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A particular problem I've been given shows two Op-Amps, which can be seen below:

enter image description here

I'm asked to compute the magnitude of the gain. Not being quite sure what this means, I decided I'd figure out the gain of Op-Amp (a) and then just use that as the "magnitude". Both op-amps supposedly produce an "equal" output signal - I chose A because it has no missing values

Computing the gain

Op-Amp (a) just looks like a simple op-amp to me. I can derive gain as follows:

  1. \$I_{feedback} = \frac{V_{out} - V_{in}}{10000}\$
  2. \$I_{input} = -I_{feedback} = -\frac{V_{out} - V_{in}}{10000}\$
  3. \$\frac{V_{out} - V_{in}}{10000} = -\frac{V_{out} - V_{in}}{10000}\$, \$V_{out} - V_{in} = -V_{out} + V_{in}\$, \$2V_{out} - 2V_{in}\$, \$2V_{out} = 2V_{in}\$, \$\frac{V_{out}}{V_{in}} = 1\$

Magnitude of the gain

Here is where I find myself stuck. The gain isn't apparently the magnitude, and the correct answer has units Ohms. What could magnitude mean in this respect?

enter image description here

Why is this the case? Is their some special meaning to magnitude in this respect?

Edit

I'm adding a sensor mentioned before which may have been needed (I didn't think it was relevant to the question originally)

enter image description here

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  • \$\begingroup\$ Gain can invert the signal if it has a negative sign while still making the amplitude larger. Magnitude means the size without the polarity. -5 and 5 are same magnitude. Gain has no units unless it is trainsimpedance and the units don't cancel. \$\endgroup\$ – DKNguyen Oct 30 '19 at 13:54
  • \$\begingroup\$ @DKNguyen Ah ok. Yet if I say magnitude = abs(gain), I still get 1. And the answer is "10Kohm" :s \$\endgroup\$ – Micrified Oct 30 '19 at 13:58
  • \$\begingroup\$ Are you sure it's not supposed to be a transimpedance amp for A? Because it would have a gain of 10000V/A. Something does seem to be missing from the problem. \$\endgroup\$ – DKNguyen Oct 30 '19 at 13:59
  • \$\begingroup\$ @DKNguyen In previous questions they talked about a sensor. But I assumed that was not to be considered when analyzing these amplifiers given. The sensor is an R and C in parallel. Do you think this is needed? \$\endgroup\$ – Micrified Oct 30 '19 at 14:00
  • \$\begingroup\$ Show us the rest. \$\endgroup\$ – DKNguyen Oct 30 '19 at 14:01
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The gain is defined as the output signal divided by the input signal.

The amplifier has a voltage output.

It's a virtual ground input configuration, the input will stay at 0v by feedback action. The input is therefore not a voltage signal, but a current signal.

The gain therefore has dimensions volts/amps, otherwise written as ohms. This configuration is usually called a transimpedance amplifier.

The magnitude of the gain is 10k, the same as the feedback resistor, as for a 1mA input, the output will have to be 10V to get a cancelling current flowing back through the resistor, to maintain zero current at the inverting input.

The main feature of a transimpedance amplifier is its (ideally) zero input impedance. This provides isolation between channels when used as a summing amplifier, for instance in audio mixers. It also short-circuits stray capacitance on the input, when used with current output devices like photo-diodes, making them able to operate much faster than with a finite resistive load.

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  • \$\begingroup\$ BTW what is the use of such an amplifier? Why not feeding a resistor with the input current without any op amp? \$\endgroup\$ – Wheatley Oct 30 '19 at 14:58
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    \$\begingroup\$ @Wheatley speed mainly. Check out figure 2 of this link to 'transimpedance amplifier' on wikipedia (that's why I put the name in the answer, to make it easy to search for the right thing). It presents a short circuit to the thing driving it, so removing the effect of its stray capacitance, thus making it much faster. The same current into the same resistor would need to develop a voltage across the resistor and the stray capacitance. It's also used in mixers, especially for audio, the s/c input isolates the channels from each other. \$\endgroup\$ – Neil_UK Oct 30 '19 at 15:04
  • \$\begingroup\$ thank you for this clear explanation \$\endgroup\$ – Wheatley Oct 30 '19 at 15:17

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