# Is this a useless resistor?

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#### AlienCircuits

##### Member level 5
I have a real circuit that someone designed, and I am curious about why there is a resistor of 10k included in the feedback path. The circuit basically takes a voltage and buffers it to be measured by a DMM or oscilloscope, and so the load represents the input impedance of a probe. When I use my circuit analysis, I see that no current should flow in this feedback resistor with an ideal opamp, and so there is no voltage drop across it. I think a feedback with 0 resistance would work just as well. I also think that this resistor could degrade the frequency response, but this is a DC application and not that important to consider either way (unless that is there to reduce bandwidth on purpose?).

I think that the other 10k resistor in series with the load is there to protect the opamp against short circuit conditions.

So, why would I want this resistor in the feedback path anyway?

The feedback resistor should match the source resistance
for least bias current input error. Resistance will also add
some peaking which can be good or bad, depending.

• AlienCircuits

### AlienCircuits

Points: 2
What is the opamp? What is the impedance of the signal source?

Keith

The feedback resistor should match the source resistance
for least bias current input error. Resistance will also add
some peaking which can be good or bad, depending.

I did not know that they should be matched, thanks for the advice.

What is the opamp? What is the impedance of the signal source?

Keith
The source signal is 800V divided down by a 100M and 1M voltage divider, so the source impedance is 990 kOhm, and the opamp is CA3140.

- - - Updated - - -

Ok, so I checked the CA3140 and found this,

"Moreover, some current limiting resistance should be
provided between the inverting input and the output when the CA3140 is used as a unity gain voltage follower. This
resistance prevents the possibility of extremely large input
signal transients from forcing a signal through the input
protection network and directly driving the internal constant
current source which could result in positive feedback via the
output terminal. A 3.9kΩ resistor is sufficient."

So, that pretty much explains it. Sorry for wasting your time Ok, so I checked the CA3140 and found this,

"Moreover, some current limiting resistance should be
provided between the inverting input and the output when the CA3140 is used as a unity gain voltage follower. This
resistance prevents the possibility of extremely large input
signal transients from forcing a signal through the input
protection network and directly driving the internal constant
current source which could result in positive feedback via the
output terminal. A 3.9kΩ resistor is sufficient."

So, that pretty much explains it. Sorry for wasting your time That's not why the resistor is in that particular circuit, though. That paragraph refers to protecting the amplifier from being driven with voltages outside its supply voltage range. But in your circuit, obviously the amplifier can't possible be capable of outputting a voltage that can exceed the supply rails.

The purpose of the resistor is to reduce input offset voltage due to input bias currents. If the noninverting input is driven with a high resistance source, then the input bias current will cause a DC error. However, if you also put the same resistance in series with the inverting input, then it will also develop the same DC offset, and the two offsets will null each other out.

So it's something that's only done if you care about DC accuracy, and the source is a very high impedance.

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The purpose of the resistor is to reduce input offset voltage due to input bias currents. If the noninverting input is driven with a high resistance source, then the input bias current will cause a DC error. However, if you also put the same resistance in series with the inverting input, then it will also develop the same DC offset, and the two offsets will null each other out.

So it's something that's only done if you care about DC accuracy, and the source is a very high impedance.
I'd agree with that for a BJT-input opamp, where the input bias currents are significant. However the CA3140 has a MOSFET input stage, so input bias currents are trivially small (in the order of pA).

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