3
\$\begingroup\$

I was searching different circuits to produce sinusoidal waves and came across one that seems simple but I do not understand how it works. It has the usual band pass filter (in this case it was a Wien bridge) and an amplification stage, which is the one I do not know how it works.

Amplification stage

In the image I have shown the circuit and a simulation in Proteus. The zener has a zener voltage of 2.2 V.

I understand that C3 and C4 are decoupling capacitors and that those set the lower cutoff of the gain plot. R4 and R7 set the bias for Q1. R5 and RV1 set the gain. And Q1 and Q2 are similar to a Sziklai Pair, but with R3 and R6+zener between the collector of Q2 and the emitter of Q1.

How does this circuit work?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ C3 and C4 are "decoupling" the AC from the DC, but that is typically not what decoupling capacitor means. Decoupling capacitor refers to a capacitor across the power pins of an active component. The term for C3 and C4 is "blocking capacitor". It's blocking the DC voltage. EDIT: (Sometimes also "coupling capacitor", because it's coupling the input to the output capacitively) \$\endgroup\$
    – MrGerber
    Commented Aug 26 at 12:59
  • 1
    \$\begingroup\$ Ah, English is not my mother tongue so maybe I have mixed coupling and decoupling without realizing, sorry :) \$\endgroup\$
    – Noxbru
    Commented Aug 26 at 13:17

1 Answer 1

0
\$\begingroup\$

Q1 is functioning as an emitter follower, but most of the current is supplied by Q2. R3 along with R5 and RV1 form a voltage divider that sets the overall gain. You can think of it as a very simple opamp, where the base of Q1 is the noninverting input, node "C" is the inverting input and the collector of Q2 is the output.

But if the output voltage rises high enough, D1 conducts, putting R6 in parallel with R3, which reduces the gain by a factor of about two. This produces a soft limiting effect that can be used to stabilize the amplitude of an oscillator.

\$\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.