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Since V+ and V- are both @ 0 volts I thought I could combine these resistors in series and parallel, but that gave me the wrong answer. (The right answer is Vout = -4IR)

Circuit

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The fastest way, I think, is just to redraw the darned thing:

schematic

simulate this circuit – Schematic created using CircuitLab

I show the virtual ground node on the bottom, lower left side there. That's the \$V_-\$ input to the opamp. But you know that there's a current, \$I\$, that must be driven through the first \$R\$ in your schematic (\$R_1\$ in mine.) So \$V_1=I R\$. Clearly, \$V_1\$ also means that there is an added \$I\$ coming in from \$R_2\$; which joins with the other \$I\$ to make for \$2I\$, now. This larger current must flow through \$R_3\$. When it does, it develops a voltage across \$R_3\$ that is equal to \$V_2=2I\cdot R_3+V_1=2IR\$. \$V_2\$ now causes a current in \$R_4\$, which is equal to \$2I\$. This adds to the current in \$R_3\$ to make a total current of \$4I\$ in \$R_5\$. This means that \$V_{OUT}=4I R_5 + V_2=4IR\$.

I think that's the simplest way to picture the answer.


Bear in mind that your output will have to slew, negatively, for a positive current into the \$V_-\$ node. So the above discussion is entirely about the magnitude, and not the sign, of \$V_{OUT}\$. (Note added per ThePhoton's comment here. Thanks!)


A completely different approach using Thevenin would be to, let's say, start at the output end and work backwards. I'll use my resistor numbering from above (since you didn't number yours.)

schematic

simulate this circuit

Starting at the output, we have the Thevenin of the obvious first divider:

$$\begin{align*} R_{TH}&=\frac{R_4\cdot R_5}{R_4+R_5}\\\\ V_{TH}&=V_{OUT}\cdot\frac{R_4}{R_4+R_5}\\\\ \end{align*}$$

\$V_{TH}\$'s series \$R_{TH}\$ impedance is also in series with \$R_3\$, so: \$R^{'}_{TH}=R_3+R_{TH}\$. This modified Thevenin output resistance now forms a divider with \$R_2\$. So:

$$\begin{align*} R^{''}_{TH}&=\frac{R^{'}_{TH}\cdot R_2}{R^{'}_{TH}+ R_2}\\\\ V^{''}_{TH}&=V_{TH}\cdot\frac{R_2}{R^{'}_{TH}+ R_2}\\\\ \end{align*}$$

\$V^{''}_{TH}\$'s series \$R^{''}_{TH}\$ impedance is also in series with \$R_1\$, so: \$R^{'''}_{TH}=R_1+R^{''}_{TH}\$.

That's about it. The sum of the currents into the \$V_-\$ node must equal zero. So:

$$I + \frac{V^{''}_{TH} - 0\:\textrm{V}}{R^{'''}_{TH}}=0$$

Just given the above, alone, you can then solve for \$V_{OUT}\$. This works out to:

$$\begin{align*} \textrm{setting }R_X&=R_3+R_5+\frac{R_3\cdot R_5}{R_4}\textrm{, then:}\\\\ V_{OUT}&=-I\cdot\left[R_X\cdot\left(1+\frac{R_1}{R_2}\right)+R_1\cdot\left(1+\frac{R_5}{R_4}\right)\right]\\\\ &=-I\cdot\left[\left(R_3+R_5+\frac{R_3\cdot R_5}{R_4}\right)\cdot\left(1+\frac{R_1}{R_2}\right)+R_1\cdot\left(1+\frac{R_5}{R_4}\right)\right] \end{align*}$$

If you plug in \$R_1=R_2=R_4=R\$ and \$R_3=R_5=\frac{R}{2}\$, then you will get the same \$V_{OUT}=-4IR\$ result.

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    \$\begingroup\$ @gabson Hopefully, that approach works for you, too. :) \$\endgroup\$
    – jonk
    Jun 5, 2017 at 6:06
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    \$\begingroup\$ Shouldn't it be -4IR? Or am I not really awake yet? \$\endgroup\$
    – The Photon
    Jun 5, 2017 at 14:45
  • \$\begingroup\$ @ThePhoton No, you are awake. I was merely trying to provide the fastest way to reach the answer sought by the OP. I'll add magnitude bars, or something. \$\endgroup\$
    – jonk
    Jun 5, 2017 at 14:49

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