Engineering Calculating Voltage Across 8 Ohm Resistor Circuit Design

[ThatGai]

Homework Statement

I am designing a circuit and trying to calculate the peak to peak voltage across the 8 ohm resistor. I'm getting 14 volts peak to peak output from the op amp. Any tips where I should start would be helpful.

I've attached my circuit to the post, Thanks

Homework Equations

The Attempt at a Solution

Since this is a design problem I am allowed to use software to play around with the circuit and measure the output, but I was wondering how would I calculate the voltage by hand.

 

[aralbrec]

Since this is a design problem I am allowed to use software to play around with the circuit and measure the output, but I was wondering how would I calculate the voltage by hand.

Q3 should be a PNP transistor.

The diodes D1,D2 and the transistors Q1,Q3 are a standard class AB output stage -- you may want to look that up if you are not familiar with it. R5 and R6 help to prevent thermal runaway on Q1 and Q3. As those transistors heat up, their VBE drops which causes more current to flow, which heats up the transistor, their VBE drops, etc. More current in R5/R6 acts as negative feedback to take away part of the VBE on Q1 and Q3 and reduce current.

Q1 and Q3 operate as voltage followers. The voltage output of the amp is directly transferred to the load through a VBE drop in Q1 and Q3. The diodes ensure that there is no cross-over distortion, ie no gap where Q1 and Q3 are both off. So the peek to peek voltage swing is going to be determined by how close the 741 can get to its rails on its output.

The negative swing sees Q3 transferring this voltage to the load less a VBE drop through the series R6/R8.

On the high swing, if the 741 can go high enough, D1 and D2 stop conducting**, and the base current of Q1 is determined by R7. So R7,Q1,R5,R8 determine how high the output can go.

** either the base of Q1 rises to 1.4V below the high supply so that the diodes can't get their VD drops in or current is diverted away from the diodes and into the base transistor to supply the load.