0
\$\begingroup\$

So I'm having a bit of trouble calculating the expression for the gain or ,\$ \frac{V_{out}}{V_{in}}\$, for the given Op-Amp circuit down below. I've gotten to the point where I've written out the KCL equations for this circuit.

What I have so far is, that the output current from the Op-Amp,\$i_0\$, splits up into \$i_1, i_2, i_3\$, which go through \$R_2\$, \$V_o\$, and \$R_3\$ respectively.

We can clearly see \$i_0 = i_1 + i_2 + i_3\$.

As far as fleshing out the KCL equations, I've calculated \$i_3\$, going through \$ R_3 \$ which I've derived as \$i_3 = \frac{V_{GND} - V_0}{R_3} = \frac{V_0}{R_3}\$ and similarly we can reduce \$i_0 = i_1 + i_2 + i_3 \rightarrow i_0 = i_2 + i_3 \$. This is since \$i_1 = i_n\$ which is the current into the negative node of the op amp, which we know that current going into either node, \$i_n\$ or \$i_p\$ must be zero.

At this point, I'm stuck producing the necessary equations for \$i_2\$ (that is dependent of the voltages \$V_0\$ or \$V_{in}\$ which we know), out into the rest of the general circuitry. Any and all help getting these equations in order would be greatly appreciated!

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
\$\begingroup\$

The trick you missed (I think) is that, since this opamp is in a negative feedback loop, you can assume that the voltage between the + and - inputs of the opamp is zero.

You can theoretically explain this by assuming that the opamp has an infinite gain so Vout = (Vin+ - Vin-) * infinite. This results in that the output voltage can have any value for input voltage = zero.

With that assumption (voltage at Vin- = zero, because Vin+ is grounded) and realizing that no current flows into or out of the opamp's inputs it is now much easier to calculate the currents and then the transfer function.

\$\endgroup\$
1
\$\begingroup\$

It sounds like you've used a sledgehmammer, but missed the nut.

Of the several things we know about an opamp (OK, an ideal opamp) when it's working linearly ...

a) the input currents are zero (so I_r1 = I_r2, in magnitude at least)
b) the voltage between the inputs is 0
c) it can deliver any output current

So we know the voltage on both ends of R1

... so we know its current

... so we know R2's current

... so we know the output voltage

job done

The nut will splatter to pulp if you hit it with KCL, but you have to hit it in the right place, at least you know the simple way to do it now to cross-check.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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