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The above circuit is a power amplifier delivering roughly 30 watts (11.7V times 2.9A) in the 4 ohm load (assuming its a speaker). When we look at \$V_{cc}\$, it looks huge when compared to electronics books providing only 9, 12 \$V_{cc}\$. Can we lower the \$V_{cc}\$ so that it still delivers the same amount of power? I'm planning to use voltage regulator ICs like LMxxx18 which regulate 18 V DC so I'm planning to make \$V_{cc}\$ equal to 18V also.

For the speakers, is the current flowing through the speakers that determine its loudness? We can still get 30 watts of power across the speaker but the sound generated might be miniature, so is it the voltage or current that determines loudness?


2 Answers 2


The problem of achieving higher peak to peak voltage than the voltage available on the rails is common in audio amplifiers, especially in car audio. The simplest solution is to configure two identical amplifiers like your schematic, in a Bridge Tied Load configuration. Put simply: Connect your speaker between the outputs of two such amplifiers, with no ground reference, and feed the two amplifiers with signals that mirror each other.

BTL amplifier (from Wikipedia)

A brief description of BTL amplifiers is here. Below is a less brief explanation of my own, if you want one. Also, this question has some additional information and links that may be useful.

A Bridge Tied Load (BTL) amplifier consists of two identical amplifier blocks, fed by an input signal in antiphase, and with the load (speaker, transducer etc) connected between the two amplifier outputs.

Since the signals at the outputs of the two amplifier blocks are mirror images of each other (180 degrees out of phase), the effective signal seen by the load is twice the signal each amplifier produces.

This is specially useful when the supply rail voltage(s) available for the amplifiers is too low to provide sufficient peak-to-peak voltage for the output power desired. A common such situation applies to car audio amplifiers, where a 12 or 14 volt supply is available, and the desired output power requires a higher peak to peak voltage than that.

Another application of bridge tied load amplifiers is when the load impedance is higher than that supported by the individual amplifier blocks. Since the effective impedance seen by each of the matched amplifier blocks is half of the actual load impedance, such an amplifier configuration allows use of loads up to twice the supported output impedance per amplifier.

As noted in Wikipedia, one common fallacy with Bridge Tied Loads is the assertion that the resultant power of a BTL amplifier is 4 times the power output of each amplifier. This is incorrect: While the voltage across the load doubles, the current remains the same. Thus, for P = V x I, while V' = 2 x V, P' = 2 x P.

Another way of looking at this is that the doubled voltage is developed across an effective load twice the load impedance seen by each amplifier. Thus, power delivered by a BTL amplifier is merely double that of a single amplifier.

However, this 4 x power myth has been propagated widely in audio DIY circles, to the extent that an opposing assertion is likely to be received with disbelief or derision. It is an old chestnut best left undisturbed when addressing a "believer".

  • 1
    \$\begingroup\$ Beware - that Wikipedia article is misleading about the 4* power increase! To double the voltage across a load without doubling the current ... the only way to do that is to also double the load impedance! And yet it's the same 4 ohm speaker... Consider it another way : if you centre-tapped the speaker in a BTL configuration, you would see 0V (AC) at the centre tap. And each half of the speaker is 2 ohms... What they gloss over in the article is that the 4* power increase is real - IF the amplifier is rated to drive 2 ohms, with the obvious change in supply current requirements. \$\endgroup\$
    – user16324
    Nov 24, 2012 at 11:20
  • \$\begingroup\$ @BrianDrummond Here's the problem with that logic: If an amplifier is rated for 10 Watts across 8 Ohms, then it'll deliver just that. To match impedance of that amp, you would not be using an 8 Ohm load in BTL, you would be using a 16 Ohm load - so the question of using the same speaker does not arise. \$\endgroup\$ Nov 24, 2012 at 11:44
  • 2
    \$\begingroup\$ Not so fast... Very few amplifiers have one and only one rating. Given 10W into 8 ohms, a perfect audio amp would drive 20W into 4, an inferior one maybe 15-18W. In any case the specified load in the problem is 4 ohms, with no implied permission to double it. Which is why I pointed out the 2 ohm drive requirement for a BTL amp driving 4 ohms. \$\endgroup\$
    – user16324
    Nov 24, 2012 at 12:56
  • 1
    \$\begingroup\$ @BrianDrummond I fail to see the point of debating this further. Opinions and math don't mix. \$\endgroup\$ Nov 24, 2012 at 18:52

To drive a 4 ohm load with a sine wave such that 30W is dissipated, the RMS AC voltage has to be about 11V, because power is $$V^2/R$$ So the peak-to-peak voltage has to be 11 * 2 / 0.707 = 31V.

If you don't have the voltage, then you need a step-up transformer.

But, careful: that of course makes the load look like less than 4 ohms!

The current through a speaker's voice coil determines the displacement. The loudness of a speaker has to do with how efficiently it converts energy into movement. That comes from many factors. For instance, given the same voice coil and everything else, a stronger magnet will make the speaker louder.

  • \$\begingroup\$ I'd like to add that what you can always do is parallelize. Get two speakers and drive them with 15W each, with the same signal and polarity. Or, you can wire speakers in parallel, then build an amp that can drive the lower impedance. Use MOSFET devices which can be paralleled for more current. Or if you build two amps, you can bridge them for more peak to peak voltage. \$\endgroup\$
    – Kaz
    Nov 25, 2012 at 0:09

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