How does BTL double the voltage swing?

[HankMcSpank]

Ok, so I read that feeding a signal to one connection of a speaker coil & the same signal but with inverted polartiy to the other end/connection of the coil doubles the voltage swing...but I don't understand how/why.

Furthermore I've yet to find an explanation on the net that tells me how....so can anyone here tell me how?!

Consider a 9V amplifier ...parking/ignoring the fact that the output signal can't swing to rail for a minute... let's assume the maximum signal voltage swing will be between 0 & 9V, therefore with AC signal at max....

coil end +ve +9V
coil end -ve 0V


when the AC signal swings the other way

coil end +ve 0V
coil end -ve +9V

total voltage 'difference' is 9V each time ....so where does the BTL 'voltage doubling' come from?

[MarcelvdG]

With an asymmetrical 9 V supply (the lower supply line at 0 V and the upper at 9 V), you could only get 4.5 V peak AC into a loudspeaker when you don't use a bridge:

0 V in the negative peaks
4.5 V DC bias
9 V in the positive peaks

Add a DC blocking capacitor to prevent a large DC current from flowing through the loudspeaker and you get:

-4.5 V in the negative peaks
0 V bias
+4.5 V in the positive peaks.

With the bridge you really go from -9 V to +9 V.

[currentflow]

The two amplifiers are driven in anti-phase, so if amplifier 1 output is at +9V, amplifier 2 output will be at -9V. The load (loudspeaker) sees the difference between the amplifier outputs as +9V - (-9V) = +18V. When the outputs swing the other way, it becomes -9V - (+9V ) = -18V.

[HankMcSpank]

Originally Posted by currentflow

The two amplifiers are driven in anti-phase, so if amplifier 1 output is at +9V, amplifier 2 output will be at -9V. The load (loudspeaker) sees the difference between the amplifier outputs as +9V - (-9V) = +18V. When the outputs swing the other way, it becomes -9V - (+9V ) = -18V.

But if the power supply rails to the amplifier is 0V and 9V, how does one side swing to -9V? Certainly scoping of a BTL output shows that both amplifier outputs seem to remain within the limits of the power rails ...therefore one side is swinging 0V>9V & so is the other albeit inverted....but the difference is always just 9V?

[MarcelvdG]

Many amplifiers have a symmetrical supply, for example +9 V, 0 V and -9 V, the difference between the highest and lowest supply voltage being +9 V-(-9 V)=18 V.

Taking a simple asymmetrical 9 V supply (the lower supply line at 0 V and the upper at 9 V), you could only get 4.5 V peak AC into a loudspeaker when you don't use a bridge:

0 V in the negative peaks
4.5 V DC bias
9 V in the positive peaks

Add a DC blocking capacitor to prevent a large DC current from flowing through the loudspeaker and you get:

-4.5 V in the negative peaks
0 V bias
+4.5 V in the positive peaks.

With the bridge you really go from -9 V to +9 V.

By the way, are there two threads with the exact same title?

[currentflow]

Originally Posted by HankMcSpank

But if the power supply to the amplifier is 9V, how does one side swing to -9V? Certainly scoping of a BTL output shows that both outputs remain within the limits of the power rails ...therefore one side is swinging 0V>9V & so is the other albeit inverted.

I assumed you were describing a symmetrical supply of 9-0-9V. If your supply is +/- 4.5V, per amplifier, the output would be +/- 4.5V maximum (as per your lossless example).

Exactly the same calculation applies:
If amplifier 1 output is at +4.5V, amplifier 2 output will be at -4.5V. The load (loudspeaker) sees the difference between the amplifier outputs as +4.5V - (-4.5V) = +9V. When the outputs swing the other way, it becomes -4.5V - (+4.5V ) = -9V.

[HankMcSpank]

Originally Posted by MarcelvdG

Many amplifiers have a symmetrical supply, for example +9 V, 0 V and -9 V, the difference between the highest and lowest supply voltage being +9 V-(-9 V)=18 V.

Taking a simple asymmetrical 9 V supply (the lower supply line at 0 V and the upper at 9 V), you could only get 4.5 V peak AC into a loudspeaker when you don't use a bridge:

0 V in the negative peaks
4.5 V DC bias
9 V in the positive peaks

Add a DC blocking capacitor to prevent a large DC current from flowing through the loudspeaker and you get:

-4.5 V in the negative peaks
0 V bias
+4.5 V in the positive peaks.

With the bridge you really go from -9 V to +9 V.

By the way, are there two threads with the exact same title?

For simplicity let's just stick with asymetrical supply...the DC midpoint of a BTL output here would be 4.5V DC...so the max BTL swing would also be 4.5V either side of 4.5V DC? Isn't that the same as you single ended example?

[Enzo]

IN a simple amp, the speaker negative terminal is grounded, and with 9v rails the hot side would go from +9 to -9 as signal demanded.

NOw add a second amp circuit of opposite polarity. Feed it the same signal as the other channel. It too will have its output go from +9v to -9v as the signal demands, but it is opposite the first channel. SO whenever the first channel goes to +9, the inverted channel is going to -9.

SO now instead of connecting the speaker from ground to an output, connect the output of channel 1 to the hot side of the speaker, and the output of channel 2 to the cold side of the speaker. Now when the channel 1 output goes to +9, the channel 2 output is going to -9. And that means there is 18v across the speaker instead of just 9.

What we are describing is what happens when stereo amps are operated in "bridge" mode.

[HankMcSpank]

Originally Posted by currentflow

I assumed you were describing a symmetrical supply of 9-0-9V. If your supply is +/- 4.5V, per amplifier, the output would be +/- 4.5V maximum (as per your lossless example).

Exactly the same calculation applies:
If amplifier 1 output is at +4.5V, amplifier 2 output will be at -4.5V

But how will amplifier 2 be at -4.5V?

[HankMcSpank]

Originally Posted by Enzo

NOw add a second amp circuit of opposite polarity. Feed it the same signal as the other channel. It too will have its output go from +9v to -9v as the signal demands, but it is opposite the first channel. SO whenever the first channel goes to +9, the inverted channel is going to -9.

Once again, how is it swinging to -9V ....when I scoped a 'centre tapped' coil (which I placed on the outputs of a small 1W BTL amp just to help with my scoping)...the DC at the centre tap is 4.5V, the AC signal signal is swingin 9V on either leg, so you've a 9V swing sitting on a 4.5V DC bias...for each leg...at no point does the -ve output go below 0V