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What would happen in a class a amplifier if I connect a capacitor in series with the load resistor (between power supply and the transistor collector)? My goal was to block the dc qiescent current to reduce energy wasted as heat.

I apologize in advance for my english (i'm italian) an for my probably stupid question.

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    \$\begingroup\$ The transistor needs a DC voltage across it in order to work, so your amplifier likely wouldn't work \$\endgroup\$ – C_Elegans Dec 3 '18 at 18:47
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The transistor needs a DC voltage between its collector and emitter or drain and source in order to work. What will happen when you put a capacitor in series with the resistor to the collector/drain is the capacitor will initially have 0v across it, but it will quickly charge to the supply voltage through the transistor, leaving 0v across the transistor.

For example, this circuit

enter image description here

produces the following output:

enter image description here

The charging of the capacitor and subsequent drop in collector voltage is visible on \$V(cap)\$, and the waveform for $\V(out)\$ is quite distorted due to the capacitor.

However, if I replace the capacitor with a short circuit, a mostly distortion free sine wave is visible on \$V(out)\$, with an amplitude of about 2V pk/pk.

enter image description here

Finally, for small signal amplifiers using BJTs, the transconductance \$g_m\$ (and input resistance \$r_e\$) is dependent on how much quiescent current is flowing through the transistor.

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