PSU capacitor design

[pacificblue]

Originally posted by megajocke
For bass frequencies where it is needed most, a non-bridged amp uses only one half of the power supply energy storage at a time. By using the whole power supply all the time this waste of power supply energy storage capacity is mitigated.

To me your explanation makes it even more clear that the smoothing capacity needs to be much bigger for a bridged amplifier.

Let me put some numbers to it. If you pass 1 A through an 8 Ohm load, a non-bridged amplifier needs to provide 8 V and 1 A. Each half of the power supply provides that current during one half cycle, so needs to supply an average of 0,5 A per half. In a bridged amplifier each half of the bridge provides 4 V and 1 A. So it takes 2 A for both bridges and the power supply has to provide that current continuously on both halves.

In a bridged amplifier you need four times the capacitance and four times the ripple current rating. To make things worse those capacitors don't have the time to cool down during every second half-cycle, so they will need much more air-flow around them or may even need to be heatsinked.

The Crown MA-600's configuration does not change anything about those figures. The Crown's big advantage is that it complies with safety regulations by connecting one side of the load to ground, thus providing a Class I protection for the speaker output. It is often overlooked, but such a measure is necessary as soon as the AC output can be bigger than 50 V or the DC output bigger then 120 V.

[megajocke]

The numbers are right but the conclusion is wrong. 4 times the capacitance is correct, but you forgot that you halved the voltage! That's the same energy storage, not more. You are right that there isn't anything special to the Crown circuit in this regard. It's just that you can do away with the center tap and dual polarity supply and use less semiconductors than standard bridged amps.

So it might seem at first that there is no advantage, but the average current simplification doesn't work at low output frequencies around and below the mains frequency. In this range the ripple is determined by instantaneous current, not average. In your example the half-bridge amp needs two 8V 1A supplies while the bridged amp only needs two 4V 1A (equal to just one 8V 1A) supplies.

For a half-bridge amplifier the ripple voltage will go up about three times when you go down in frequency (peak-to-average ratio of power supply current is pi), keeping output level constant, assuming the transformer is large. For a bridge amp it will only go up 1.5 times, half of that. (peak-to-average ratio is pi/2)

[pacificblue]

Originally posted by megajocke
4 times the capacitance is correct, but you forgot that you halved the voltage! That's the same energy storage, not more.

4 x 0,5 = 2, not 1. You need four times the capacitance and double the energy storage.

Originally posted by megajocke
It's just that you can do away with the center tap and dual polarity supply and use less semiconductors than standard bridged amps.

The schematic shows a split supply and a standard bridged amp. The connection from one speaker terminal to ground and PE does not change anything about the way it works.

Originally posted by megajocke
the half-bridge amp needs two 8V 1A supplies while the bridged amp only needs two 4V 1A (equal to just one 8V 1A) supplies.

The current that flows in through the high-side flows out of the low-side. That means both sides consume 1 A, which is a total of 2 A for the power supply at 100 % duty factor from both halves of the power supply. The non-bridged amplifier consumes 1 A with 50 % duty factor for each half of the power supply. 2 A x 100 % = 2 A, while 1 A x 50 % = 0,5 A for each half of the power supply. A factor of 4:1 against the bridged amplifier.

Originally posted by megajocke
For a half-bridge amplifier the ripple voltage will go up about three times when you go down in frequency (peak-to-average ratio of power supply current is pi), keeping output level constant, assuming the transformer is large. For a bridge amp it will only go up 1.5 times, half of that. (peak-to-average ratio is pi/2)

Maybe you can expand on this a bit, because I am probably not the only one to whom this is news.

Until then I stick to what I learned and that is Vpp=I/(2*f*C). If the current draw is four times as big, you need a capacitor that is four times as big to maintain the ripple voltage the same. Anyhow, the ripple current will be higher for a bridged amp and the capacitors need an accordingly higher ripple current rating.

[megajocke]

Originally posted by pacificblue

4 x 0,5 = 2, not 1. You need four times the capacitance and double the energy storage.

Energy storage is proportional to the square of the voltage. 4 * 0.5^2 = 1.

Originally posted by pacificblue
The schematic shows a split supply and a standard bridged amp. The connection from one speaker terminal to ground and PE does not change anything about the way it works.

The split supply is for the op-amp bits. T100, D117 and C126 make up the power supply for one channel. I never said there is anything fundamentally different from a standard bridged amp, just that it doesn't need a full-blown amplifier circuit for the low side and that the PSU doesn't need a center tap.

Originally posted by pacificblue
The current that flows in through the high-side flows out of the low-side. That means both sides consume 1 A, which is a total of 2 A for the power supply at 100 % duty factor from both halves of the power supply. The non-bridged amplifier consumes 1 A with 50 % duty factor for each half of the power supply. 2 A x 100 % = 2 A, while 1 A x 50 % = 0,5 A for each half of the power supply. A factor of 4:1 against the bridged amplifier.

Yes, the capacitance for equal HF properties is 4 times as large per side of the bridged amplifier's power supply, ie. it has the same amount of energy storage.

Originally posted by pacificblue
Maybe you can expand on this a bit, because I am probably not the only one to whom this is news.

Until then I stick to what I learned and that is Vpp=I/(2*f*C). If the current draw is four times as big, you need a capacitor that is four times as big to maintain the ripple voltage the same. Anyhow, the ripple current will be higher for a bridged amp and the capacitors need an accordingly higher ripple current rating.

Well, I guess this is the interesting part. The ripple formula you showed is correct of course, but what does the I in it mean? Instantaneous or average current? If the load current waveform is of high frequency it might just as well have been DC of its average value so there you can use the average current.

Let's take an extreme example of low frequency: A rectangle wave current is pulled from a PSU. It looks like this: 100A for 1 second, 0A for 99 seconds, repeat. You can't just put in the average current of 1A in this case of course. That should be obvious. The same is true for the amplifier PS at low frequency, although not as obvious - it's not average current that matters any longer.

The higher peak-to-average ratio in the half-bridge amplifier is what makes power supply ripple degrade more towards low frequencies than it does in the bridged one.

It's true that the ripple current in the capacitor(s) is higher, but it's not 4 times as high like you say, it's only sqrt(2) times higher. As an example, take an amplifier which delivers 1A RMS to the load:

Half bridge: each power supply delivers 1 / sqrt(2) A RMS.
Full bridge: each power supply side delivers 1A RMS, which is sqrt(2) times higher.

Capacitor ripple current is (although a bit simplified) proportional to power supply RMS current. If you keep the two capacitors from your half-bridge amp and put them into an equivalent size bridge amp which doesn't use the center tap, like the Crown topology for example, between plus and minus rail in parallell you have:
* the same energy storage as the original
* better bass performance, only half the ripple at LF
* lower ripple current by a factor of 1/sqrt(2) in the capacitors

You could remove one of the capacitors and get back the LF ripple properties of the old one, with half the energy storage. Ripple current would be sqrt(2) times higher than the non-bridged one though.

[pacificblue]

Originally posted by megajocke
Energy storage is proportional to the square of the voltage. 4 * 0.5^2 = 1.

0,5^2 = 0,25, not 2, but the assumption is wrong anyway, so it does not matter.

I have to admit that I had not looked at the power supply section before, but by skipping the center-tap the great advantage of using half the voltage rating is gone. There is no factor of 0,5. The capacitor has to be rated for the same voltage as in a non-bridged split power supply.

Originally posted by megajocke
it doesn't need a full-blown amplifier circuit for the low side and that the PSU doesn't need a center tap.

The amplifier circuit for the low side looks very much the same as the one for the high side to me. On the contrary it seems that the high side is more complex than normal, because it needs, what they call a 'Voltage Translator Stage', which is more or less a virtual ground.
How much does a center-tap cost?

Originally posted by megajocke
Well, I guess this is the interesting part. The ripple formula you showed is correct of course, but what does the I in it mean? Instantaneous or average current? If the load current waveform is of high frequency it might just as well have been DC of its average value so there you can use the average current.

The average current is a function of peak current and wave form. Nothing changes in the relationship.

Originally posted by megajocke
The same is true for the amplifier PS at low frequency, although not as obvious - it's not average current that matters any longer.

You stated that in a previous post. Please explain, why that should be so.

Originally posted by megajocke
The higher peak-to-average ratio in the half-bridge amplifier is what makes power supply ripple degrade more towards low frequencies than it does in the bridged one.

And please explain also, why a non-bridged amplifier has a lower ripple degradation than a bridged one. Ripple depends on the current draw. If the ripple remains more constant across different load situations, that implies the current draw rises proportionally less in a bridged amplifier than in a non-bridged. Do you assume that the bridged amplifier has a higher quiescent current?

Originally posted by megajocke
It's true that the ripple current in the capacitor(s) is higher, but it's not 4 times as high like you say, it's only sqrt(2) times higher.

Originally posted by megajocke
Capacitor ripple current is (although a bit simplified) proportional to power supply RMS current.

Which is it now? Proportional to the current draw or sqrt(2) times higher for 4 times the current? My vote remains for proportional.

Originally posted by megajocke
As an example, take an amplifier which delivers 1A RMS to the load:

Half bridge: each power supply delivers 1 / sqrt(2) A RMS.
Full bridge: each power supply side delivers 1A RMS, which is sqrt(2) times higher.

A bridged amplifier draws 2 A from the power supply to deliver 1 A to the load.

[megajocke]

On the original topic, I agree with RocketScientist, Steve Dunlap et. al. The transformer may be the single most expensive part so using one you have will save you money if it is suitable. It is suitable in this case, for a bridged amp. The 400W should be possible to meet unless the amplifiers used have unusually high rail loss.

I second the idea of using standard amplifiers and bridging them; the Crown circuit is just something I brought up when someone said you MUST build two complete amplifiers, which is, demonstrably, not true.

Originally posted by pacificblue

0,5^2 = 0,25, not 2, but the assumption is wrong anyway, so it does not matter.

Of course it's 0.25. And how much is 0.25 times 4, which was the question? One.

Originally posted by pacificblue

[...]but by skipping the center-tap the great advantage of using half the voltage rating is gone. There is no factor of 0,5. The capacitor has to be rated for the same voltage as in a non-bridged split power supply.

Having a center tap or not does not matter. Don't you know what happens to the capacitance of seriesed capacitors? It's the same energy storage either way. For example, a half bridge amp has one 100V 10mF capacitor per rail. The corresponding bridge amp would have one 40mF 50V per rail. That's equivalent to one 20mF 100V capacitor across the rails, ie. the two original caps in parallell.

Originally posted by pacificblue

How much does a center-tap cost?

200€+ over here if I already have a suitable 80V 2000VA transformer, without centertap, which I wouldn't be able to use otherwise.

Originally posted by pacificblue

The average current is a function of peak current and wave form. Nothing changes in the relationship.

The current waveform is different in the both cases, that's the whole point. It's the different peak-to-average ratio of the different waveforms that makes the difference in power supply utilization at low load current frequency. The peak-to-average ratio of PSU current is double in a half-bridge amplifier compared to a full-bridge.

Originally posted by pacificblue

You stated that in a previous post. Please explain, why that should be so.

If it isn't obvious why you can't use average current in ripple calculations, when the frequency of the load current is low, I can't help you. Please look at the stupidly exaggerated example again.

Originally posted by pacificblue

Which is it now? Proportional to the current draw or sqrt(2) times higher for 4 times the current? My vote remains for proportional.
[...]
A bridged amplifier draws 2 A from the power supply to deliver 1 A to the load.

Please show some respect for the difference between peak values, instantaneous values, average values and RMS values. They are totally different quantities. Moreover, you can't just add and subtract RMS and peak values if waveforms are different. Also, please don't try to repeal the first law of thermodynamics. That the current draw (measured in some arbitrary way) is double in the bridge amp is irrelevant as the voltage (measured in a corresponding way) is halved. It's the difference in peak-to-average ratio of the current that gives you the advantage in a bridged amp.

Absolute current, voltage and capacitance values don't say anything in themselves. The other two needs to be known too to be able to draw conclusions.

[megajocke]

pacificblue, here are some waveforms.

The transformer for both cases is 35-0-35 (50V peak), practically ideal. Putting in parasitics gives an even greater advantage to the bridged amplifier.

Capacitors are 2x15mF, one per 50V rail in the half bridge amplifier. The full bridge amplifier has only one 15mF capacitor on its single 50V rail. The rectifier is a center-tap two-diode full wave in the bridge amplifier, not that it matters.

The load is 5A peak at 25Hz output from the amplifier. (for example 8 ohms driven to 40V peak) Mains frequency is 50Hz.

Here it can be clearly seen that peak-to-peak ripple voltage is the same for both amplifiers even though the half bridge needs double the amount of energy storage. If parasitics, mainly those of the transformer, are modeled too the advantage of the bridge amplifier grows even greater.

The ripple on the negative rail looks the same as on the positive , just out of phase, so its plot was omitted for clarity.

RMS ripple current in the capacitor of the full bridge is about 150% of the ripple current in a single capacitor of the half bridge. (Or 75% of the total ripple current if you want to look at it that way)

[pacificblue]

Originally posted by megajocke
200€+ over here if I already have a suitable 80V 2000VA transformer, without centertap, which I wouldn't be able to use otherwise.

Originally posted by megajocke
The peak-to-average ratio of PSU current is double in a half-bridge amplifier compared to a full-bridge.

You keep repeating that. Is there any chance that you support it with an explanation, a link, a reference, anything?

Originally posted by megajocke
If it isn't obvious why you can't use average current in ripple calculations, when the frequency of the load current is low, I can't help you.

Which means, there is obviously no chance that you support that with an explanation, a link, a reference, anything.
In this document from the Panasonic website it says on page 1 "Ripple current is the rms value of alternating current flowing through a capacitor".

Originally posted by megajocke
Also, please don't try to repeal the first law of thermodynamics.

As far as I understand that law, if you want 8 V and 1 A at a speaker, you can get that through an 8 V and 1 A power supply from a non-bridged amplifier or through a 4 V and 2 A power supply from a bridged amplifier. It does not matter, if you use peak, average, instantaneous or what-not-ever values. 4 x 1 just is not 8.

Originally posted by megajocke
Here it can be clearly seen..

that the current consumption is doubled for the bridged amplifier, because it shows two sequential half-waves, while the non-bridged shows one followed by an equally long time along the 0-axis.

Originally posted by megajocke
The ripple on the negative rail looks the same as on the positive , just out of phase, so its plot was omitted for clarity.

Looks like the same peak value and half the rms value on the non-bridged amplifier to me. Do you agree that the ripple factor is Vripple(rms)/Vdc as defined e. g. on Wikipedia? Then the capacitor needs to be double the size for the single supply bridged amplifier to achieve the same ripple factor. And to answer your question, I do know, what happens with series-connected capacitors, which makes me suspect that you come out at four times the capacitance with a split supply bridged amplifier.

Originally posted by megajocke
RMS ripple current in the capacitor of the full bridge is about 150% of the ripple current in a single capacitor of the half bridge.

And with double the capacitance for the single supply bridged amplifier? And with four times the capacitance for a split supply bridged amplifier?

I suggest, we stop this discussion here, so that others can remain on topic. If you want to keep the topic alive, we should have the thread split or open another one.

[megajocke]

I have used the "report" button and requested that our posts be split off to their own thread

that the current consumption is doubled for the bridged amplifier, because it shows two sequential half-waves, while the non-bridged shows one followed by an equally long time along the 0-axis.

Power consumption is the same. The half bridge amplifier had two such 50V supplies providing 5/pi = 1.6A average each. The full bridge has one 50V supply providing double the current of one such supply, 5 * 2 / pi = 3.2A. You could look at it like two 25V supplies in series providing 6.4A total, but that's just confusing and contraproductive.

"The peak-to-average ratio of PSU current is double in a half-bridge amplifier compared to a full-bridge." You keep repeating that. Is there any chance that you support it with an explanation, a link, a reference, anything?

In both cases the peak of the PSU current (see that diagram again) is 5A. The bridge amplifier draws 3.2A average while the half-bridge needs 1.6A average from each side of the supply. There's your higher peak-to-average ratio in the half bridge.

Looks like the same peak value and half the rms value on the non-bridged amplifier to me. Do you agree that the ripple factor is Vripple(rms)/Vdc as defined e. g. on Wikipedia? Then the capacitor needs to be double the size for the single supply bridged amplifier to achieve the same ripple factor. And to answer your question, I do know, what happens with series-connected capacitors, which makes me suspect that you come out at four times the capacitance with a split supply bridged amplifier.

More importantly, it's the same ripple valley voltage, and this is what the amplifier cares about as the amplifier clips instantly if supply voltage drops under what it's needing. The amplifier doesn't care about Wikipedia's definition of "ripple factor".

[pacificblue]

Originally posted by megajocke
Power consumption is the same...

There you have it. And double the current needs double the capacitance for the same performance. That is, what AndrewT and Jan Dupont pointed out in the original thread and what I agree to.

Originally posted by megajocke
In both cases the peak of the PSU current (see that diagram again) is 5A. The bridge amplifier draws 3.2A average while the half-bridge needs 1.6A average from each side of the supply. There's your higher peak-to-average ratio in the half bridge.

Yes. And where is the connection to the frequency that you claimed in post #2?

Originally posted by megajocke
For a half-bridge amplifier the ripple voltage will go up about three times when you go down in frequency (peak-to-average ratio of power supply current is pi), keeping output level constant, assuming the transformer is large. For a bridge amp it will only go up 1.5 times, half of that.

Originally posted by megajocke
it's the same ripple valley voltage, and this is what the amplifier cares about as the amplifier clips instantly if supply voltage drops under what it's needing.

And that depends on when it clips.
The bridge amplifier passes through that valley twice as often as the non-bridged. That means both amplifiers will show the same clipping behavior during 50 % of the time. During the remaining 50 % the non-bridged amplifier will clip at a higher voltage than the bridged amplifier. Advantage for the non-bridged configuration?
If you increase the capacitance for the bridged amplifier, you can level that out. The ripple factor will reach the same level as for the non-bridged, but the valley will not dip down as much, so the bridged will keep on clipping earlier then the non-bridged during 50 % of the time, but will clip later during the remaining 50 %.

Originally posted by megajocke
The amplifier doesn't care about Wikipedia's definition of "ripple factor".

It is not Wikipedia's definition. Wikipedia is only one of many sources, where you can find the definition. You can derive the importance of the ripple factor from your own definition of, when an amplifier clips. Lower ripple factor -> lower supply rail dip -> higher clipping level.