Amplitude control circuit problem

Question

I have a problem with understanding this particular circuit given above, just to make it clear, this is the circuit I am talking about (part of this circuit is amplitude control though, but still).

Now, it turns out that, when we are looking for voltage Vo, we do it using superposition principle, so for example, when we are considering only Vcc generator we have:

$$V_o=V_{cc} \frac{R_4+R_5}{R_2+R_3+R_4+R_5}$$

NOTE: I know that this is not the complete expression, V0 will eventually be the same as if we had inverting opamp (in case that R2=R5 and R3=R4)

But, this expression confuses me because it implies that current is the same through all of the resistors R2,...,R4 (R1 is shorted in this case.)

I don't know how is that possible since we have resistor Rf and output of OPamp, shouldn't both of them have a current flowing through, thereby, we would have different currents on resistors R2 and R5 for example.

How come then this expression still happens to be correct in all textbooks, what am I missing?

EDIT: Values of Va and Vb are such that these diodes are off.

Answer 1

If \$V_i = 0\$ and the opamp is ideal, then there is no reason for the current through \$R_2\$, \$R_3\$, \$R_4\$ and \$R_5\$ to be identical. Due to negative feedback, the voltage over \$R_1\$ will be 0, and there will therefore be no current flowing through \$R_f\$. The output will be fixed to 0V by the opamp regardless of the resistor values.

Maybe the solution you found is meant for a different circuit, maybe it is not used in the context of superposition, maybe the situation they are trying to describe is something else entirely, or it could also be just wrong. I can only say that the expression you gave does not give you the output voltage for \$V_i = 0\$ as you would do with superposition.

Answer 2

Cdummie, the shown diagram contains one of the classical methods for soft-limiting the output amplitude of a sinusoidal oscillator. For describing the function, we will assume that Vi is a sinusoidal signal with rising amplitudes. Without the diode networks, the output amplitude would be clipped (hard-limited) as soon as this amplitude is as large as the power rail Vcc.

Concerning the 4-resistor chain: The given formula for Vo (as a function of Vcc) does not apply because the opamp is enforcing Vo=0 V (because dual supply voltages are shown). The formula applies only in case of singe supply. More than that, the DC voltage between R2 and R3 (R4 and R5) will be +Vr (-Vr) and is determined by the voltage divider rule.

What happens when Vi is a sinus with rising amplitudes?

There will be a sinusoidal voltage at the opamp output (centered at 0 volts DC). As soon as the positive (negative) amplitude of Vo is able to slighly open D2 (D1) there will be an additional negative feedback path (parallel to Rf) and the gain will be, therefore, reduced. Hence a soft-limiting effect takes place and the output signal will have lower distortion (THD) if compared with the hard-limiting case.