that the input impedance is only around 1 or 2 hundred of kilo ohms.

I want to ask here, how can I estimate the input impedance of this voltage follower from the datasheet of the opamp?

I made this figure with simulink. Hopefully, it can show my measurment method. I can connect or disconnect the "breakpoint" so that the influence from the buffer (voltage follower) can be noticed by measuring the voltage at this breakpoint. The result: with 1.5V VCC over the two resistors, the voltage at the breakpoint was 1.326V without connecting the buffer. It dropped to 0.3V after connecting the buffer. So I can estimate this impedance by calculating the parallelled resistor. It's around 26k Ohm.

The input current can also be measured by the multimeter: it rised from 1.76 uA to around 12 uA. This result can also show that the buffer input impedance is about 26k.

Now I'm not sure if I can ignore the effect of the ADC.

Input impedance measurement

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    \$\begingroup\$ Your question is specifically not about a 741 opamp, so why did you use that tag? \$\endgroup\$
    – Marcus Müller
    Commented Aug 19 at 18:43
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    \$\begingroup\$ I bet the high input impedance of the opamp causes the multimeter's tiny test current to result in a voltage higher than 3V3 between the probes, which would make the ESD diodes in the opamp conduct. You'll need another multimeter (or a spec sheet) to check voltage between the multimeter probes in ohmmeter mode. \$\endgroup\$
    – bobflux
    Commented Aug 19 at 19:23
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    \$\begingroup\$ OK, now this is weird. Have you used solder flux which contains lots of ions like water soluble? These can create suspiciously high leakage current if not cleaned. \$\endgroup\$
    – bobflux
    Commented Aug 20 at 16:16
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    \$\begingroup\$ Did you actually measure the impedance of a physical system, or is that all in simulation? Please edit the question to make it clear. The schematic you have looks like something in a simulation. \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented Aug 21 at 10:41
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    \$\begingroup\$ I would suggest that you use a real schematic editor (there is one built in to ee.se, in the post edit box), not something completely different with non-standard symbols and elements. \$\endgroup\$
    – MrGerber
    Commented Aug 21 at 11:36

2 Answers 2

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The input impedance at low frequencies is essentially infinite (GΩ) for input voltages within the power supply range. Numerically, it could change as much as +20pA to -20pA over the input range of (say) 5V, so that would be the equivalent of 125GΩ. At higher temperatures it can be typically (40 unit test in Figure 7-5) closer to 10GΩ at 125°C, still very high.

If you go outside the power supply range, then the internal protection networks can start to conduct. As your linked datasheet illustrates:

enter image description here

That's probably why you got a measurement like that by applying some voltage from a multimeter in resistance mode to the input.

At higher frequencies the input impedance will decrease due to the pF input capacitance. For example, with a 0.1V 10MHz input (low enough that the OPA365 is not slew-rate limited), a simulation shows an input current of 37uA peak, so the equivalent impedance is about 2.7kΩ (capacitive), equivalent to about 6pF.

At frequencies around 25kHz or lower the input impedance of the follower will be more than 1MΩ (not counting the contribution from the stray capacitance of your layout, and for a sinusoidal input).

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  • \$\begingroup\$ Thank you for your answer! But I only use that for DC. I went for an alternative way to measure its input impedance. I used a constant DC voltage supply (1 ~ 2V), and measure the (average) input current with the multimeter. The calculated input impedance is also very low. \$\endgroup\$
    – Student
    Commented Aug 20 at 14:55
  • \$\begingroup\$ I don't know if there is any protection diode broken. Or maybe the open-loop voltage gain has any effect on that? \$\endgroup\$
    – Student
    Commented Aug 20 at 14:57
  • \$\begingroup\$ Show the exact circuit you used (edit into the bottom of your question). Including the meter and the settings on the meter. Maybe the IC is damaged or maybe there is some issue with the methods used. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 20 at 15:08
  • \$\begingroup\$ I have updated the figure and the measurement result. You will find it now. \$\endgroup\$
    – Student
    Commented Aug 21 at 9:43
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    \$\begingroup\$ I don't see any problems with your method. Maybe the part is damaged or counterfeit. You're actually measuring bias current rather than input resistance, of course. If the bias current was constant (for changes in input voltage only) then the datasheet would say the input resistance was infinite. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 21 at 12:20
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The DC impedance is so high it is not even mentioned. AC impedance is 1 or 5 pF depending on which of them you are interested in.

Bias current should be less than 20 pA.

So it depends if you want DC or AC impedance.

Also you can't expect to measure input impedance with a multimeter. What the multimeter does is it injects a test voltage/current and measures the resulting current/voltage, and current and voltage at DC are only related for resistors and the op-amp input is not a resistor. The multimeter just knows there is curretly some voltage and some current so since it is set to resistance measurement mode, it has to assume the load is a resistor, and it calculates you a resistance for displaying it.

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  • \$\begingroup\$ Only DC impedance is interesting for me. I have chosen another way to measure its input impedance. I used a constant DC voltage supply (1 ~ 2V), and use the multimeter to measure the (average) input current. It still turned out that the impedance is only some kilo ohm. What I found very interesting was that the input impedance is highly nonlinear at different voltage points. \$\endgroup\$
    – Student
    Commented Aug 20 at 14:59
  • \$\begingroup\$ @Student The input impedance can't be kilo-ohms. It would mean at 1V you have 1mA going in. Which should be impossible. The current should 4 decades less. \$\endgroup\$
    – Justme
    Commented Aug 20 at 15:47
  • \$\begingroup\$ Yeah. I also hope it's my fault / the problem of the device. But even if I change to different multimeters, the result is still not very nice. It shows that the impedance is roughly 26k Ohm. \$\endgroup\$
    – Student
    Commented Aug 21 at 9:44
  • \$\begingroup\$ @Student I still don't understand the method what you are measuring and how. You can't use any multimeter in ohms measurement mode. Also for current measurement, you need a really sensitive and expensive device capable of sub-microamps, i.e. picoamps. \$\endgroup\$
    – Justme
    Commented Aug 21 at 11:14
  • \$\begingroup\$ The multimeter measures voltage. If the voltage doesn't change too much, then I know the impedance is really high and not much current flows in that loop. However, the voltage drops quite a lot, so it means that the buffer loop has certain current there. \$\endgroup\$
    – Student
    Commented Aug 21 at 11:57

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