Electronic – Non-linear amplification circuit

What if we start with:

$$U_{a,eff,max} = \frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C}$$

Then we get:

$$P_L = \frac{(U_{a,eff,max})^2}{R_L}=\frac{(\frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C})^2}{R_L} = \frac{U^2_0R^2_L}{8R_L(R_L+R_C)^2}=\frac{U^2_0R_L}{8(R_L+R_C)^2}$$

And with \$R_C=R_L\$ we get for\$P_{L,max}\$:

$$P_{L,max}=\frac{U^2_0}{32\cdot R_L}$$

For the circuit power we have to consider the power loss of \$R_C\$ and the losses of the transistor itself.

$$P_{R_C} = P_{R_C,DC}+P_{R_C,AC}$$ $$P_{R_C,DC} = \frac{U_{R_C,DC}^2}{R_C}=\frac{(\frac{1}{2}(U_{R_C,max}+U_{R_C,min}))^2}{R_C}=\frac{(\frac{1}{2}(U_0+U_{a,max}))^2}{R_C}$$

If we set again \$R_C=R_L\$ then \$U_{a,max} = \frac{1}{2}U_0\$ and \$U_{R_C,DC}=\frac{3}{4}U_0\$ we get:

$$P_{R_C,DC} = \frac{9U^2_0}{16R_C}$$

For the AC-Part follows:

$$P_{R_C,AC}=\frac{(U_{R_C,eff})^2}{R_C}=\frac{(\frac{U_{R_C,max}-U_{R_C,DC}}{\sqrt{2}})^2}{R_C}=\frac{(\frac{U_{0}-\frac{3}{4}U_{0}}{\sqrt{2}})^2}{R_C}=\frac{U^2_0}{32R_C}$$

Finlay we get:

$$P_{R_C} = \frac{9U^2_0}{16R_C} + \frac{U^2_0}{32R_C} = \frac{19U^2_0}{32R_C}$$

Next how do we get \$P_{transistor}\$ ?

The way I would approach this is to observe that Rf/R3 = 2 (by inspection)

The gain from the non-inverting op-amp input is thus 2 + 1 = 3.

So you need a divider with gain 1/3 from V1 and 2/3 from V2.

Set R1||R2||R4 = R

Then R/R2 = 2/3 and R/R1 = 1/3

Since 1/R4 = 1/R - 1/R2 - 1/R1 = 1/R - 2/3R - 1/3R = 0 So R4 = \$\infty\$

You can pick two more values arbitrarily- say R = 10K which leads to R2 = 15K, R1 = 30K

It would make sense to constrain RF||R3 = 10K to balance the offset from bias currents, so RF = 30K R3 = 15K.

Edit:-

General principle for a voltage divider with n resistors R1...Rn

Vout/Vin = R * (V1/R1+V2/R2+...+VN/RN)

Where R = R1||R2||..||RN

(It's easy to prove this by superposition)