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I'm really inexperienced when it comes to operational amplifiers, but I'm trying my best to learn more about them. That being said, I'm currently having trouble understanding the transresistance amplifier shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

With regards to the circuit, I'm having trouble figuring out the output impedance of the op-amp circuit. My textbook tells us to use a generic 1 A current source to figure out the voltage at the output to calculate the output impedance, but I'm confused as to the results with the setup shown below:

schematic

simulate this circuit

Turning off all other independent sources, the negative terminal acts as a virtual ground, so Vx = Ix * R1, and Zout = Vx / Ix. Therefore, from what I see, Zout = R1. However, according to my textbook, zero output resistance appears at the output terminal of the ideal op-amp. How does this make sense if using a test source to calculate the output impedance?

EDIT: Assuming an ideal op-amp model envisioned below, the internal output resistance of the op-amp is zero if I remember correctly. But let me try and work this assumption out. Using the model below of a differential amplifier, Vd (the difference between the positive and negative terminals) is 0 since both terminals are 0, so the dependent source provides 0 volts, so it 'acts' as if it is 'shorted' to ground. Since Ro is really tiny, close to 0, is that how we get the overall output resistance of the circuit being 0?

schematic

simulate this circuit

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  • \$\begingroup\$ The TRA is expected to produce a fixed voltage for a given input. So you want it to have as low an output resistance as possible. If you want to know the actual value, you need to know the equivalent output resistance of the op-amp itself. \$\endgroup\$ – The Photon Apr 2 '19 at 20:15
  • \$\begingroup\$ You are neglecting the fact that it is an ideal op-amp. The amplifier will hold the voltage at \$V_-\$ to zero -- what does it have to do with \$v_x\$ to make this happen? \$\endgroup\$ – TimWescott Apr 2 '19 at 21:19
  • \$\begingroup\$ Don´t forget the opamps internal output resistance...what is it for an IDEAL opamp? \$\endgroup\$ – LvW Apr 3 '19 at 6:41
  • \$\begingroup\$ @LvW Ideally, it would be 0, if that's the case, wouldn't every ideal op-amp circuit (at least with ones involving only one amplifier) have an ideal 0 ohms output resistance? \$\endgroup\$ – user101402 Apr 3 '19 at 14:00
  • \$\begingroup\$ Yes, of course...even for a real opamp we have a very small output resistance: Rout/LG. With: Rout=open-loop output resistance (data sheet) and LG=Loop gain. \$\endgroup\$ – LvW Apr 3 '19 at 18:14
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The first thing you can do to simplify things a little bit is, transform the dependent voltage source \$A\cdot V_D\$ into a current source using Source Transformation. The equivalent circuit will look as shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

Now we need to find: $$R_{output} = \frac{V_{output}}{I_x}$$ Calculating the sum of the currents at \$V_{output}\$: $${\frac{A\cdot V_d}{R_o}} + I_x - \frac{V_{output}}{R_o} - \frac{V_{output}}{R_i + R_2} = 0$$
Using a current divider: $$-\frac{V_{output}}{R_i + R_2}\cdot R_i = V_d$$ Replacing \$V_D\$ in the first equation yields: $$-\frac{V_{output}}{R_i + R_2}\cdot R_i \cdot \frac{A}{R_o} + I_x - \frac{V_{output}}{R_o} - \frac{V_{output}}{R_i + R_2} = 0$$
Rearranging: $$V_{output} \left[ \frac{R_i}{R_i + R_2}\cdot \frac{A}{R_o} + \frac{1}{R_o} + \frac{1}{R_i + R_2} \right] = I_x$$
Thus: $$\frac{V_{output}}{I_x} = \frac{1}{\left[ \frac{R_i}{R_i + R_2}\cdot \frac{A}{R_o} + \frac{1}{R_o} + \frac{1}{R_i + R_2}\right]}$$
Rearranging: $$R_{output} = \frac{R_o}{\left[ \frac{R_i}{R_i + R_2}\cdot A + \frac{R_o}{R_i + R_2} + 1 \right]}$$ The input resistance \$R_i\$ is usually very high so as \$R_i \rightarrow \infty\$ the equation above can be written as: $$\frac{R_o}{1 + A}$$ Which is the output resistance divided by the loop gain thanks to the feedback provided by \$R_2\$.

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An opamp has a vast amount of voltage gain and there is lots of negative feedback in your circuit. But to avoid oscillation it has a compensation capacitor which cuts frequencies above about 10Hz. Therefore its output impedance at DC and low frequencies is zero but its maximum output current is fairly low, about 20mA for many opamps.

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