The datasheet for the AD625 instrumentation amplifier shows an example circuit with 'Auto Zero' (Figure 36), below. In this circuit an op-amp AD711 is being used to give an offset voltage to zero the instrumentation amplifier.

Figure 36

But surely the op-amp is configured as an integrator not as a sample and hold? or am I missing something. Is the data sheet just plain wrong?



The circuit indeed is an integrator. It integrates the non-zero output of the opamp and immediately also offsets it, until the opamp outputs zero, at which point the integrator's output won't change anymore.

If you have any function \$y_i(t)\$, which represents the signal you want to integrate, the function you integrate with the opamp will look like this: $$ y_s(t) = \left\{ \begin{array}{ll} y_i(t) &\mbox{if } t_1 < t < t_2\\ 0 & otherwise \end{array} \right. $$

Where \$t_2 - t_1\$ is the time the switch of the AD7510DIKD is closed.

If you integrate over \$y_s(t)\$ now, you'll get the following:

\$\int_{-\infty}^{\infty}y_s(t)\mathrm{d}t = \int_{-\infty}^{t_1}0\mathrm{d}t + \int_{t_1}^{t_2}y_s(t)\mathrm{d}t + \int_{t_2}^{\infty}0\mathrm{d}t = \int_{t_1}^{t_2}y_s(t)\mathrm{d}t\$

As you can see the output signal will remain constant once the input is zero.

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  • \$\begingroup\$ Ah! yes that's the point I have been missing. Of course, the output of the integrator reduces its own input, during the auto zeroing. Thank you \$\endgroup\$ – Icy Sep 25 '15 at 13:13

When A3 is open the AD711 will 'hold' the output value (in reality it will slowly move away from the initial value due to leakage and bias currents, like any real analog S&H). There will also be a slight transient when A3 switches due to charge injection, but that won't be many volts because 0.1uF is large compared to the charge injection of the analog switch.

When the auto zero loop is closed, the output is disconnected, the op-amp acts as an integrator with a time constant 1K * 100nF, the input is disconnected from the outside world and shorted, so it will be very close to 0uV. The output of the AD711 changes to drive pin 10 of the AD625 close to 0V (same as its non-inverting input terminal give or take a bit of offset voltage). If the AZ loop is not given enough time it will at least drive towards zero, and will get closer every time.

Edit: Ignoring the gain factor between the AD625 balance input and output, the time constant is 100usec. If that gain factor was 1, the output would approach zero exponentially Vos = \$V_0(1 - e^{t/\tau})\$ (ignoring offset, noise) where \$\tau\$ = 100usec. So the remaining offset voltage after each pulse would ideally be reduced by a factor 0.13:1. If we assume it started with an input-referred offset of 5mV, then the offsets would ideally be after each 200usec pulse:

0 5mV

1 670uV

2 92uV

3 12uV

4 1.6uV (probably as low as we can get)

5 230nV (definitely as low as we can get)

So you can reduce the input offset voltage of the AD625+AD67502 to something approaching the input offset voltage of the AD711 divided by the closed-loop gain from input to output.

There are modern 'zero drift' amplifier chips that do things like this internally and (almost) transparently to the user, give or take weird stuff like enormous (relatively speaking) current pulses coming out of the inputs and strange noise peaks at certain frequencies.

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  • \$\begingroup\$ But, lets say the output of the integrator is x volts, when the auto-zero input is activated it will change from this voltage at a rate proportional to the output offset voltage - it won't at any time settle at the offset voltage. \$\endgroup\$ – Icy Sep 25 '15 at 13:10
  • \$\begingroup\$ Mathematically it will not, however in practice it will be as close as it will ever get (due to offset and noise- practical factors) within 5-10 time constants. Since the time constant of the integrator is 100usec and the pulses are 200usec it may take several pulses to settle to that value, depending on where it started. Of course that assumes a stable offset of the AD625+AD67502, so this only works if the offset voltage of the front end is changing relatively slowly (the most important cause is thermal- which can be slow if the amplifier is lightly loaded and shielded from air currents). \$\endgroup\$ – Spehro Pefhany Sep 25 '15 at 13:30
  • \$\begingroup\$ This type of auto zero circuit uses the OUTPUT to zero things up. Zero-drift circuits like Chopper amps use the INPUT OFFSET of the amp, or difference between V+ and V-. This type of arrangement is for when there's a real offset in the input SIGNAL that needs to be removed, such as an electrode/skin potential in a biopotential amp. \$\endgroup\$ – Scott Seidman Sep 25 '15 at 14:03
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    \$\begingroup\$ Never mind --- there's a nightmare switching circuit on the input side too! I had tunnel vision on the output side. \$\endgroup\$ – Scott Seidman Sep 25 '15 at 14:26
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    \$\begingroup\$ Yup, it's just like a chopper amp. \$\endgroup\$ – Scott Seidman Sep 25 '15 at 14:56

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