Here is a simple audio amp. Is there an easy way to calculate the power it gives?

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Resistor value, input level and speaker impedance changed.


2 Answers 2


This stinks.

The two 150k\$\Omega\$ resistors bias the transistor at a 14\$\mu\$A base current. At an \$H_{FE}\$ of 15000 that's 210mA quiescent current. So far so good, if it weren't that that DC current flows through the speaker. You don't do that. Speakers should only get AC. Anyway, your quiescent point will be at 3.3V, which gives you a maximum \$V_P\$ of 1.7V, or 175mW RMS in 8\$\Omega\$.

The 10\$\mu\$F cap has an impedance of 16\$\Omega\$ at 1000Hz. With two base-emitter resistances of 26\$\Omega\$ that's a 4.4mA peak base current at 300mV. Even at the lowest specified \$H_{FE}\$ of 4000 that would give 17.6A collector current. With the appropriate power supply that gives 2500W in 8\$\Omega\$. The 5V supply and the 1.5V \$V_{CE(SAT)}\$ will limit that to 440mA, but it does mean that the transistor goes completely in saturation.

edit (after changes in schematic)
The changes don't make much difference. Base current is now 5.9\$\mu\$A, collector current 89mA, which, with a 16\$\Omega\$ speaker gives a quiescent point at 3.6V. With an 8\$\Omega\$ speaker it's 4.3V. The results are the same.

  • \$\begingroup\$ @Stevenh - see my link. the original writer knows what he is about - but warns of the shortcomings of this circuit. He also offers many others. \$\endgroup\$
    – Russell McMahon
    May 28, 2012 at 10:44

This is NOT a good beginner's circuit. The circuity designer does know what he is doing but his intention is that the circuit be played with by people who understand what is happening - see web page ref below.

The maximum possible power output from an amplifier of this sort is given by
Vo^2 / Rload where
Vo is AC RMS voltage output and
Rload is the load resistance.

Vout_peak_to_peak max is the power supply voltage.
AC swing = _/1 Vpowersupply/2

If you have a sine wave output that just touches both ground and V+ then the RMS value is Vdc/2 / sqrt(2) (based on the definition of RMS voltage).

In practice slightly more than rail-to-rail will be achievable. Lets use the above on this amplifier:

I'm preparing an independent answer on this subject. BUT the easiest way to tell is to read what the person said who published it - he tells you. The original is here at www.techlib.com - a superb website and a magnificent resource.

With a 5V supply the max possible Vout AC = +/- 2.5 VAC peak to peak.
5 VAC peak to peak = 5 x 1/2 x 0.7071 V RMS = 1.77 V RMS.
Say about 1.5V RMS to allow some "headroom".
Power = V^2/R = here 1.5^2/8 = 280 mW.
He says "about 250 mW" and as "about 250" is exactly equal to 280 (at least for engineers) the answer seems liable to be about correct :-). See below for his comments.

This circuit is vastly superior to the circuit based on it which was posted in a question about 1 day ago. It appears that somebody took this circuit and tried to simplify it and make it work on 9V, without understanding how it worked.

Unlike the prior one this has 5V supply , no resistor in series with the speaker, different bias and an input capacitor. The text with it on the above webpage explains its good and bad points.

This circuit can be greatly improved by
- connecting a resistor of about 1 ohm (say 0.47 ohm at high currents and 2 ohm at low currents) from the transistor emitter to ground and

  • connecting a 100 uF capacitor from emitter to ground.

  • You could also change the upper 150k resistor to 180k.

The added 1 ohm resistor plus capacitor will improve output quality and will reduce output very very slightly.

You can operate this circuit on 9V for more power (about 1 Watt) by changing the upper 150 k to 390 kohm. This increases the collector voltage to about 4.5V (half supply). You may need to change the upper resistor slightly above and below this to limit current. Use the added emitter resistor and cap as above - with maybe 0.33 ohm as emitter resistor.

Changing to a larger TO220 package would be wise at 1 Watt.
eg this TIP122 in stock at Digikey for 68 cents
The TIP120, TIP121, TIP122 are all equally good ij this application.


  • The 5 volts should be provided by a regulated power supply. The efficiency is below 25% and significant DC current flows in the speaker and that additional power should be figured in to the power rating of the speaker. But look how simple it is! The voltage gain is only about 20 and the input impedance is about 12k. The schematic shows two values of bias resistor to be used with the corresponding speaker impedance. With the 150k bias resistor and 8 ohm speaker, the circuit draws about 210mA (1 watt) and can deliver about 250 mW to the speaker which is plenty of volume for most small projects. The speaker should be rated at 500 mW or more and should exhibit a DC resistance near 8 ohms (perhaps 7 ohms). Check the candidate speaker with an ohmmeter; much below 7 ohms will cause excessive current draw. With the 220k resistor and 16 ohm speaker, the circuit draws about 100 mA (500 mW) and delivers about 125 mW to the speaker. The 16 ohms speaker should be rated at 200 mW or more and exhibit nearly 16 ohms of DC resistance. (Most small speakers have a DC resistance near the rated impedance and that resistance is used to set the quiescent current level in this circuit.) Other NPN darlington transistors will work but choose one that can dissipate 1 watt minimum. Most power types don't need a heatsink but tiny TO92's might overheat.
  • \$\begingroup\$ True, 250 == 280 :-). But both you and the author seem to forget about the Darlington. Collector won't go lower than 1.5V, so that's 3.5Vpp, not 5V. I get 175mW (a bit less than the theoretical 190mW because the quiescent point is not quite in the middle. \$\endgroup\$
    – stevenvh
    May 28, 2012 at 11:20
  • \$\begingroup\$ @stevenvh - False :-). You've forgotten about phase of the moon, which way the wind is blowing and bank holidays. ie this circuit is made to be critiqued and generally understood - not actually used :-). || Darlington collector with enough drive can fall to approaching Vbe of the bottom transistor as Vce upper can approach zero. Call this 0.6V under good drive - or 0.8 or whatever. I allowed significant headroom in RMS voltage at both side. At no headroom and 0.8V sat = ((5-0.8)/2 x 0.7071)^@ / 8 =~ !!! 280 mW :-). (Just happened that way). ie it's all too rough to put good figures on. \$\endgroup\$
    – Russell McMahon
    May 28, 2012 at 11:59
  • \$\begingroup\$ True, I forgot about bank holidays. I went shopping this morning and stood at a closed supermarket :-/. Phase of the moon was accounted for. \$\endgroup\$
    – stevenvh
    May 28, 2012 at 12:19

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