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I used the summing-point constraint and KVL to get \$V_{R_1<<}=V_{\text{in}}\$. Followed by a voltage divider for the node left of \$R_L\$; yielding:

$$V_{R_1<<}=V_{\text{in}}=V_o\left(\frac{R_1}{R_1+R_2}\right)^2\iff\frac{V_o}{V_{\text{in}}}=\frac{(R_1+R_2)^2}{R_1^2}=\boxed{1+2\frac{R_2}{R_1}+\frac{R_2^2}{R_1^2}}$$

However, textbook claims it's: $$1+\color{red}{3}\frac{R_2}{R_1}+\frac{R_2^2}{R_1^2}$$

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  • \$\begingroup\$ The textbook is right- a quick spice simulation showed 11V output with 1V input and R1 = 1k and R2 = 2k. \$\endgroup\$ – SunnyBoyNY Oct 9 '15 at 20:13
  • \$\begingroup\$ Heh, almost the same circuit as in electronics.stackexchange.com/questions/193937/… Although resistor values are different there (and given numerically), the method for solving is basically the same. \$\endgroup\$ – SX welcomes ageist gossip Oct 9 '15 at 21:00
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The textbook is correct.

Let \$R_{1\text{L}}\$ and \$R_{2\text{L}}\$ refer to the leftmost \$R_1\$ and \$R_2\$, respectively, and \$R_{1\text{R}}\$ and \$R_{2\text{R}}\$ refer to the rightmost \$R_1\$ and \$R_2\$, respectively.

The voltage at the inverting input of the op amp is \$V_- = V_{\text{in}}\$, so the current through \$R_{1\text{L}}\$ is \$V_{\text{in}}/R_{1\text{L}}\$. Since there is ideally no current into the op amp's input, the current through \$R_{2\text{L}}\$ is also \$V_{\text{in}}/R_{1\text{L}}\$.

The voltage across \$R_{2\text{L}}\$ is

$$\frac{V_{\text{in}}}{R_{1\text{L}}}R_{2\text{L}}$$

by Ohm's Law.

The voltage \$V_M\$ at the middle node (at the T intersection of the resistors) is therefore

$$V_M = V_{\text{in}} + \frac{V_{\text{in}}}{R_{1\text{L}}}R_{2\text{L}} \tag1$$

The current through \$R_{1\text{R}}\$ is \$V_{M}/R_{1\text{R}}\$. The current through \$R_{2\text{R}}\$ is this current plus the current through \$R_{2\text{L}}\$:

$$\frac{V_{M}}{R_{1\text{R}}} + \frac{V_{\text{in}}}{R_{1\text{L}}}$$

so the voltage across it is

$$\left(\frac{V_{M}}{R_{1\text{R}}} + \frac{V_{\text{in}}}{R_{1\text{L}}}\right)R_{2\text{R}}$$

This voltage plus \$V_M\$ is \$V_{\text{out}}\$:

$$V_{\text{out}} = V_M + \left(\frac{V_{M}}{R_{1\text{R}}} + \frac{V_{\text{in}}}{R_{1\text{L}}}\right)R_{2\text{R}} \tag2$$

Substituting \$(1)\$ into \$(2)\$ and dropping the L and R from the subscripts:

$$\begin{split}V_{\text{out}} &= V_{\text{in}} + \frac{V_{\text{in}}}{R_{1}}R_{2} + \left(\frac{V_{\text{in}} + \frac{V_{\text{in}}}{R_{1}}R_{2}}{R_{1}} + \frac{V_{\text{in}}}{R_{1}}\right)R_{2} \\ &= V_{\text{in}}\left(1 + \frac{R_{2}}{R_{1}} + \frac{R_{2}}{R_{1}} + \frac{R_{2}^2}{R_{1}^2} + \frac{R_{2}}{R_{1}}\right) \\ &= V_{\text{in}}\left(1 + 3\frac{R_{2}}{R_{1}} + \frac{R_{2}^2}{R_{1}^2}\right)\end{split}$$

$$\boxed{\frac{V_{\text{out}}}{V_{\text{in}}} = 1 + 3\frac{R_{2}}{R_{1}} + \frac{R_{2}^2}{R_{1}^2}}$$

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  • \$\begingroup\$ Gee, your solution looks nicer than mine! :) I better work on my formatting! \$\endgroup\$ – SunnyBoyNY Oct 9 '15 at 20:51
  • \$\begingroup\$ @SunnyBoyNY The important thing is that they are correct. :) \$\endgroup\$ – Null Oct 9 '15 at 20:54
  • \$\begingroup\$ You got my +1 too :) \$\endgroup\$ – SunnyBoyNY Oct 9 '15 at 20:55
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Not all steps are shown but hopefully enough to guide one through this interesting circuit!

Assumptions:

  1. No current flows into the opamp negative input terminal.

  2. The voltage difference between the negative and positive opamp input terminals is zero, hence the voltage on the negative input opamp pin is \$V_{in}\$.

Current "I" can be calculated as:

\$ I = \frac{V_{out}}{R_2+R_1||(R_1+R_2)} \$

Current through the left resistor \$ R_1 \$ is:

\$ I_a = I \frac{R_1}{2R_1+R_2} \$

Voltage drop across \$ V_{in} \$ due to \$ I_a \$ is:

\$ V_{in} = I_a R_1 = I \frac{R_1^2}{2R_1+R_2}\$

Hence:

\$ \frac{V_{out}}{V_{in}} = \frac{(2R_1+R_2)(R_2+R_1||(R_1+R_2))}{R_1^2}\$

\$ \frac{V_{out}}{V_{in}} = \frac{2R_1R_2+R_2^2+R_1^2+R_1R_2}{R_1^2}\$

\$ \frac{V_{out}}{V_{in}} = 1+3\frac{R_2}{R_1} + (\frac{R_2}{R_1})^2\$

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  • \$\begingroup\$ Good solution! You got my +1. \$\endgroup\$ – Null Oct 9 '15 at 20:54

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