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Here's the circuit:

enter image description here

Here's my attempt:

enter image description here

I already have the values for Ad. Is this right? I need to choose the values for R1 and R2 wisely I think.

What I don't understand is the point of having R3 in there, and what would the value for R3 have to be taking into consideration what I said about R1, R2 and the gain. It is said to be ~R2//R1 but I dont know why.

What should I do compute the input impedance of that circuit (op-amp with negative feedback?) Can you guys give me any tips?

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  • \$\begingroup\$ By assuming 1). openloop gain >> closedloop gain, and 2). Opamp is not saturated, you can assume that V- = V+ = 0V, and thus simplify your equations. Assuming V- = Vout/Ad is not correct. \$\endgroup\$
    – EvertW
    May 19, 2020 at 8:55

3 Answers 3

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You know there's a problem since your resistor value for R2 is lower than that from an ideal op-amp calculation, and we know that with non-infinite gain it must be higher than the ideal value of 50K.

In your second from last line, the gain is -10, not +10.

I get:

\$R_2 = \frac{R_1\cdot G (1+1/A_0)}{1-G/A_0}\$ where G is the desired gain (times minus 1) and \$A_0\$ is the open-loop gain of the op-amp.

The resistor R3 is to eliminate offset caused by input bias current. If the input bias currents are equal, then making R3 = R1||R2 will do that.

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The gain of an inverting opamp, assuming the open loop gain >> closed loop gain, is given by \$- \frac {R_f} {R_i}\$ and in your case that is \$- \frac {R2} {R1}\$. You can read about the derivation at this question. (The - shows phase inversion).

A practical consideration is to keep the feedback resistance in a reasonable range (typically below 100k and above perhaps 1k or so although lower feedback resistances can be found - the reason is to minimise the amount of current in the feedback loop while avoiding a low frequency pole caused by a very high feedback resistance).

The impedance 'seen' by the inverting input is R2 || R1 and to prevent (or at least minimise) voltage offsets caused by input bias current, R3 should be equal to that value. The || means 'in parallel'.

The input impedance as 'seen' by the signal source is simply R1 as the inverting input is a virtual ground.

Taking standard values, we could select R2 = \$100k \Omega \$ and R1 = \$10k \Omega\$ for a gain of -10.

R3 should be equal to the parallel equivalent of those which is \$ \frac {R2R1} {R2 + R1}\$ = \$9.09k \Omega \$. A standard value that is close would be \$9.1k \Omega \$

The input impedance is simply R1 (because assuming the opamp is within it's common mode range, the inverting input is at 0V) and is therefore \$10k \Omega\$ for this example.

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Due to R3 holding the + input to ground, the - input becomes a virtual ground, as such your signal input impedance is just R1

Next up what R3 is there for, normal op amps are not quite perfect, some current flows in to or out from the input pins, so a common approach to balance out the influence of this is to make both op amp inputs have the same resistance,

R1/R2 you can think of cancelling currents to keep that - input at ground, so if your input is 1V, and you want the output to be 10x larger, it needs to be 10x the resistance, so 10x the voltage gives the same current. so if R1 was 1K, R2 would be 10K, and you can scale up or down this ratio to better suite your input impedance requirements, too low and the op amp wont be able to supply enough current, too high and it gets very noisy

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