Electrical – Squaring Circuit

I put your circuit into LTspice but it didn't work very well even when I upped the supply voltage to +/- 12V, so I Googled "precision full wave rectifier" and then clicked on the "More images..." header.

I found the circuit below, which looked pretty good, and tried it.

Unfortunately, I got oscillations on the rectified negative half-cycles and, suspecting the phase response through part(?) of the circuit to be the reason, I plugged in a couple of LT1007s to see if I could speed the thing up and brute force the oscillation out of there, and it cleaned up everything nicely!

If you want to play with the circuit, here's the LTspice circuit list.

The left op-amp is working in inverting mode.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Positive going input.

• If the INPUT voltage goes positive then the - input will start to rise above the + input and the output will start to go negative.
• D1 will then conduct pulling the - input back down to zero.
• The voltage on R2/R4 junction will be -Vin.
• The voltage on OA2 + input will be 0 V since it's tied to the virtual zero of OA1.
• OA2 is then operating in inverting mode so the output will be -(-Vin) = Vin.

^{simulate this circuit}

Figure 2. Negative going input.

• When the INPUT voltage goes negative the cleverness of the circuit comes into play.
• OA1 - input will be virtual zero so I1 = 0.1 mA.
• If OA2 is in equilibrium both inputs will be at the same voltage. Therefore \$V_B = V_C\$.
• Since there is 20k in the upper circuit and only 10k in the lower circuit \$i_3 = \frac{2}{3}I_1\$.
• Therefore \$V_B = -\frac {2}{3}V_{IN}\$ as is \$V_C\$.
• OA2 is then operating in non-inverting mode so the output $$ V_{D} = V_B (1 + \frac {R_f}{R_i}) = V_B (1 + \frac {10k}{20k}) = 1.5 \cdot V_B = 1.5\cdot\frac{2}{3}(-V_{IN}) = -V{IN} $$

It took me a while to see what was happening. I hope that helps.