1. Question

What is the maximum output over $R_L$ and the maximum efficiency $\eta$ for the amplifier circuit shown below: Below you will find my solution. Is this done correctly? What would you have done differently?

2. Voltage over $R_L$

Assumed is an ideal transistor with infinite current amplification $\beta$ and saturation voltage $U_{ CE_{ sat } }$ to be 0. Therefor the base current is 0 and the voltage drop over the transistor is considered constant for the whole output range. Furthermore is the output voltage $u_a$ sinusoidal.   This means for $$u_a(t)=\hat{U}_a\sin(\omega t)$$ the amplitude $\hat{U}_a$ is the voltage divider over $R_L$, when the maximum of the dynamic range of the transistor is reached, that no current flows over the transistor.  $$U_{a,\,max}=U_0\frac{R_L}{R_L+R_C}$$ $$\hat{U}_{a,\,max}=\frac{U_{a,\,max}}{2}$$ $$U_{a,\,eff,\,max}=\frac{\hat{U}_{a,\,max}}{\sqrt{2}}$$ The root mean square leads to:  $$U_{a,\,eff,\,max}=\frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C}$$

3. Maximum output $P_L$ over $R_L$

$$P_L=\frac{1}{2}\frac{U_{a,\,eff,\,max}^2}{R_L}$$ $$P_L=\frac{U_0^2}{16}\frac{R_L}{(R_L+R_C)^2}$$ For $\frac{dP_L}{dR_L}=0$ leads $R_L=R_C$ to the maximum output over $R_L$.  $$P_{L,\,max}=\frac{1}{64}\frac{U_0^2}{R_L}$$

4. Maximum efficiency $\eta$

The total circuit power is the voltage over both resistors which is $U_0$:  $$P_{in}=\frac{1}{2}\frac{U_0^2}{R_L+R_C}$$ With the given condition that $R_L=R_C$: $$P_{in}=\frac{1}{4}\frac{U_0^2}{R_L}$$ The maximum efficiency $\eta$ is: $$\eta=\frac{P_L}{P_in}=\frac{1}{16}=6.25\%$$

1. Question

What is the maximum output over \$R_L\$ and the maximum efficiency \$\eta\$ for the amplifier circuit shown below:

Below you will find my solution. Is this done correctly? What would you have done differently?

2. Voltage over \$R_L\$

Assumed is an ideal transistor with infinite current amplification \$\beta\$ and saturation voltage \$U_{ CE_{ sat } }\$ to be 0. Therefor the base current is 0 and the voltage drop over the transistor is considered constant for the whole output range. Furthermore is the output voltage \$u_a\$ sinusoidal. 

This means for $$u_a(t)=\hat{U}_a\sin(\omega t)$$ the amplitude \$\hat{U}_a\$ is the voltage divider over \$R_L\$, when the maximum of the dynamic range of the transistor is reached, that no current flows over the transistor. 

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

The root mean square leads to: 

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

3. Maximum output \$P_L\$ over \$R_L\$

$$P_L=\frac{1}{2}\frac{U_{a,\,eff,\,max}^2}{R_L}$$ $$P_L=\frac{U_0^2}{16}\frac{R_L}{(R_L+R_C)^2}$$

For \$\frac{dP_L}{dR_L}=0\$ leads \$R_L=R_C\$ to the maximum output over \$R_L\$. 

$$P_{L,\,max}=\frac{1}{64}\frac{U_0^2}{R_L}$$

4. Maximum efficiency \$\eta\$

The total circuit power is the voltage over both resistors which is \$U_0\$: 

$$P_{in}=\frac{1}{2}\frac{U_0^2}{R_L+R_C}$$

With the given condition that \$R_L=R_C\$: $$P_{in}=\frac{1}{4}\frac{U_0^2}{R_L}$$

The maximum efficiency \$\eta\$ is: $$\eta=\frac{P_L}{P_{in}}=\frac{1}{16}=6.25\%$$