Unfortunately, the calculated peak input voltage for 2.02 V peak out is 2.099418493 V, so 4.198836987 V peak-peak in for 4.04 V peak-peak out, ratio of output to input voltage 0.962171195.
I guess the variations of the current through the diodes are too large. When you have to take that into account, you probably end up with transcendental equations that have to be solved iteratively. IQ depends on the 1.9th power of the current through the diodes due to their emission coefficient of 1.9.
Maybe you can approximately calculate the effect of current variations through the diodes when you realize that the output current is mostly determined by the NPN in the positive peaks and by the PNP in the negative peaks. Hence, the NPN base has to swing 2.099418493 V above its bias voltage during the positive peaks and the PNP base 2.099418493 V below its bias voltage during the negative peaks. Node A then has to swing more than 2.099418493 V in the positive and 2.099418493 V in the negative direction. To be continued.
I guess the variations of the current through the diodes are too large. When you have to take that into account, you probably end up with transcendental equations that have to be solved iteratively. IQ depends on the 1.9th power of the current through the diodes due to their emission coefficient of 1.9.
Maybe you can approximately calculate the effect of current variations through the diodes when you realize that the output current is mostly determined by the NPN in the positive peaks and by the PNP in the negative peaks. Hence, the NPN base has to swing 2.099418493 V above its bias voltage during the positive peaks and the PNP base 2.099418493 V below its bias voltage during the negative peaks. Node A then has to swing more than 2.099418493 V in the positive and 2.099418493 V in the negative direction. To be continued.
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You need about 86 mV more in the positive peaks when you take into account the current variations through the diodes. 4.285 V still isn't quite 4.36 V.
After some pondering and playing with numbers, I think I figured out a rather simple way to compute the 'apparent' re' -- indicated below as a-re' -- good enough that it can be plugged into Av = Rl/(re' + Rl) and provide a usable ballpark figure with a somewhat acceptable accuracy.
1, figure out Ip-p for a given output level
2. average Ip-p over the full cycle (divide by pi) -> Ia (mA)
3. 25/Ia = re'
4. multiply re' by 2 (since we got 2 transistors) to get a-re'
5. Then simply use Av = Rl/(a-re' + Rl)
I ran some simulations and also computed Av using the above derivation:
simulated, computed
1. 0.91743, 0.92721 (Rl = 100, Re = 0)
2. 0.94129, 0.93941 (Rl = 330, Re = 0)
3. 0.88327, 0.88102 (Rl = 330, Re = 22)
4. 0.88038, 0.86421 (Rl = 220, Re = 22, Vcc = 36V)
Looks like some refinements might still be needed (maybe a diode drop somewhere) but it's much better than what I had originally.
1, figure out Ip-p for a given output level
2. average Ip-p over the full cycle (divide by pi) -> Ia (mA)
3. 25/Ia = re'
4. multiply re' by 2 (since we got 2 transistors) to get a-re'
5. Then simply use Av = Rl/(a-re' + Rl)
I ran some simulations and also computed Av using the above derivation:
simulated, computed
1. 0.91743, 0.92721 (Rl = 100, Re = 0)
2. 0.94129, 0.93941 (Rl = 330, Re = 0)
3. 0.88327, 0.88102 (Rl = 330, Re = 22)
4. 0.88038, 0.86421 (Rl = 220, Re = 22, Vcc = 36V)
Looks like some refinements might still be needed (maybe a diode drop somewhere) but it's much better than what I had originally.
