0
\$\begingroup\$

I was going through BJT Power Amplifiers and the efficiency is given as

Efficiency = (AC power at Load)/(Input DC Power)

I have two different questions for the above expression:

  • Why are we not considering AC Output power/AC input power as overall efficiency?

  • Why are we not considering AC input power in above expression? I mean the efficiency have to be = (AC power at Load)/(Input DC Power + Input signal Power)

Can you please explain ? Thanks in advance...

\$\endgroup\$
  • 1
    \$\begingroup\$ The assumption is that the BJT amplifier uses DC input power. There is no AC input power. \$\endgroup\$ – Kevin White Nov 13 '16 at 19:13
  • 1
    \$\begingroup\$ (1) because the AC power in is the signal so that would be the power gain. (2) Because the AC power in is the input signal and has nothing to do with the efficiency. The AC power at the output comes only from the DC power supplied. \$\endgroup\$ – JIm Dearden Nov 13 '16 at 19:13
  • 1
    \$\begingroup\$ Thanks for your reply.. but where does the input ac power go..?? \$\endgroup\$ – METALHEAD Nov 13 '16 at 19:14
  • 1
    \$\begingroup\$ Some of the voltage/current goes into the first stage of the amplifier, some may go through an input resistor. Remember its an amplifier so the input power is usually very small. It does not contribute to output power. \$\endgroup\$ – JIm Dearden Nov 13 '16 at 19:19
3
\$\begingroup\$

To be strictly correct the power taken from the input source would be fair game for calculating efficiency however, most folk neglect that because a push-pull (for example) amplifier's signal input power is probably less than 1% of the output power.

This is because a push-pull stage needs another transistor in front of it (usually common emitter) to set up the biasing for the push pull stage and if ten percent of the power to the push pull stage is provided by this transistor (via the power rails), then the input ac power to this front end transistor might be one tenth lower again and probably a lot more.

So, it's a fair point but a trivial one IMO.

\$\endgroup\$
1
\$\begingroup\$

All the power amps I can think of start with are powered from a fixed DC supply (usually fixed voltage). Power transistor(s) are controlled to convert this supply to generate output AC power. Different Classes of power amplifiers have a theoretical maximum efficiency - for simplicity, this maximum doesn't include driving power, because that would require knowledge of transistor characteristics:
Class A max efficiency: 50%
Class B max efficiency: 75%
Class D max efficiency: 100%
These figures are never achieved in practice because no transistors are rail-to-rail (in either voltage or current). And overall efficiency as you propose is even worse, because no transistor has infinite power gain.
For example, I have built radio frequency amplifier of Class E type with RFpower/(DCpower) efficiency of about 95%. But the overall transmitter efficiency, including oscillator, buffer, driver, and power stage only had 50% efficiency.

\$\endgroup\$
0
\$\begingroup\$

Considering audio power amps, the input signal power is negligible, compared to the amp's output power. Consider a closed-loop voltage gain of maybe 10x to 50x, and you're also comparing an output load impedance of 4-8 Ohms to an actual input impedance in double digit kiloOhms at the very least (let's skip the debate about noise figure vs. signal source impedance for the moment). Thus, the power gain can be 10000x or above.

If you should mean an RF line amp, a fairly normal power gain of +20 dB still means 100x amplification. Does 1% difference in energy efficiency make any practical difference?

In that context, overall energy efficiency is a much more useful figure: power output vs. power consumption. Note that class AB linear amps have a typical efficiency of 50 to 65 % (71% theoretical maximum for a sinewave signal and DC power rails), "switch-mode" class D amps have "switch-mode" efficiency levels (80 to 90% or maybe more). The figures above are for "very little distortion", often specified as 1% THD. One particular aspect, where knowledge of energy efficiency of the circuit comes useful, is thermal design (heat dissipation).

=> it's rather a question of definition and practical utility.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.