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I've been given the following circuit to build for an education lab. The values of the resistors aren't accurate; they're only to illustrate that the gains of the two op-amps are different.

The aim is to have two separate op-amps of different gains connected to the same input signals, and then a SPDT switch connects the required output to \$V_{out}\$. The two op-amps will be in the same chip (LM353).

Since the output of AMP1 is fed back, and in some sense is connected to Node2, will the first op-amp effect the gain of the second op-amp and vice-versa? Or will the two op-amps behave as if they are separate and just happen to have the same input signals?

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT: Thank you for the replies. The input will be provided from a MPX2100 thin film pressure sensor which the datasheet indicates should have an output impedance of 1400-3000\$\Omega\$. Is this too large?

If so, perhaps it would be smarter to switch the inputs, or use a buffer op-amp like an instrumentation amplifier? The problem is the PCBs for the circuit were made long ago, and I'm now trying to reverse engineer the circuit from effectively no documentation. So making changes is difficult.

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    \$\begingroup\$ That depens on the source impedance of the input signals. If the are sufficiently 'stiff' (low impedance) the two circuits will not influence each other. (Note that even if you had only one section, its behaviour would depend on the source impedance!) \$\endgroup\$ Aug 16 '15 at 12:41
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    \$\begingroup\$ Why? Why not just switch the feedback resistors? Make before break of course. \$\endgroup\$
    – user207421
    Aug 16 '15 at 13:12
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To ensure the two circuits have the correct intended gain, a strong voltage source should feed node 1 and node 2. In other words the output impedance of the voltage source has to be very low or the amplifiers will not have the gain as expected.

Because of this, it's reasonable to say that any loading (due to adding the number of amplifiers in parallel) will have negligible effects.

If the voltage source is not considerably low in its impedance then neither op-amp circuit will work with the gain they are expected to deliver.

Regarding the effect of a feedback resistor, are you aware of what a "virtual" earth means and the spin-off it has on differential amplifiers?

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  • \$\begingroup\$ I'm not fully sure. I believe the general idea is that the very large open loop gain of the op-amp tries to force the Voltage of the positive and negative input to be the same. Since the positive input is grounded the negative input also acts as though it were a ground, or at leas at a constant voltage level. I'm not quite sure what specific spin-off this would have in diff amplifiers. \$\endgroup\$
    – Rohan
    Aug 16 '15 at 14:05
  • \$\begingroup\$ Ahh - perhaps you mean since the negative input acts as a virtual ground there will not be interference due to both the inputs of both op-amps always 'trying' to be at ground. Hence, issues only come up if the voltage source is not sufficiently 'stiff'. Why do low impedances correspond to 'stiff' voltage sources? Is it due to an Ohmic relationship? Hence, regardless of current draw output voltage will be constant? Wouldn't a sensor already account for this? \$\endgroup\$
    – Rohan
    Aug 16 '15 at 15:09
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    \$\begingroup\$ @Rohan, the signal on the -input will be, by op-amp action, the same as the one on the +input - not quite a virtual earth in this case because of the differential nature but, what it means is that there are equal currents down each input resistor and things balance out. Yes, a stiff source is one with low output impedance and no, you have to be careful with some sensors and read the data sheet - check on the output impedance etc.. \$\endgroup\$
    – Andy aka
    Aug 16 '15 at 16:22
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The feedback resistors you are using are very low in value, especially for the LM353, which is only reliably capable of a few mA output current. Maybe okay if your signal levels remain quite small.

As Andy said, at DC I'm expecting negligible effects if the driving source impedances are low (as you'd expect from the output of another op-amp with suitable drive capability).

I'm guessing you might see some distortion effects with AC signals, especially if you drive one of the op-amps into saturation or (more likely) current limit because the output impedance of the driving source will not be ~0 at higher frequencies. Just a guess though.

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