Push-Pull Amplifier Homework: Calculate Vcc, R Input & Gain

[Unskilled]

Homework Statement

1. Calculate Vcc and Vee(should be around 14-16).
2. Calculate the input resistance. (You should get a high value).
3. Determine the voltage gain. No calculation is needed.

http://img210.imageshack.us/my.php?image=effectampqe2.jpg


Ucesat = 1 V
Poutmax = 10 W
Rl = 8 ohm
Icq9=Icq11=40mA
Re=0.47 ohm
R8=R9=1000 ohm
Vbe=0.7V


2. The attempt at a solution
1. I used P=U^2/(2*R)=>U=sqrt(P*2*R)=sqrt(10*2*8)=12.65V
Vcc=U+Ucesat=12.65+1=13.65 but this is less than 14.

2.Here i don't know how to start.

3.I don't know how to solve this either.

 

[unplebeian]

Can you please repost a clear circuit. It's hard to see. Thanks.

 

[piercas]

I would first like to commend you for attempting to solve this problem on your own. It shows that you are actively engaged in your learning and are willing to put in the effort to understand the material. However, I would like to provide some guidance to help you better understand this concept.

1. To calculate Vcc and Vee, we can use the formula P = V^2/R. Since we are given the maximum output power (Poutmax) and the load resistance (Rl), we can rearrange the formula to solve for V. This gives us V = sqrt(Poutmax*Rl) = sqrt(10*8) = 8.94V. However, this is the voltage across the load, so we need to add the voltage drop across the collector-emitter junction (Ucesat) to get the total supply voltage. Therefore, Vcc = V + Ucesat = 8.94 + 1 = 9.94V. Similarly, Vee = V - Ucesat = 8.94 - 1 = 7.94V. These values are within the range of 14-16V mentioned in the problem statement.

2. The input resistance can be calculated using the formula Rinput = Vbe/Ib, where Vbe is the base-emitter voltage and Ib is the base current. In this case, Vbe is given as 0.7V and the base current can be calculated using Ohm's Law (Ib = V/R). The resistance R is the parallel combination of R8 and R9, which can be calculated as R = (R8*R9)/(R8+R9). Substituting all the values, we get Rinput = 0.7/(0.7/(1000*1000)/(1000+1000)) = 2M ohm. This is a high value, as expected for an amplifier.

3. The voltage gain can be determined by looking at the circuit diagram. This is a push-pull amplifier, which means the output voltage is the sum of the voltages from two transistors (Q9 and Q11). Since they are in opposite phases, the output voltage is amplified. Therefore, the voltage gain is approximately equal to the number of turns in the transformer (N = 1 in this case). No calculation is needed, but it is important to understand how the amplifier