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Can somebody please explain in both cases the biasing method?

1.Signal connected between biasing diodes with single decoupling capacitor:

Signal connected directly to each transistors base with separate capacitors:

I don't understand the following points:

  • It's OK that we need ~0.7V for PNP and NPN base-emitter voltage, that's why a diode is used for it, because diode voltage is also ~0.7V, but how does it work exactly?
  • For an NPN transistor, the base voltage should be 0.7V higher than the emitter voltage, for a PNP transistor base voltage should be 0.7V lower than the emitter voltage. How do we know that the emitter voltage of the NPN transistor will be lower than the base voltage but this point is also higher than the PNP base? Because it is grounded via RL? But the capacitor blocks DC, so what is the voltage level on the connection of the 2 emitters? Maybe if somebody can explain it with exact VCC, Vin values it would be more helpful.
  • What is the voltage level of TR2 base and what is the voltage level of TR1 base?
  • How is the input signal added to the base? Why do the diodes not block it in case of the first figure?
JRE
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Darko
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  • `Because it is grounded via RL?` Nope! Look at the output coupling cap which is basically an open-circuit at DC. And we are biasing the transistors at DC. And also, take a look at [here](https://electronics.stackexchange.com/questions/601641/class-ab-bjt-biasing-diodes/601682). – Rohat Kılıç Jan 04 '22 at 13:46
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    Does this answer your question? [Class AB BJT biasing, diodes](https://electronics.stackexchange.com/questions/601641/class-ab-bjt-biasing-diodes) – TooTea Jan 04 '22 at 13:57
  • Signal diodes must match Vbe to track thermally but this either uses bipolar V+/V- or huge caps on output. (unlike fig 2 which is wrong, not to show -Vcc) but for AC; gnd and -Vcc are the same, so I'm assuming there was a reason for showing this way – Tony Stewart EE75 Jan 04 '22 at 13:59
  • https://electronics.stackexchange.com/questions/502552/class-ab-power-amplifier-output-signal/502583#502583 and here https://forum.allaboutcircuits.com/threads/compensating-diodes-used-on-push-pull-amplifier.101749/#post-766917 – G36 Jan 04 '22 at 14:42

1 Answers1

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Bipolar drivers must be biased at middle of entire supply range, which is usually bipolar supplies.

Then the Class AB uses the input diode pair to control the output Vbe pair which controls the output DC current. Since the diode is smaller it also has slightly higher internal resistance such that it's current and thus PN voltages are controlled by the equal R1,R2 drops into the total supply voltage remaining. (+V + -V )-2* VBE)/ (R1+R2) .

Then for thermal stability the diodes must be mounted on the same heatsink so that when Vbe of power transistors drops the current increases unless the diodes drop in voltage at the same time approx -4mV/'C. This is to prevent "thermal runaway". Then negative feedback is normally used to reduce the previous stage voltage gain and further reduce crossover distortion. Ideally you want PNP current to decrease while NPN increases during AC signal crossover. This occurs when there is sufficient DC quiescent current in the output stage, but not excessive so that it heats up from VI drop.

It is possible to use an Op Amp driver with negative feedback coming from the DC output, but you may need an extra buffer stage to boost output current only if you have a low impedance load. The DC reference for the Op Amp must again be in the middle of the total supply range to null the DC error voltage. Limiting the gain of say a 1MHz GBW OA is determined by desired BW . E.g. 1e6 / 1e4 Hz (BW) = Av=100 max

Tony Stewart EE75
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