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-   -   How does BTL double the voltage swing? (https://www.diyaudio.com/forums/solid-state/212307-btl-double-voltage-swing.html)

HankMcSpank 7th May 2012 07:48 PM

How does BTL double the voltage swing?
 
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 7th May 2012 08:03 PM

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 7th May 2012 08:05 PM

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 7th May 2012 08:09 PM

Quote:

Originally Posted by currentflow (https://www.diyaudio.com/forums/solid-state/212307-btl-double-voltage-swing-post3015508.html#post3015508)
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 7th May 2012 08:14 PM

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 7th May 2012 08:19 PM

Quote:

Originally Posted by HankMcSpank (https://www.diyaudio.com/forums/solid-state/212307-btl-double-voltage-swing-post3015515.html#post3015515)
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 7th May 2012 08:19 PM

Quote:

Originally Posted by MarcelvdG (https://www.diyaudio.com/forums/solid-state/212307-btl-double-voltage-swing-post3015531.html#post3015531)
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 7th May 2012 08:20 PM

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 7th May 2012 08:21 PM

Quote:

Originally Posted by currentflow (https://www.diyaudio.com/forums/solid-state/212307-btl-double-voltage-swing-post3015540.html#post3015540)
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 7th May 2012 08:23 PM

Quote:

Originally Posted by Enzo (https://www.diyaudio.com/forums/solid-state/212307-btl-double-voltage-swing-post3015542.html#post3015542)
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


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