How does this offset measurement work?

[Yuu]

I was reading pdfs from Analog Devices on op-amps and this circuit to measure offset voltage is making me uncomfortable.

Arbitrarily placing the offset voltage \( V_{OS}\) at the inverting input of the DUT, the gain of the DUT is \( V_{o1} = a_1 \cdot (b V_o - V_{OS}) \) where \(b = 1/1000\) is the feedback factor.

The gain of the integrator is just \(V_o = -\frac{1}{j \omega C 100k} V_{o1}\) which is infinite at DC. Basically, it's just going to increase your already very high open-loop gain. That's fine and even good because we have feedback but I think what is confusing me is we are feeding back to the NON-inverting input and not the inverting input. We don't have negative feedback.

So, instead of our regular negative feedback equations which tell us \(V_o = \frac{1}{b}\frac{1}{1 + 1/L} V_i\). We have the case that \(V_o = \frac{1}{b}\frac{1}{1 - 1/L} V_{OS}\). That's making me uncomfortable. Is this okay? Why does this work feeding back into the non-inverting input?



Another question, the analog devices article says "Negative feedback forces the output of the DUT to ground potential. (In fact, the actual voltage is the offset voltage of the auxiliary amplifier—or, if we are to be really meticulous, this offset plus the voltage drop in the 100-kΩ resistor due to the auxiliary amplifier’s bias current—but this is close enough to ground to be unimportant)".

Wouldn't input bias current through a 110kOhm resistor be large and definitely not unimportant? Is it because net offset effects at the input of the integrator are divided by the gain of the integrator and can therefore be said to be unimportant?

[TimFox]

The measured output at TP1 will be (99,900 + 100)/(100) = 1000 x V_OFF the the DUT.
That's the voltage at TP1 that is required to give a compensating voltage at the DUT input, reducing its output to (very close to) zero.

[Yuu]

Sure but we have positive feedback. The output is feeding back into the non-inverting input.  That's what is throwing me off. That's what I'm confused on. When this happens can you still apply usual negative-feedback reasoning? I would imagine you cannot.

[MasterT]

The fact you've missed that integrator is also Inverter. So negative sign already in the feedback path, and there non-inverting input.
If feedback goes as usual to inverting input, than we 'd have positive feedback

[TimFox]

Rather than use the feedback fraction equations, just do the algebra assuming an offset voltage on the DUT and see what the equilibrium voltages will be.
The overall feedback is negative, since the integrator inverts, as MasterT stated.

[MrAl]

You can get an idea what you have by actually measuring the output of the first op amp in the actual application.
If you have a gain of say 100 and with zero input you measure 0.1vdc output, then the input offset is about 0.1/100=1mv, plus or minus as applicable.  If you were nulling, you would look at the output of the op amp anyway.

[Yuu]

The fact you've missed that integrator is also Inverter. So negative sign already in the feedback path, and there non-inverting input.
If feedback goes as usual to inverting input, than we 'd have positive feedback

Ahhh yep this is what I was missing. Thank you.

[MrAl]

I was reading pdfs from Analog Devices on op-amps and this circuit to measure offset voltage is making me uncomfortable.

Arbitrarily placing the offset voltage \( V_{OS}\) at the inverting input of the DUT, the gain of the DUT is \( V_{o1} = a_1 \cdot (b V_o - V_{OS}) \) where \(b = 1/1000\) is the feedback factor.

The gain of the integrator is just \(V_o = -\frac{1}{j \omega C 100k} V_{o1}\) which is infinite at DC. Basically, it's just going to increase your already very high open-loop gain. That's fine and even good because we have feedback but I think what is confusing me is we are feeding back to the NON-inverting input and not the inverting input. We don't have negative feedback.

So, instead of our regular negative feedback equations which tell us \(V_o = \frac{1}{b}\frac{1}{1 + 1/L} V_i\). We have the case that \(V_o = \frac{1}{b}\frac{1}{1 - 1/L} V_{OS}\). That's making me uncomfortable. Is this okay? Why does this work feeding back into the non-inverting input?



Another question, the analog devices article says "Negative feedback forces the output of the DUT to ground potential. (In fact, the actual voltage is the offset voltage of the auxiliary amplifier—or, if we are to be really meticulous, this offset plus the voltage drop in the 100-kΩ resistor due to the auxiliary amplifier’s bias current—but this is close enough to ground to be unimportant)".

Wouldn't input bias current through a 110kOhm resistor be large and definitely not unimportant? Is it because net offset effects at the input of the integrator are divided by the gain of the integrator and can therefore be said to be unimportant?

Hi,

You already know now that the second op amp inverts the signal so there is actually negative feedback so i'll address the second question.

Apparently, the bias current does not matter much.  Changing the 110k resistor from 60k to 160k has very little effect on the output voltage.  For a 1mv input offset the output of the second op amp becomes for either extreme:
0.9999999399mv (60k)
0.9999998399mv (160k)

so the theoretical difference is very small meaning it shouldn't affect the measurement.

This is using the other values as shown in the schematic, and also making the open loop gain of both op amps equal to 100000 for DC.

[magic]

Current through 110kΩ is simply the input bias current of the AUX opamp, there is no other place for (DC) current to go. This doesn't vary significantly with voltage (within normal operating range) and is less than 1μA for most bipolar opamps and less than 1nA for JFET/CMOS. Hence there is almost no voltage difference (<1mv) between DUT output and AUX inverting input. Feedback loop keeps AUX IN- close to IN+, so DUT output is close to ground.

[David Hess]

Why does this work feeding back into the non-inverting input?

The integrator inverts the output of the operational amplifier under test, so the function of the inverting and non-inverting inputs are swapped when the output of the integrator is used for feedback.