# class A power amplifier

Discussion in 'Homework Help' started by lihle, Feb 21, 2011.

1. ### lihle Thread Starter Active Member

Apr 12, 2009
83
4
Why the current I should be greater than I (peak) for proper operation of the circuit.

lihle

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Are you asking why the power source needs to be capable of delivering at least the peak load current at rated output?

3. ### bitrex Active Member

Dec 13, 2009
79
4
Consider the simplest case of a class A amplifier, a single transistor with a resistor collector load, biased such that the collector is at 1/2 the supply voltage and capacitively coupled to the load.

At the limits, the single transistor can only do one of two things - it can saturate, or it can cut off. In saturation it can sink lots of current, so there's not a problem with the negative swing at the load. However, when the transistor cuts off, the only source for current to the load is through the collector resistor. To ensure that you get the maximum current through the load on the positive swing the collector resistor has to be small enough, hence the quiescent current large enough, that the voltage drop across that resistor doesn't compromise the output swing.

For example, let's say you have a 5 volt supply and want a swing of 1 volt into 1 ohm at the output, for an output current of 1 amp. Assume the circuit is set up such that the transistor is fully turned off at that point. As the transistor turns off, the collector voltage rises from 2.5 volts to 3.5 volts, and that AC current goes into the load. For it to happen properly, the collector load resistor must be less than (5-3.5 volts)/ 1 amp = 1.5 ohms.

In the quiescent state, a 1.5 ohm collector resistor against a 2.5 volt drop implies a collector current of 1.67 amps. So with a resistive load it seems one needs significantly more than the desired output current to achieve sufficient output swing. This is obviously pretty inefficient, as the quiescent power dissipated in that resistor is over 2.5 watts, for a 1 watt peak output.

A somewhat better way to do it is use a current source collector load and set it for the desired peak output current. Then the transistor is essentially "steering" the current from the current source between itself and the load, and the transistor takes twice the peak current on the negative swing (from both the load and the current source), and no current on the positive swing when the current from the current source goes into the load.

Last edited: Feb 21, 2011