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?
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?
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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.
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.
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?
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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?
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?
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.
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?
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.
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.
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
Oops, posted while posts were posting...
Doesn;t matter what voltage rails you decide on, the principal is the same.
If you want a single 9v supply then yes, the output goes from 4.5v at rest to +9 or to 0. SO if you have two channels doing that and one is opposite in polarity from the other. Then connecting a speaker from the hot lead of one to the hot lead of the other results in the +9 on one end of hte speaker with 0 on the other, so you get the full 9v instead of just the 4.5.
Doesn;t matter what voltage rails you decide on, the principal is the same.
If you want a single 9v supply then yes, the output goes from 4.5v at rest to +9 or to 0. SO if you have two channels doing that and one is opposite in polarity from the other. Then connecting a speaker from the hot lead of one to the hot lead of the other results in the +9 on one end of hte speaker with 0 on the other, so you get the full 9v instead of just the 4.5.
Because with a single supply rail and a single output the output is cap coupled. The output sits at some voltage potential and cannot be connected directly to a loudspeaker through a standard ground returned system. If you had a single supply amp powered by a 9 volt supply what you really see happening is the output swinging + and - about 9/2 volts. In other words you're power is defined by a waveform that swings +- 4.5 volts.
In the case of the BTL two outputs are tied together. Both of them sit at exactly 1/2 the rail voltage. This means that the + and - inputs to the loudspeaker are both connected to two individual amplifier channels and are held at the same potential (Vcc/2) and no current flows.
In this case as the +ve output swings high, up to +9V, the -ve output swings low down to 0 volts, giving you a positive going voltage peak of +9v. However this time around when the +ve output swings low, to zero volts, the -ve output swings high up to +9V and in this case gives you a negative going voltage peak of -9V. Here we end up having the potential to swing +9v and -9 volts.
Edit - It's like in the second case you're connecting a 9V battery up to the loudspeaker and the cone moves forwards. Then to get the cone to move backwards you reverse the polarity of the battery. In this case you're loudspeaker cone is moving as if it were energised one way by 9volts, then the other way by 9 volts, to a total peak to peak value of 18 volts.
In the case of the BTL two outputs are tied together. Both of them sit at exactly 1/2 the rail voltage. This means that the + and - inputs to the loudspeaker are both connected to two individual amplifier channels and are held at the same potential (Vcc/2) and no current flows.
In this case as the +ve output swings high, up to +9V, the -ve output swings low down to 0 volts, giving you a positive going voltage peak of +9v. However this time around when the +ve output swings low, to zero volts, the -ve output swings high up to +9V and in this case gives you a negative going voltage peak of -9V. Here we end up having the potential to swing +9v and -9 volts.
Edit - It's like in the second case you're connecting a 9V battery up to the loudspeaker and the cone moves forwards. Then to get the cone to move backwards you reverse the polarity of the battery. In this case you're loudspeaker cone is moving as if it were energised one way by 9volts, then the other way by 9 volts, to a total peak to peak value of 18 volts.
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Because the amplifier is biased to the mid-point of the rails by necessity. It is undesirable to apply DC to the load under a no-signal condition. Consider a single-ended supply amplifier which is AC coupled to the load. The steady state output of the amplifier (with no signal) is biased at +4.5V DC. AC coupling this to the load via a capacitor means that the load sees zero volts when no signal is present. A swing of between 0 and +9v at the amplifier output becomes transposed into a swing of -4.5V to + 4.5V in the load. When amplifier 1 is at +4.5V (instantaneous signal), amplifier 2 is at -4.5V. Amplifier 2 is fed from an inverted signal being fed to amplifier 1.
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For the purposes of calculating current through a speaker coil for single ended vs BTL
(assuming just DC resistance & ignoring impedance for simplicity)...
Coil = 8 ohm
peak to peak AC signal voltage swing = 8V (as measured with a scope on one leg)
To calculate the current through speaker coil, I need to convert the peak to peak swing to RMS, but since this is BTL, do I therefore double the swing ...therefore 16V peak to peak?
So for single ended driving amplifier, 8V peak to peak = 2.828 RMS...therefore divided by 8 Ohms = 354mA through the speaker coil.
For BTL I'm figuring it's 16V peak to peak therefore RMS = 5.657 RMS = 707mA through the speaker coil.
Are those calculations correct?
(assuming just DC resistance & ignoring impedance for simplicity)...
Coil = 8 ohm
peak to peak AC signal voltage swing = 8V (as measured with a scope on one leg)
To calculate the current through speaker coil, I need to convert the peak to peak swing to RMS, but since this is BTL, do I therefore double the swing ...therefore 16V peak to peak?
So for single ended driving amplifier, 8V peak to peak = 2.828 RMS...therefore divided by 8 Ohms = 354mA through the speaker coil.
For BTL I'm figuring it's 16V peak to peak therefore RMS = 5.657 RMS = 707mA through the speaker coil.
Are those calculations correct?
Remember that voltage must be a potential between 2 points. Don't use peak-to-peak voltage for the single-ended output - the "single-end" can't be both positive and negative at the same time. So per your example, the voltage would swing 4V positive, then 4V negative. Use 4V to convert and calculate.
In bridged mode, one end of the speaker is sent 4V positive while the other end is sent 4V negative... at the same time. The voltage is doubled, so the current is doubled.
In bridged mode, one end of the speaker is sent 4V positive while the other end is sent 4V negative... at the same time. The voltage is doubled, so the current is doubled.
The rule for bridging and balanced amplifiers is simple.
A pair of amplifiers in bridged format will deliver twice the power into twice the load impedance.
eg.
two 100W into 4r0 amplifiers will deliver 200W into 8r0 when bridged.
That shows quite well that nothing is gained by bridging.
The total power from the two amplifiers is identical.
There is no magic 4times the power !!!!!
Having seen how easy it is, then it becomes obvious that to consider a single unbridged amplifier as the basis for all the investigations avoids all the confusion.
A pair of amplifiers in bridged format will deliver twice the power into twice the load impedance.
eg.
two 100W into 4r0 amplifiers will deliver 200W into 8r0 when bridged.
That shows quite well that nothing is gained by bridging.
The total power from the two amplifiers is identical.
There is no magic 4times the power !!!!!
Having seen how easy it is, then it becomes obvious that to consider a single unbridged amplifier as the basis for all the investigations avoids all the confusion.
For the sake of clarity (since they're were two duplicate threads running...and unfortunately I reopened my latest line of questioning in the wrong one!), here's what I asked...
So in my above example, for a BTL amplifier for a signal measuring 8V peak to peak on one of the amplifier outputs (and ignoring impedance just for simplicity of discussion(!), for a coil with a resistance of 8 Ohms what would be the RMS AC current through the coil?
Since I'm scoping 8V peak to peak on one amplifier output, is the actual voltage scoped doubled? .....therefore 16V peak to peak, which is 8V peak, therefore 5.657V RMS, there 5.657V/8Ω = 0.707125A?
So in my above example, for a BTL amplifier for a signal measuring 8V peak to peak on one of the amplifier outputs (and ignoring impedance just for simplicity of discussion(!), for a coil with a resistance of 8 Ohms what would be the RMS AC current through the coil?
Since I'm scoping 8V peak to peak on one amplifier output, is the actual voltage scoped doubled? .....therefore 16V peak to peak, which is 8V peak, therefore 5.657V RMS, there 5.657V/8Ω = 0.707125A?
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