1
\$\begingroup\$

I was working on this problem:

Problem Screenshot

Find values for \$\{R_2, R_3, R_4\}\$

  • Assume ideal op-amp
  • Gain of 100
  • Input resistance of \$1M \Omega\$
  • No resistor is larger than \$1M \Omega\$

And I build the following system of equations to find an expression for the resistor values.

$$ \begin{equation} \begin{cases} \dfrac{0-V_x}{R_2} +\dfrac{0-V_x}{R_3} = \dfrac{V_x-V_{\text{out}}}{R_4} &&\text{KCL on } V_x \\ \dfrac{V_{\text{in}}}{10^6} = \dfrac{0-V_x}{R_2} &&i_1 = i_2 \\ \dfrac{V_\text{out}}{V_\text{in}} = -100 &&A_v = \dfrac{V_\text{out}}{V_\text{in}} \end{cases}\, \end{equation} $$

I believe this system of equations to be incorrect for an expression in terms of \$\{R_2, R_3, R_4\}\$. I think I have a misunderstanding of the principles of op-amps or I've just contradicted something somewhere.

Where have I gone wrong and how do I solve this correctly?

\$\endgroup\$
2
  • \$\begingroup\$ I haven't worked through your equations to see how this affects them, but the current directions for I2, I3, and I4 are reversed. \$\endgroup\$
    – AnalogKid
    Apr 6 at 3:05
  • 1
    \$\begingroup\$ I think the most simple method is to use the star-triangel transformation for the three resistors R2...R4. From the resulting feedback circuitry two resistors have no influence on the gain and can be neglected. Hence, we have a simple inverter with two gain-detrmining resistors only. \$\endgroup\$
    – LvW
    Apr 6 at 13:29

2 Answers 2

3
\$\begingroup\$

The negative reaction causes the inverting input to be at virtual ground. The current in \$R_1\$ is \$I1=\frac{V_i}{R_1}\$ and is the same as that circulating in \$R_2\$, so the voltage across \$R_2\$ is \$V_2=R_2 \cdot I1\$ , enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ Thanks @Franc, just wondering how the voltage on \$R_2\$ is inverted on \$R_3\$? Is that because there is no current flowing through \$i_4\$ in the analysis? \$\endgroup\$ Apr 6 at 9:49
  • \$\begingroup\$ Also, do you have any hints as to why my original attempt failed? I still don't understand. Thank you \$\endgroup\$ Apr 6 at 9:50
  • \$\begingroup\$ @Freddy I wanted to say that the positive is on the ground side. \$\endgroup\$
    – Franc
    Apr 6 at 9:54
4
\$\begingroup\$

It's all about how you approach the problem and hardly anything about using math. You can use math of course but, failing to recognize how an op-amp works is just asking for trouble and making the problem bigger than it needs to be.

Consider this redrawn image and temporarily assume we have a "new" output: -

enter image description here

At our "new output" we can arrange it to have negative unity gain by making R1 = R2

Additionally, we can imagine that R4 is the natural output impedance inside the op-amp and, that R3 is an output load. For those reasons and this: -

Input resistance of 1 MΩ

We can justify that R1 and R2 are 1 MΩ.

But, we also want R3 to be a lot smaller than R2. When we do this we find that the loading effect of R2 on the R3/R4 attenuator is trivially small (making the values for R3 and R4 easy to find).

So, with R1 at 1 MΩ we fulfil the input resistance requirements (because the inverting input is a virtual ground).

Then, because we know that we have negative unity gain at our "new output", the signal levels at the true op-amp output need to be 100 times higher to meet this requirement: -

Gain of 100

I expect that it actually means that the gain magnitude is 100

So, making R3 and R4 form an attenuator of 100 fulfils our requirements but, we should ensure that small values are used compared to R2. Typically, we could use 10 Ω for R3 and 990 Ω for R4.

So, what does it look like in a simulator: -

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer Andy, and yes the gain was meant to be the magnitude as it's inverting anyway. I appreciate your intuitive working out and though process but I just have 2 questions. 1, is your deduction based on the idea that \$R_4\$ (internal resistance) and \$R_3\$ (load resistance) when assumed to be 0 \$\Omega\$ converts the circuit to this? 2) is there a way this method expands to an equation? I need to find one with of all the resistors to solve this \$\endgroup\$ Apr 7 at 9:24
  • 2
    \$\begingroup\$ @FreddyMcloughlan (1) my answer is about showing the circuit relationships so that you can take it to the next level and, if you want, mathematically formulate a solution based on proper circuit understanding. For instance, R3 needn't be trivially small compared to R2 but, then R2 has to take account of the Thevenin source impedance presented to it by R3 and R4. Currently in my answer the Thevenin source impedance is a fraction under 10 ohms and this can be ignored (one solution from many). (2) due to what I've said you can see that there is a mathematical method of generalizing the circuit. \$\endgroup\$
    – Andy aka
    Apr 7 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.