# Tips on analyzing circuit with multiple op-amps and bjts

In a homework exercise I need to sketch the Vout-vs-Vin graph, having it first obtained analytically. I know that Ur and Vin are both nonnegative, and the circuit is shown here:

It seems like there is a lot of symmetry in the circuit, and that this apparenntly complicated circuit can be broken down into a few smaller ones. All four of the op-amp stages resemble the classical logarithmic amplifier, however I feel that we can somehow bypass invoking the $\ln$ function in this analysis. UPDATE: Here is an Lt-spice dc sweep graph I did for R=0.5k, and Ur=5V. Seems, like Q3 was at first saturated, then went off.

• Wow, Kinda tricky. I'd start with op3 and op4, summing all the current as a function of U, Vin and Vout (and the R's) That will get you the current through Q3 and Q4. Then I'd sum the currents into OP1... It gets a bit hazy here...but that will give the current through Q1. But you'll have some expression for the output voltage of op1... and then find the current through Q2. Looking forward to a more complete answer. – George Herold Jan 7 '16 at 21:04
• @GeorgeHerold Thanks for the hints. I will now update my question, though. – Emir Šemšić Jan 7 '16 at 21:12
• It's important to know what the power supplies to the Op-Amps are. If the negative supply is at ground, some BJTs are not even going to turn on. If it's powered by symmetrical supplies it much more complicated. – jpcgt Jan 12 '16 at 18:09

When analyzing complex circuits, you need to be able to break down the overall circuit into sub-circuits which you have already done. Then you need to understand how each sub-circuit functions. You can do this by simulation or do some research and find similar circuits. The next step is to find some equations if possible to describe the sub-circuits.

Superposition is your friend, remove and add different parts of the circuit or substitute voltages and currents. Observe how the summing circuit would function without Q2. Then add it in and see what happens. Also simulate the "green" log circuit with a sin wave input. Run frequency sweeps or an AC analysis if the design concerned with the frequency domain. This circuit has a fixed voltage which makes it nice because half of the circuit is running at a DC fixed value, that makes an equation analysis easier.

Here is some info I've found on log circuits: Maxim integrated Log IC

Start by redrawing the circuit in a way that you can identify sub-circuits and feedback (Note: I used Vdd instead of Ur):

You can see 4 sub-circuits, all with just 2 pins. For all of them you can write a Iin/Vout or Vin/Vout transfer function.

It's also useful to identify certain currents and their directions given that the polarity of some voltages is given, and that you know there is a virtual ground on all negative inputs to the Op-Apms.

Finally observe the feedback path, which takes the voltage from the output and feeds back a current of 3*Vout/*(2R).

Now you have all the tools you need to build your analytical solution. Beware though that the feedback could be positive under certain conditions, which will make your circuit rail.

Some extra tips:

Analyze the circuit without feedback first. I.e. cut the line that says feedback. Then you can know if the feedback is positive (adds to the input when the input increases) or negative.

Consider OP3 Q3 to start with, now if the opamp is in its linear region the two input voltages will be near enough identical, and the non inverting input is tied to 0V, so the output will be riven such that Ic (Q3) = Ur/R + Iin (The input to the section), and the input to the section is a virtual earth node.

Ebbers moll will then give you the opamp output voltage ( = Vbe) in terms of Ic and hence in terms of input current.

The same sort of reasoning applies to OP4/Q4, but here the input current is Vout/2R.

Next consider OP1/Q1, in the linear region again Ic (Q1) must equal Ur/R (for the opamp inputs to be equal) So ebbers moll will give you the required Vbe, note that the base is now biased to the same (slightly negative voltage as the emitter of Q3, so the optput of OP1 must be more negative then that).

Ic (Q2) can now be computed (ebbers moll again) as we know Q4s Vbe.

• Isn't there also Vin/R entering the node at the collector of Q3? – Scott Seidman Jan 12 '16 at 17:07
• Ic(q3) = Ur/R + Iin, where Iin is clearly Vin/R + Vout/R, so Ic(Q3) = (Ur + Vin + Vout)/R, but I was trying to come up with a transimpedance model of the section (current input, voltage output) as that expression is probably more useful for finding a solution. – Dan Mills Jan 13 '16 at 13:14