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Old 31st December 2004, 12:51 AM   #21
hermanv is offline hermanv  United States
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The schematic above is conceptual. Some math is needed to work out actual values. The one's I used are sort of close but the 10 Ohm or the 470uF needs work.

In the real world probably one of those programable zeners (actually a shunt regulator with a built in reference) on the pass element base could be made to work. The gain of this device would eliminate the 470uF and the resistor to ground. The actual value of the 10 Ohm depends on voltage, current and beta of the pass element. Design for about 1/2 of specified minimum gain to leave some headroom and to make it work at frequencies higher than DC.

I assumed from your first drawing you knew how to do much of this kind of thing.
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Old 31st December 2004, 01:16 AM   #22
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I personally wouldn't bother with regulation to the output stage, it's unlikely to make any difference. Most high end amps don't have it. Yes a few of them do, but they don't perform any better than the ones that don't.

Just regulate the voltage gain stages. If you go to http://www.bryston.ca.schemprod, they have their schematics, and you can see how they do it.
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Old 31st December 2004, 02:12 AM   #23
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Most likely I won't go with regulation for the final output stage. Regulating up to 15 amps would be a real pain in the butt. I'd rather go the route of caps. Although I am planning on regulating all other parts, though that shouldn't be hard.

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Old 31st December 2004, 04:14 AM   #24
Stocker is offline Stocker  United States
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I am contemplating doing this same thing on an upcoming project of mine. Remember to account for the extra voltage drop across the output transistors when selecting a heatsink.
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Old 31st December 2004, 05:22 AM   #25
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And that brings up one more question: if I'm using a design where I know the power supply rejection ratio, how can I use that to determine how much of an affect a voltage ripple will have (assuming an unregulated supply)?

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Old 31st December 2004, 05:39 AM   #26
Stocker is offline Stocker  United States
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You will not kick yourself if you have an excessively quiet power supply. You may say bad words if you have to tear it all up once constructed to add filtering. Overkill will only hurt the sound of the amplifier to some people. Trial and error will be called for if you feel that to be the case with too quiet of a supply.
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Old 31st December 2004, 06:10 AM   #27
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Well, the main reason I ask is because I have a limited budget, so I'm curious how much I should invest in caps.

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Old 31st December 2004, 08:15 AM   #28
PRR is offline PRR  United States
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> I don't fully agree with that...

It does not matter if we agree; reality is the only truth.

I believe the truth is that you are working against a fallacy. Reality may prove me wrong, but I'm pretty sure. If so, all this talk of SOA and tranny-spikes, which are of course just solvable engineering problems, misses the mark completely.

> With the capacitors on the AC line, I was getting about a 2.5 volt ripple at 84VDC, so that wasn't really bad.

It's 3% ripple, a perfectly reasonable value for the first stage.

> The signal coming out of the transformer would have been squarish, but that's all the better as rectified full wave that would have produced a near perfect DC on it's own, even without the 4 10,000uf caps

Right, almost. And for the moment, for analysis, ignore the added capacitors. The square wave avoids the dip at zero-crossing (actually, from one peak to the other peak) that we have with a sine. With a perfect square and a perfect rectifier, the output is perfect DC.

But the square is not perfect. It has 3% of 120Hz ripple! Assuming perfect transformation and no further capacitance, the "DC" will have 3% ripple.

Still assuming perfect transformation and rectification, your 20,000uFd of amp-rail capacitance simply reflects back onto the line-caps.

Hmmmm... if V*uFd/$ is a constant, there is some advantage in filtering at higher voltage and lower current. 100V and 1A needs 1,000uFd for 1% ripple. 200V and 0.5A needs 250uFd for 1% ripple. Since V*uFd/$ of suitable commercial capacitors is approximately constant, there is some advantage in doing your filtering at higher voltage, if you can transform cheaply.

You won't have perfect transformation. In one sense, that is good: the transformer impedance makes this a 2-stage filter. But a resistor could do that cheaper, and a choke could do it better.

When you need DC cleaner than around 2% ripple, you should consider a multi-stage filter. That used to be common: when caps cost a lot (in 1938) all radios had chokes. The advent of incredibly cheap big-caps changed that: more aluminum is cheaper than iron/copper. But consider a design where you are given just two caps. And say one cap gives 3% ripple. Adding the second cap in simple parallel gives 1.5% ripple. Building a C-R-C filter with 2% loss in the R will give about 14dB reduction on the 3% ripple, or around 0.7% ripple. Stuffing dead resistance in the power leads goes against the grain. It "clearly reduces output". But hi-fi amp clean output is limited by the dip of the ripple. Taking a little loss to get less-deep dips doesn't hurt much. And in Class AB speech/music, it hurts very little: loss at low output is negligible and peaks will reproduce well, only test-tone power is reduced.

If you want a radical clean-AC/DC supply: put a 4" pulley on a 3600RPM AC motor and belt it to a car alternator. A car alternator makes about 1V/100RPM: above engine-idle, a regulator kills the field coil to hold the output down to 13V. Booger that regulator, you can get 60-70V at large Amps. And at somewhat higher frequency. AND the output is 6-phase power: even without a cap, the output never drops to zero! IIRC it is around 15% ripple unfiltered, and a whole lot easier to clean-up than single-phase. However noise (acoustic and brush-hash) may be a problem.

> if I'm using a design where I know the power supply rejection ratio, how can I use that to determine how much of an affect a voltage ripple will have (assuming an unregulated supply)?

100V supply. 5% ripple. There is 5V ripple on the rail. Say amp has 40dB or 100:1 ripple rejection. Ripple on the output is 5V/100 or 0.050V. 0.05V in 8Ω is 35dB below 1 Watt. Assuming 90dBSPL/Watt speakers, ripple is 90-35= 55dB SPL of buzz a yard in front of the speakers (and about 52dB SPL everywhere in an average residential room). That's a lot of buzz. (Real-world PSRR is never as simple as this computation, but it will tell you if you are in trouble.)

We can usually design for less than 5% ripple. And a secret advantage of Class AB operation is that supply current and ripple are low at low output, when buzz is most audible. Many amps buzz when pushed up near clipping, and some show it on test, but it is often negligible to the ear. Class A pulls full power (and ripple) all the time, and more supply power than a Class AB amp working to max.

And 40dB PSRR is pretty crummy for a power amp. Low-level stages sometimes get away with 6dB PSRR, but their power demand is so low we can use "wasteful" power filtering. That's awful expensive in a POWER stage. Fortunately a BJT's Collector, which is "an input", is about 1,000 times less sensitive than its Base. So if you can get clean power to the driver, and hold the output devices' Bases at signal voltage without ripple, the output stage has nearly 60dB PSRR without any special care. Combine that with Class AB operation (low ripple at idle) and some NFB, any buzz on the output is probably due to layout, not power supply. (And I've been guilty of throwing time and money into a power supply when the buzz was elsewhere.)

> I have a limited budget,

An amp that drinks 15 Amps seems "big" to me. Can't fight the laws of physics or factories. Big amps cost big bucks.
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Old 31st December 2004, 10:08 AM   #29
MikeB is offline MikeB  Germany
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Originally posted by hermanv

Driving transformers with square waves can be made to work fine. Its done all the time
Hi !
Yes, SMPS does this, but with very small transformers and high freqs.
But this is not too easy, if switching is not done carefully, you get
some loud bangs. And the calculations with number of windings from
primary to secondary are only valid for sinus.
I don't know how exactly these issues are adressed in an SMPS,
maybe EVA can say something ?

For the case of switching, SOA is not really relevant, as the transistor
only sees on/off, you have only max voltage with 0 amperes, and
max amperes with nearly 0volts vce. Also heating is very small.
Only the transient during switching needs to be observed, and the
energy fighting back from the inductive load is critical.

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Old 31st December 2004, 01:27 PM   #30
Eva is offline Eva  Spain
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rjon17469 :

You are effectively trying to design an offline SMPS and your circuit concept is right except for inadequate components and some mistakes

When rectifiying mains AC you can't use earth as a reference, no current at all should flow between mains and earth [actually a few mA are allowed and sometimes required for proper EMI filtering]. Also note that 120Vrms AC rectification produces about 170V DC [1.4142*120V]

Direct mains rectification has some advantages, one of them is that high voltage electrolytic capacitors tend to store more energy in less space at less cost. This produces less ripple and uses less space for the same price. You can compare .5*C*V^2 / volume [joules/cm^3] for different capacitors

Transformers may be driven with any waveform, even with DC if the core is gapped. For power conversion applications, the best efficiency happens when transformers are driven from square waves. Driving transformers with sine waves produces inferior efficiency and requires an oversized transformer. A transformer rated at 500VA for 120Vrms AC sinewave will be rated at 1000VA for +-120V of squarewave amplitude [240Vpp] [volts*turns/seconds product is the same in both cases so the same flux is generated in the core]

This happens because square waves transfer the maximum power during the entire cycle while sine waves only transfer the maximum power during 30% of the cycle. Doing the math : The peak current for a given power output is 1.4142 times higher in a sine wave and the copper losses are twice as high as for a square wave, given the same transformer

Transformer efficiency is particularly improved with square waves when the secondary side is rectified, since a peak-rectified sine wave does not transfer power at all for 70% of the time and has to transfer 3 times the output average power during only 30% of the time, causing 3 times higher copper losses and 9 times higher voltage drop in comparison with a resistive load

So driving transformers with square waves requires less copper saving space and reducng weight...

But the most smart feature of driving transformers with a squarewave generated from a switched DC source is that you can change the frequency and the duty cycle

For a given transformer core, the number of primary turns required to avoid saturation is proportional to the maximum allowed voltage and inversely proportional to the minimum allowed frequency. Also, for a given number of turns, the mass of iron or ferrite required by the transformer core is proportional to the maximum allowed voltage and inversely proportional to the minimum allowed frequency

So theoretically : A transformer operating at 50Hz requires 100 times more turns than one operating at 5Khz. Or : A transformer operating at 50Hz requires 100 times more core mass than one operating at 5Khz. Or : A transformer operating at 50Hz requires 10 times more turns and 10 times more core mass that one operating at 5Khz. Etc...

In practice, there is some dissipation in the core that increases with frequency and limits the maximum operating frequency for that core material

In fact, one of these E42 ferrite transformers, measuring 42 x 42 x 40mm, weighting less than 500g and costing less than 5 euro [raw core and former without windings] ...
Click the image to open in full size.
Click the image to open in full size.
Is happily capable of transferring 1KW of continuous power when used at 35Khz, a pretty small switching frequency for today's standards. At 100Khz [maximum recomended operating frequency for that material] it can transfer more than 2KW

But controlled square waves have another advantage : Output regulation. If you place a LC lowpass filter at the output of the transformer just after the diodes, then the output voltage will be the square wave amplitude multiplied by the duty cycle. Also the LC lowpass filter effectively filters the square wave and attenuates the switching transients down to a few milivolts

Regulation allows to convert a high voltage with lots of ripple into a lower voltage with milivolts ripple

But switching high voltages optimally, designing optimum transformers and LC filters, and making all them operate inside a stable closed regulation loop is somewhat harder than making an audio amplifier work

If you want to go on with this project, you should start reading and learning about SMPS and experimenting with low voltages, and also get an oscilloscope if you still haven't one

[Disclaimer : Never try to build anything that switches high voltages and/or currents before completing and optimizig some projects using lower voltages and currents. Allways test the high-voltage circuits with lower voltages first. Never use mains as a direct high voltage supply for experimentation, use 1:1 isolation transformers. Use a light bulb in series with the power supply when first testing with high voltages. Follow proper mains isolation rules when building transformers and tracing PCBs. Do everything at your own risk. Etc...]
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