2
\$\begingroup\$

I was attempting to implement figure Figure 34 on the AD734 datasheet in multisim (the national instruments SPICE simulator) Figure 34

Based off what I know of op amps, the output of U2b should be saturated to rail voltage (in my simulation it is supplied by +15 V to -15 V), meaning that the diagram in the datasheet is incorrect (ie., it doesn't output the rms of V_in).

My problem is that within NI multisim, with the circuit as depicted in the datasheet, it works perfectly. Within the simulation the output of U2b is the same as the input to the negative pin. Within the simulation if the negative pin input is replaced with a voltage source, the output saturates to the rail voltage as you would expect. If the AD708 pair is replaced with a pair of virtual ideal opamps, the output also hits the rail as I would expect.

I suspected that Figure 34 is incorrect, but multisim is backing it up, and I don't see how this isn't a bug. Would this circuit work in reality?

\$\endgroup\$
3
  • \$\begingroup\$ Perhaps if you walked us through why you think it should be saturated we could let you know what you've done wrong. \$\endgroup\$ Commented Mar 6, 2015 at 6:13
  • \$\begingroup\$ U2b is in a positive feedback configuration with no resistors. if the voltage on the negative leg is positive, the op-amp will increase its output voltage until it runs out of power (saturated). If it was feeding back into the negative input, and the circuit's output was connected to the positive input, I would expect it to simply output whatever voltage is on the positive input \$\endgroup\$
    – JamesL
    Commented Mar 6, 2015 at 6:18
  • \$\begingroup\$ If U2b is replaced by a single ideal op-amp, it behaves exactly as I said in the above comment within multisim \$\endgroup\$
    – JamesL
    Commented Mar 6, 2015 at 6:20

1 Answer 1

1
\$\begingroup\$

Your suspicion is correct. The figure is wrong. U2b is intended to be a voltage follower and should be connected as

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

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.