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enter image description here

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.
enter image description here

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?

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    \$\begingroup\$ There's current through R1, so voltage divider doesn't apply. \$\endgroup\$ – Chu Jul 26 '17 at 7:15
  • \$\begingroup\$ Use nodal on V1, then substitute your own equation for V1 and solve for the output voltage. \$\endgroup\$ – jonk Jul 26 '17 at 7:34
  • \$\begingroup\$ 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 \$\endgroup\$ – Utshaw Jul 26 '17 at 7:40
  • \$\begingroup\$ @Utshaw I saw that you placed an answer and then deleted it. Let me show you what I mean. See answer below. \$\endgroup\$ – jonk Jul 26 '17 at 7:54
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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.

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  • \$\begingroup\$ 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 \$\endgroup\$ – Utshaw Jul 26 '17 at 8:12
  • \$\begingroup\$ @Utshaw Yes, that would work as well. \$\endgroup\$ – jonk Jul 26 '17 at 8:19
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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:

schematic

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.

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

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