I am reading through some school notes on audio amplifiers and have come to the section on calculating power when there is no signal present. My own calculations disagree with the answer given (highlighted) and I would be grateful if someone could confirm whether I am correct. If I am not correct, I would be very grateful if someone could explain how the given formula is derived and where I am going wrong.

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Image source https://resource.download.wjec.co.uk/vtc/2017-18/17-18_3-3/eng/gce-electronics-book-chapter-6.pdf (29 June 2021)

As an aside, I would like to point out the bias voltage (X) will not be exactly half the rail voltage as this ignores the loading effect of the transistor amplifier. I am ignoring this point for now.

Assuming Vbe equals 0.7 V I agree with the calculation (Vout = Vx – Vbe) for output voltage across the load, 6.8 V

I also agree with the calculation for power dissipated in the loudspeaker (P = V^2 / R), 0.578 W

However, I do not agree nor understand how they have calculated power dissipated by the transistor.

By my calculations:

Vce = Vs – Ve = 15 – 6.8 = 8.2 V

I will assume the base current is negligible and can be ignored when calculating collector current.

Ic = loudspeaker power / loudspeaker voltage = 0.578 / 6.8 = 0.085 A

Transistor power + loudspeaker power = Vs * Ic = 15 * 0.085 = 1.275 W


Power dissipated by transistor = total power – loudspeaker power = 1.275 – 0.085 = 1.19 W and not 0.697 W as stated.


In the previous calculation I accidentally used the wrong value (current and not power). My corrected formula should be: Power dissipated by transistor = total power – loudspeaker power = 1.275 – 0.578 = 0.697 W which matches the value provided in the booklet.

  • \$\begingroup\$ Your last equation confuses power and current. Substitute loudspeaker power and try again. (And yes that is a hilariously bad exercise) \$\endgroup\$ Jun 29, 2021 at 20:22
  • \$\begingroup\$ Thanks for spotting that - it now works. \$\endgroup\$
    – Pzy
    Jun 29, 2021 at 20:50
  • \$\begingroup\$ What's good is that you have two methods, which produce the same result. Now you can examine theirs and see how it is exactly equivalent to yours, via Ohm's Law. \$\endgroup\$ Jun 29, 2021 at 20:52


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