# Solving a problem with an op-amp in saturation

I am solving this problem regarding a circuit with two op-amps.

The voltage sources $$\u_1\$$ and $$\ u_2 \$$ are ideal DC sources, so the capacitor is simply an open-circuit.

The op-amps are ideal and equal, and their saturation voltages are $$\ \pm 6 V \$$ with $$\R_0= 10 k\Omega\$$ and $$\R= 20 k\Omega\$$.

Ok so assuming linear region, I can understand that basically each op-amp configuration is the superposition of an inverting and non-inverting input, namely

$$u_{10}=\left( 1 + \frac{R}{R_0} \right)u_1-\frac{R}{R_0}u_2=3u_1-2u_2$$

$$u_{20}=\left( 1 + \frac{R}{R_0} \right)u_2-\frac{R}{R_0}u_1=3u_2-2u_1$$

Ok, now if $$\u_1=3V\$$ and $$\u_2=1V\$$ the expressions would lead to $$\u_{10}=7 V\$$ and $$\u_{20}=-3 V\$$, so the first op-amp is in saturation.

That means these expressions are not valid.

I have replaced $$\u_{10}\$$ by the saturation voltage of $$\6V\$$. Now I don't know how I can calculate voltage $$\u_{20}\$$ in these conditions. The answer is apparently -7/3 V according to the textbook. Can someone give me an hint. I have tried KVL/KCL but I think I keep applying ideal op-amp conditions when they are not valid outside the linear region (such as zero currents and zero voltage across terminals).

So, following Jonathan suggestion I'll remove the op-amp A1 from the circuit while forcing $$\u_{10}\$$ equal to $$\6V\$$. In that case

$$u_{20}=\left( 1 + \frac{R}{R+R_0} \right)u_2-\frac{R}{R+R_0}u_{10}= 5/3 u_2- 2/3 u_{10}$$

And that leads to $$u_{20}=5/3-12/3=-7/3$$

So it is solved!

• Only A1 is saturated, A2 is not. So create an expression for the output voltage of A2 with normal opamp function. You can assume that the input current to A1 is negligible even though it is in saturation, so it won't load the voltage at its input. Feb 26, 2023 at 1:49