Calculating the power of a simple amplifier

Question

I found this circuit on internet. It says it's a 1w amplifier. There is no explanation about how that is calculated. How could the power of this circuit calcuted? (input is a radio)

And is there any design error at this circuit?

Answer 1

This circuit was 'designed' by somebody who does not have a good idea of what they are doing. As well as what has been said by others, there is worse to come.

The two x 100k resistors set Vbase at half Vcollector.
As Vbase mean = 2 x Vbe diode drops =~ 1.2V the collector DC voltage is ~= 2.4V.
V collector can swing to 0 Volt and to 4.8V for an undistorted sine wave - although you will get distortion - see below. The RMS Voltage at the collector = Vpeak /2/sqrt(2) = 4.8 / 2 / 1.414 =~ 1.7VAC RMS.

It gets worse.
With an 8 ohm load the most power you could get = V^2/R = 0.36 Watt.
BUT the 'designer' has added a 50 ohm resistor in series with the speaker - presumably to limit the current that otherwise flows. This resistor IS the main load. The most power you can get with that resistor as the load = V^2/R = 1.7^2/50 =~ 58 mW.

It get worse:
The transistor "wants" to maintain Vbe = 2 x Vbe diode drops ~= 1.2V. Input signal that drives the base away from this point "for no reason"* will produce distortion. If you try and drive Vcollector to near ground the distortion will be severe. Adding a small value emitter resistor with a capacitor across it to pass AC will help somewhat, but the circuit is fatally flawed and this is not worthwhile. [* re: "for no reason" - Vbe will vary in order for base current to alter as required but the transistor will "find its own operating point" and does not need or appreciate being driven by voltage.

It gets worse:
Finally - the connection of the 2 x 100k bias resistors to the collector is not an effective and stable way of biasing silicon transistors. this is a "trick" from the long long long ago days of Germanium transistors and worked OK enough for them - for silicon their are different characteristics which tend to make this not wise. Also, if you did do this, it would usually be desirable to split the top resistor in two and add a decoupling capacitor to ground between the two to stop AC negative feedback which you do not want (probably) in this case.

Overall, a bad circuit that should not be used BUT an excellent learning opportunity.

Answer 2

To determine the power this amplifier can deliver into the speaker, you have to know the speaker's impedance. Most speakers are 8 or 4 Ohms. In this case considering the 50 Ω in series with the speaker, higher will get more power, so pick 8 Ω.

Now find the minimum and maximum current thru the speaker. Since we're talking about audio power, this is the maximum peak to peak current amplitude of a sound wave. Assume a sine shape, so the RMS of the AC part of that current will be the peak to peak value divided by 2 square roots of 2, or 2.83. Remember that power is the current squared times the resistance, so square the RMS amps and multiply by 8 Ω.

I don't like this circuit because it puts a DC bias current thru the speaker. That will require that the speaker be rated for more than just the audio power this amplifier can deliver since it will be operating over only half its range at best. If this amp puts out 1 W of audio power, then it won't work with a 1 W speaker. Even a speaker that can handle the extra power may not sound quite right with a DC current thru it added to the audio current.

Answer 3

The idea here is likely that if the output of the amplifier is biased halfway between the 9V power rails, it can have a maximum swing of +/- 4.5 volts. From 4.5 peak we get 0.707 * 4.5 = 3.2V RMS. Problem is, there is a 50 ohm resistor in series with the transistor's collector and the speaker. So we actually have a load which is at least 50 ohms, plus the impedance of the speaker. 3.2V RMS into 50 ohms is only 0.2W using \$P = V^2/R\$. The designer was probably incorrectly applying \$V^2/R\$ using the impedance of the speaker only: from that we get about 1.2W.

Another problem is that it is a bad idea to drive speakers this way. When there is no signal flowing, there is a DC voltage across the speaker. This kind of a circuit can burn out the speaker voice coils. Also by energizing the voice coil, it basically places tension on the speaker, moving it in or out.

A proper audio amplifier does not move any current through a speaker when the signal is quiescent allowing the cone to be at rest and not dissipating heat through the voice coil.

This means that for a class A design along these lines, the speaker should be coupled via a capacitor which blocks the DC.

I would skip this circuit.