I have a basic Class AB push-pull amplifier that uses diode biasing for the transistors. The circuit operates from a single-ended supply (e.g., +VCC and ground), and the input is a sinusoidal signal with a DC offset so that it remains positive. I would like to understand how this circuit works in detail and whether it will only function as intended with a dual power supply. Additionally, how can I calculate the average power dissipation in each transistor under these conditions?
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\$\begingroup\$ Try to read this electronics.stackexchange.com/questions/309936/… and this electronics.stackexchange.com/questions/450576/…, and this also may be useful electronics.stackexchange.com/questions/537846/…. The average power dissipation in the transistor will be around \$P_{D} ≈\frac{0.025 * V_{CC}^2}{RL} \$ plus a DC current looses \$P_{DC} = 0.5*I_Q*V_{CC} \$ \$\endgroup\$G36– G362026-01-01 07:25:34 +00:00Commented 21 hours ago
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1\$\begingroup\$ Whether it will "function as intended", which you ask, depends on what is intended. The title and description only tell us that it should work as class AB push-pull with a signal that remains positive, and that does not require a dual power supply at all! It will work fine as it is. (Bootstrapping R3 and R4 would make more sense as an improvement.) \$\endgroup\$Jos Bergervoet– Jos Bergervoet2026-01-01 07:36:29 +00:00Commented 21 hours ago
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\$\begingroup\$ electronics.stackexchange.com/a/660775/61398 \$\endgroup\$Circuit fantasist– Circuit fantasist2026-01-01 07:36:55 +00:00Commented 21 hours ago
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1\$\begingroup\$ You have drawn the circuit into a simulator, LTSpice by the looks of it. What is the intent of asking the question as you could simply simulate the circuit yourself already, with the simulator providing you power consumption etc in any condition you want? \$\endgroup\$Justme– Justme2026-01-01 11:07:31 +00:00Commented 17 hours ago
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\$\begingroup\$ @user641035 please put in the schematic embedded as an image \$\endgroup\$Voltage Spike– Voltage Spike ♦2026-01-02 03:07:58 +00:00Commented 1 hour ago
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1 Answer
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Your single supply push-pull stage works the same way as an "ordinary" push-pull stage supply from a split supply:
How does this Push-Pull amplifier work?
The diodes are here to reduce "deadzone".
Of course, for your circuit, you need to add an input capacitor as well.
simulate this circuit – Schematic created using CircuitLab
The circuit is working in the following way:
In short. For positive half-cycles, the upper NPN transistor delivers current to the load by charging the output capacitor. For negative half-cycles, the lower PNP transistor discharges the capacitor. And this is why we have a negative voltage at the load. The base current is coming from R1 and R2. Vin only provides a current to D2 and R2 for the positive half-cycle. And to D1, R1 for the negative half-cycle.
The maximum load current will be equal to:
$$I_{Lmax} = \frac{V_{CC} - V_{C2} - V_{BE}}{\frac{R_1}{\beta + 1} + R_{E1} + R_L}$$
Where \$V_{C2}\$ is a DC voltage across C2 capacitor (Vcc/2).
The theoretical average power dissipation in the transistor is equal to: $$P_D = \frac{V_{CC}^2}{4 \pi^2 R_L} \approx 0.025 \frac{V_{CC}^2}{R_L}$$
Plus a DC current looses \$ P_{DC} = I_{EQ} *(V_{CC} - V_{C2}) \$
