DC Operation of the preamp stage
- for clarity I removed the components involved in AC signals.
I am posting this for the purpose of clarifying my previous posts regarding not recommending to jumper out the 2.2K resistor. If the ideas presented here are aligned with that of some people it is not for the purpose of siding with him. If some ideas are not aligned with that of some people, it is not for the purpose of opposing him. I am posting to share what I think is useful for everybody. It can also be used to analize the tweaks presented by Tony.
The preamp compose by Q1 and Q2 must be able to drive the power amp with enough output swing so that the power amp can deliver its rated 60W output power.
To deliver 60w into 8 ohms the output of the power amp must be able to put out 60/8 = 2.738Arms with a voltage output of approx. 22Vrms. Since the gain of the power amp is approx 47K/2K2 = 21 , We need 22Vrsm/21 = 1.05Vrms at the input of the differential stage. That’s 2.96V pk-pk or approx 3V pk-pk.
The preamp stage therefore must be able to deliver 3Vpk-pk at its output (collector of Q2). This means the voltage at the collector of Q2 must be able to swing 1.5V up and 1.5V down from its nominal voltage during no signal conditions.
Tony previously presented an optimum operating point for the preamp stage but that approach is quite complicated from a designer standpoint. The most common design approach is to set the collector of Q2 at ½ the supply voltage or 10V in this case. We will see if this approach will allow the preamp circuit to deliver the 3V pk-pk signal required by the power amp. If not then we can consider Tony's approach.
When the collector of Q2 is at 10V, The current through R5 and R6 (ignoring the current through R3) will be 10V/(2k2+10K) = 0.82mA. Ignoring the base current of Q2, we can say current through R4 will also by 0.82mA. Under this condition, the voltage at Pt. B will be 5.57V and the voltage at Pt. A will be 18.2V.
If the voltage at the collector of Q2 swings 1.5V down to 8.5V, the current through the resistors will increase from 0.82mA to 0.943mA. The voltage at Pt. B will be 6.4V. The difference between 8.5V and 6.4V is 2.1V. This means Q2 is far from saturation yet. If the voltage at the collector of Q2 swings 1.5V up to 11.5V, calculation will show that there is plenty of room for this positive swing. Since this shows that the 3V pk-pk signal required at the collector of Q2 is possible without driving Q2 into saturation or cut-off (meaning no clipping), the common design approach of setting collector of Q2 at ½ the supply voltage presents no problem at all.
Both Q1 and Q2 must be biased such that changes in temperature which affects the Vbe of transistors is compensated. Otherwise, the operating point of Q1 and Q2 could drift resulting in clipping of the signals especially when the amp is operated at high power levels.
For Q1, bias stability is achieved by takings its base bias from the emitter of Q2 via R1. For example ambient temp rises such that the Vbe of Q1 decreases. This will cause Q1 to conduct more resulting in its collector voltage to drop. This drop in collector voltage is unwanted. If left unchecked, Q1 could go into saturation and cause distortion. But since the collector of Q1 drives the base of Q2, when the voltage at the collector of Q1 drops, the base drive of Q2 is reduced. As a result Q2 conducts less and its emitter voltage drops. This reduction in emitter voltage reduces the base drive to Q1 forcing Q1 to conduct less - thus opposing the effect of temperature on the operating point of Q1.
For Q2, bias stability is achieved by taking its base bias from the junction of R6 and R5 or pt. A via R3. If ambient temp rises, Vbe of Q2 will decrease causing it to conduct more. Again if left unchecked, Q2 could eventually go into staturation. But as Q2 starts to conduct more as a result of temperature rise, the voltage drop across R6 increases. This causes the voltage at Pt. A to drop which in turn reduces the base drive to Q2 and forcing Q2 to conduct less. Again, the process opposes the effect of temperature on the operating point of Q1.
For Q1 and Q2, the opposite scenario will occur when temperature drops. For both Q1 and Q2, negative DC feedback are used to compensate for temperature effects. It is also important to note that the same negative DC feedback compensates for hfe variations.