# Determining Output Resistance of Transresistance Amplifier

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:

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:

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?

simulate this circuit

• 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. Commented Apr 2, 2019 at 20:15
• 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? Commented Apr 2, 2019 at 21:19
• Don´t forget the opamps internal output resistance...what is it for an IDEAL opamp?
– LvW
Commented Apr 3, 2019 at 6:41
• @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? Commented Apr 3, 2019 at 14:00
• 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.
– LvW
Commented Apr 3, 2019 at 18:14

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:

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$$
By inspection: $$-\frac{V_{output}}{R_i + R_2} = \frac{V_d}{R_i}$$ Thus: $$-\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\$$.

• This helped me, but why you can ignore the input current in the expression for Vd? Commented Oct 15, 2023 at 8:49
• @sandyscott it’s not being ignored. The input current is the fourth term in the second equation. Commented Oct 15, 2023 at 12:22
• I mean in the voltage divider expression you used to calculate Vd. ie. Vd = -Vout . Ri / ( Ri + R2). The voltage divider equation is only valid if the current in both resistors is same - ie. no current in or out at the midpoint between them, which in this example in the input current. Commented Oct 16, 2023 at 7:15
• @sandyscott I updated the answer. A current divider wasn't being used. What is being used is equating the currents since they should be the same given that there's no input current (the source is open). Commented Oct 16, 2023 at 11:49

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.