# Relation between Output voltage and current in a transresistance amplifier I need to show that for the above current to voltage converter,

$\frac{V_0}{i_s} = -R_1(1+\frac{R_3}{R_1}+\frac{R_3}{R_2})$

assuming that the op-amp is ideal,
Voltage at the negative input terminal = $V_n$
Current through the negative input terminal = $i_n$
Current through the positive input terminal = $i_p$
Voltage at the positive input terminal = $V_p$
$V_p$ = $V_n = 0V$
$i_p = i_n = 0A$
Using voltage divider rule, $V_1$ = $\frac{R_2}{R_2+R_3}V_0$
$i_s = \frac{0-V_1}{R_1}$, So , using these two equations ,
$\frac{V_0}{i_s} = -R_1(1+\frac{R3}{R2})$
Why my answer is wrong ?

EDIT: I think have figured out the error on my previous calculation . The voltage divider rule still works here like this. Let, $R_p$ be equivalent for R1 and R2
$R_p = \frac{R_1R_2}{R_1+R_2}$
$V_1 = \frac{R_p}{R_p+R_3}V_0$
$I_s = \frac{0-V_1}{R_1}$
$I_s = \frac{0-\frac{\frac{R_1R_2}{R_1+R_2}}{\frac{R_1R_2}{R_1+R_2}+R_3}V_0}{R_1}$
After solving this the proof comes.Is there any discrepancy in this?

• There's current through R1, so voltage divider doesn't apply. – Chu Jul 26 '17 at 7:15
• Use nodal on V1, then substitute your own equation for V1 and solve for the output voltage. – jonk Jul 26 '17 at 7:34
• Can I assume that $R_1$ and $R_2$ are parallel as they have same voltage between them? I have edited my question and using that approach I have got the proof . @jonk – Utshaw Jul 26 '17 at 7:40
• @Utshaw I saw that you placed an answer and then deleted it. Let me show you what I mean. See answer below. – jonk Jul 26 '17 at 7:54

You already know $V_1$. And given your edited/added approach to solving the problem, which works too, I've no problem adding the follow-up to my earlier suggestion that you use nodal analysis.

So just do the nodal for $V_1$:

\begin{align*} \frac{V_1}{R_2}+\frac{V_1}{R_3}&=i_s+\frac{v_o}{R_3}\\\\ V_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)&=i_s+\frac{v_o}{R_3} \end{align*}

That's the nodal for $V_1$. But you also know that $V_1=-i_s\cdot R_1$. (You already said so.) So:

\begin{align*} -i_s\cdot R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)&=i_s+\frac{v_o}{R_3}\\\\ -i_s-i_s\cdot R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)&=\frac{v_o}{R_3}\\\\ -i_s\cdot\left[1+ R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)\right]&=\frac{v_o}{R_3}\\\\ v_o&=-i_s\cdot R_3\cdot\left[1+ R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)\right]\\\\ \frac{v_o}{i_s}&=- R_3\cdot\left[1+ R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)\right]\\\\ \frac{v_o}{i_s}&=- \left(R_3+ \frac{R_1 R_3}{R_2}+R_1\right)\\\\ \frac{v_o}{i_s}&=- R_1\cdot\left(1+\frac{R_3}{R_1}+ \frac{R_3}{R_2}\right) \end{align*}

Which amounts to what you said you needed to prove.

However, it wouldn't hurt to go one more step:

\begin{align*} \frac{v_o}{i_s}&=- R_1\cdot R_3\left(\frac{1}{R_1}+ \frac{1}{R_2}+\frac{1}{R_3}\right)\\\\ &=-\frac{R_1\cdot R_3}{R_1\:\mid\mid\: R_2\:\mid\mid\: R_3} \end{align*}

Since all three resistors are attached to voltage sources, and a common node, you'd expect that they are in some way parallel to each other. The above equation makes that fact explicit.

• Thanks ,I have understood it.But have you seen my **Edited ** section of question ?Can I consider that R2 and R1 are in parallel ?As this question has come across my mind after solving that way I deleted the answer and posted it in a separate section inside the question . @jonk – Utshaw Jul 26 '17 at 8:12
• @Utshaw Yes, that would work as well. – jonk Jul 26 '17 at 8:19

To solve the circuit you may try to apply delta-wye (triangle-star)transformations, if you know them. The three resistors are in a star (aka wye) configuration.

If you substitute them with the equivalent triangle (aka delta) configuration, you get a circuit like this: simulate this circuit – Schematic created using CircuitLab

Then you can convert the input current source with Ra in parallel to a voltage source and you get the classic inverting amplifier circuit.

• But as $V_n$ is 0V, I think , voltage divider still works here . – Utshaw Jul 26 '17 at 7:31