bridging two amps x4 power ?

[djdamix]

I was wandering if bridging two amps give as a results 4 times more power. I mean with two 30V rms (= 30 * 30 / 8 = 112Wrms) bridged amp (double output swing) I'll get 60 x 60 /8 = 450 W rms. Am I right ? Do I forget some losses that appear in this mode ?

David

[Cortez]

Yes, you are right !

[darkfenriz]

yes, absolutely right
but remember the power dissipation is also 4x higher, which is 2x per amp.
to drive a 8ohm speaker in bridge, a single amp should be able to drive 4ohm in 'normal', mode.
cheers

[djdamix]

Great !!! Thanks. To bridge, I plan to invert the input of one of the amp using a op amp with -1 gain. Is there any special thing else to do ?

David

[the_monster]

hi

A bridge tied load (BTL) amplifier applies a normal signal to one terminal of the speaker, and an inverted signal to the other. If a single amp is capable of producing 20V RMS across the speaker, this equates to P = V2 / R, so in this case, 20^2 / 4 = 100W.

When connected in BTL, the same speaker "sees" 20V at one terminal, and an inverted 20V signal on the other - a total of 40V RMS (I shall leave the proof of this to the reader :-) Using the same formula, 40^2 / 4 = 400W - four times the power. But ... each amp now sees only half the load impedance (think of an imaginary centre tap in the voice coil, connected to ground). The amplifier must be stable into 2 ohms, or this method will not work. Of course, you can use an 8 ohm speaker and still get 200W if the amplifier cannot drive 2 ohms safely.

ferds

[darkfenriz]

some time ago I wanted to invert phase using opamp (for n-channel amp) and several guys warned me about considerable delay through opamp. Maybe you should use non-inverting follower based on the same opamp for the non-inverting amp to equalize delay. But i would rather use:

[Cortez]

"delay through opamp"
hmm...