The separate amplifiers are designed with a half-bridge topology. Two of these are placed in series with the load, which is no longer ground referenced, to make a full bridge. In other words, the load is wired to each separate amplifier's output instead of being wired between one amplifier and the common power supply return. The full bridge connection can produce twice the voltage of the half bridge. Real power = output voltage squared / load resistance. The full bridge connection can produce up to four times the power of the half bridge.


Notice the "up to" in the above sentence. Often, the forward-biased-safe-operating area (FBSOA) of the output transistors in the amplifier are not stressed for 4 X the power. They may not be stressed for 1 watt more than the half-bridge can produce. Then the full-bridge results in only double the power. Other amplifiers may be stressed for powers in between 2x and 4x.


The other parts of the amplifier that are expensive to build are the power supply, and the heat rejection capacity of the amplifier. Either or both of these can limit ultimate power. In fact, it is generally true that FBSOA, power supply and heat rejection are all designed to be simultaneously limiting at some power level, and that will be the number you see.

 

A little more specific answer to your question: In the case where 150 W + 150 W = 250 W. The full bridge connection will naturally double the voltage. If the load resistance also doubles to 16 Ohms, then there is no more stress on the output transistors than the half bridge connection at 8 Ohms, and the full bridge connection produces twice the power into 16 Ohms as a half bridge amplifier produces into 8 Ohms. When the full bridge is connected to the 8 Ohm load, the stress on the output transistors is higher, at any power level, than it is for the half bridge case.


Here are the numbers for the 150 W + 150 W = 250 W case: One amplifier requires 49 volts, peak, to produce 150 W, sine wave power, into an 8 Ohm load. Therefore, the 150 W amplifier must have a power supply bus of +/-51 VDC or greater. Into a resistive load, the current peaks at the peak voltage. The peak current into the 8 Ohm load is 6.1 A peak at 49 V peak. At the peak output voltage, with 49 V across the load, the transistor has only 1 V across it. 6.1 A * 1 V = 6.1 W of instantaneous dissipation in the output transistor. Now this is not the power dissipation of the output transistor for the whole cycle, just what occurs at the peak output voltage. With the full bridge connection, at 250 W output, 63.3 V peak, and 7.91 A peak are required. Now since each amplifier in the full bridge is producing only half the output voltage, the voltage drop across each power transistor is 51 VDC - (63.3/2) = 19.4 V, at the peak of the output, sinewave, waveform. The output current is 7.91 A, so the power dissipation in the output transistor at this point on its load line is 19.4 V * 7.91 A = 153 W. It is worth noting at this point that the popular misconception that a more powerful amplifier works less than a less powerful amplifier is actually the opposite.


With a limited FBSOA in the power transistors, the manufacturer using a regulated or tapped power supply may reduce the power supply voltage for full bridge operation, to reduce FBSOA stress in the outputs. The 250 W full bridge amplifier would require a power supply voltage of +/- 33 VDC.

 

wow, thanks...sort of.

http://www.merriam-webster.com/dictionary/layman


see #2 above.

 



Quote:
Originally Posted by dookie1 /forum/post/17440081


wow, thanks...sort of.

http://www.merriam-webster.com/dictionary/layman


see #2 above.

Well dookie, I tried, so I stayed away from the real issues- reactive loads with resonances, as one finds in real speakers. Oh, and I didn't quote any obscure scriptures either, see #1, above.

 

Because the bridged rating is dependent on the amp's power at lower impedances.


8 ohm bridged power = 4 ohm watts per channel rating x 2, and


4 ohm bridged power = 2 ohm watts per channel rating x 2


Use the following amp ratings for an example...


100wpc @ 8 ohms

150wpc @ 4 ohms

190wpc @ 2 ohms


For this amp, 8 ohm bridged power would be 300 watts, 4 ohm bridged power would be 380 watts. As you see, if the amp performance at lower impedances isn't known, the bridged power ratings will be mostly speculative as to how they came to be.


Note that many amps aren't rated for two channel operation with 2 ohm loads, therefore they also don't like 4 ohm bridged operation.