Switching Regulator for Audio Amplifier

[Jan Panteltje]

Pictures, what pictures?

Oops, seems I forgot the link:
ftp://panteltje.com/pub/PA3000/

What is this equation called V * T = C * I, and what is is supposed
to do?

Well, we say 'U' and you say 'V', but apart from that, it says
that if you charge capacitor C with current i, then after t seconds the voltage on that
cap will be U (or V ;-) ) volts.
Actually it is Q = C .U = i . t, where Q is the charge.

I is supposed to make sure the output maintains a certain amp
capacity?

I only used it to see how big the voltage drop would be between 2 consequentive
mains tops in your rectifier for a given line frequency and a given cap, and
a given current.

I've been reading that calculations of ripple current a really just a
technical exercise, because the ESR of both the filter caps and the
transformer will only be know after build time. The rule of thumb
that was presented was to size the filter caps with an ESR rating of
about 3.5 to 4.5 the output current depending on the size of the
transformer.

I dunno about rule of thumb, you can simply calculate it, or, if you
want a visual, run it in spice, or stick a scope on it.
That last thing rules.
Electrolytic caps have tolerances that can be quite large anyways, say 20%.

On the capacity side, were the sightly contradictory suggestions of
aminimum capacity to reduce ripple voltage to <= to 10% of supply
voltage, and 2000uf per ampere of output current.

It is up to you, again if you want to run 100% sine waves all the time,
then you need more C, I would think, then for normal music.
100% sine waves is _not_an audio application, but you could be driving a motor or
whatever.

Somebody suggested thermal protection in the transformer, I think that is a cool idea.

 

[Fred]

Oops, seems I forgot the link:
ftp://panteltje.com/pub/PA3000/


Well, we say 'U' and you say 'V', but apart from that, it says
that if you charge capacitor C with current i, then after t seconds the voltage on that
cap will be U (or V ;-) ) volts.
Actually it is Q = C .U = i . t, where Q is the charge.


I only used it to see how big the voltage drop would be between 2 consequentive
mains tops in your rectifier for a given line frequency and a given cap, and
a given current.


I dunno about rule of thumb, you can simply calculate it, or, if you
want a visual, run it in spice, or stick a scope on it.
That last thing rules.
Electrolytic caps have tolerances that can be quite large anyways, say 20%.


It is up to you, again if you want to run 100% sine waves all the time,
then you need more C, I would think, then for normal music.
100% sine waves is _not_an audio application, but you could be driving a motor or
whatever.

Somebody suggested thermal protection in the transformer, I think that is a cool idea.

I've got the chip up and running again with the old transformer:
384VA 10% reg -> 2x24.

I resoldered the wiring and removed some caps. The voltage now is
about +/-37 volts. it drops to about +/-35 under load.

A interesting thing I noticed is that it only puts out about 7.5V into
a 5R power resistor before the positive side starts to clip. If I
raise the input signal, the top rail will continue to clip, while the
bottom rail will go to about -18V pk before the spike protection
starts activating.

Also I reviewed the thermal considerations, and realized that my 3C/W
heat sink is only half of what it should be. I can see on the scope
the thermal protection activate as the heat sink gets hot.

The data sheet states at +/35V the max dissipation of the chip is
65W. I found a spice model for it and have been studying various
transformer models and capacitor values with it. I modeled the
transformer I actually have and the chip dissipation was 56W at max
output of 88W. The data sheet says 68W continues into 4 ohms with
+/-35 Volts. I can plainly see that thermal management is a priority
here.

I have to deal with over a hundred watts of dissipation with two
chips. I've been looking at heatsinks and fans. I found a neat
control circuit that provides linear voltage response in regard to
temperature with a floor that will just keep the fan turning with the
heat sink is cool.

Fred

 

[Fred]

Oops, seems I forgot the link:
ftp://panteltje.com/pub/PA3000/


Well, we say 'U' and you say 'V', but apart from that, it says
that if you charge capacitor C with current i, then after t seconds the voltage on that
cap will be U (or V ;-) ) volts.
Actually it is Q = C .U = i . t, where Q is the charge.


I only used it to see how big the voltage drop would be between 2 consequentive
mains tops in your rectifier for a given line frequency and a given cap, and
a given current.


I dunno about rule of thumb, you can simply calculate it, or, if you
want a visual, run it in spice, or stick a scope on it.
That last thing rules.
Electrolytic caps have tolerances that can be quite large anyways, say 20%.


It is up to you, again if you want to run 100% sine waves all the time,
then you need more C, I would think, then for normal music.
100% sine waves is _not_an audio application, but you could be driving a motor or
whatever.

Somebody suggested thermal protection in the transformer, I think that is a cool idea.

I forgot to mention, that's neat that capacitors hove linear
relationship with voltage when charged with constant current.

Fred

 

[Jan Panteltje]

I forgot to mention, that's neat that capacitors hove linear
relationship with voltage when charged with constant current.

Fred

Same goes for inductors, the current rises linear with time for a fixed aplied voltage.
i = U . t / L

So if you have 1 Henry, and apply 1 Volt, then after 1 second the current is 1 Ampere,
after 2 seconds 2 Ampere, etc...

For a non-ideal coil the resistance will cause non-linearity, and the curve will begin
to flatten, finally the current will be U / Rs.

 

[Fred]

Same goes for inductors, the current rises linear with time for a fixed aplied voltage.
i = U . t / L

So if you have 1 Henry, and apply 1 Volt, then after 1 second the current is 1 Ampere,
after 2 seconds 2 Ampere, etc...

For a non-ideal coil the resistance will cause non-linearity, and the curve will begin
to flatten, finally the current will be U / Rs.

I must have damaged that chip too. I am glad I started with such a
simple project to learn the basics with.

I replaced it with a LM3876 that I have two, and it sounds much
better, makes about 14pk into 5 ohms before SPiKe activates. Now I
just have to figure out how to keep it cool. Junction to heat sink is
aout 1.5C/W at 65W dissipation at 50W that leaves very little room for
the heat sink 0.5C/W max really should be less.

Maybe I will liquid cool them. Maybe I put the whole thing in a
bucket of oil!

Fred

 

[Fred]

Same goes for inductors, the current rises linear with time for a fixed aplied voltage.
i = U . t / L

So if you have 1 Henry, and apply 1 Volt, then after 1 second the current is 1 Ampere,
after 2 seconds 2 Ampere, etc...

For a non-ideal coil the resistance will cause non-linearity, and the curve will begin
to flatten, finally the current will be U / Rs.

I find it fasinating how the energy storage devices, capacitors and
inductor inversely parallel there action in the electro-static and
magnetic ways.

Fred

 

[Jan Panteltje]

I replaced it with a LM3876 that I have two, and it sounds much
better, makes about 14pk into 5 ohms before SPiKe activates. Now I
just have to figure out how to keep it cool. Junction to heat sink is
aout 1.5C/W at 65W dissipation at 50W that leaves very little room for
the heat sink 0.5C/W max really should be less.

Tjunction max is 150 C, so with 50C environment, you have 100 degrees play room.
65 W in 100 C makes 1.53 C/W, that leaves .3 for the heatsink.
Those exist, but are big and expensive.
Maybe get a processor cooler with 12V fan....
Those are common items, cheap.
Bit noisy, add some temp sensor, you won't hear that fan at full volume...

You can make a simple temp sensor with a normal si diode, or old TO220
transistor (use be junction) that you can screw on the heatsink, and an opamp.

 

[Jan Panteltje]

PS that cannot be right?
14 V peak is about 9.9 V eff, makes only 19.6 Watt in 5 Ohm.
Check your circuit!

 

[Fred]

PS that cannot be right?
14 V peak is about 9.9 V eff, makes only 19.6 Watt in 5 Ohm.
Check your circuit!

Since the unit was a proto type some of the connections use wire
nuts. The chip is on a PC board. the output leads are kinda thin
maybe 20-18 awg stranded. There might be high resistance some where.
I was thinking that might have been the reason for the abnormally high
voltage in the first place. When I just rewired the PS, that high
voltage went away.

I will check into 10 ohms too.

I donno about forced air cooling, the heat sink would still have to be
big and the air flow would have to be substancial, I have a half
dozen of these heat sinks, and they would require 1000ft/min to get to
0.3C/W.

http://www.aavidthermalloy.com/cgi-...gth=3&airflow=57.2&CType=Natural&AirUnits=LFM

I'm still thinking liquid cooled heat sink. Especially if I use the
LM4780, that make 165W to deal with. thats about 30W/cm^2

Fred

 

[Jan Panteltje]

I donno about forced air cooling, the heat sink would still have to be
big and the air flow would have to be substancial, I have a half
dozen of these heat sinks, and they would require 1000ft/min to get to
0.3C/W.

Those processor coolers, that you can buy in any computer shop,
cool a 90W thermal design power Pentium XX to safe temperature.

I'm still thinking liquid cooled heat sink. Especially if I use the
LM4780, that make 165W to deal with. thats about 30W/cm^2

That chip is specified at max 125 W in any case, less if your heat sink
environment dictates that (Note 3 page 4 of the data sheet).

I would not drive it to that extreme.

Get one of those big processor coolers perhaps.

 

[Fred]

Those processor coolers, that you can buy in any computer shop,
cool a 90W thermal design power Pentium XX to safe temperature.


That chip is specified at max 125 W in any case, less if your heat sink
environment dictates that (Note 3 page 4 of the data sheet).

I would not drive it to that extreme.

Get one of those big processor coolers perhaps.

That waa 14Vrms.

I checked the circuit and found the mute resistor was too high of a
value. I replaced with lower value and now I get 17Vrms clean, then a
little something at the top of the wave form from there to 19.2Vrms
where clipping starts.

Amp sound Much better now. Thermal limitation of heat sink plain
limits power, I can hear it start to crackle a bit after just a few
second of high output.

I put fan on the heat sink, maybe 20-25cfm for a little improvement.
Sound is very clear even through my cheap 93db/w drivers. Can hardly
wait for better thermal management and good speakers!

Fred

 

[Fred]

Those processor coolers, that you can buy in any computer shop,
cool a 90W thermal design power Pentium XX to safe temperature.


That chip is specified at max 125 W in any case, less if your heat sink
environment dictates that (Note 3 page 4 of the data sheet).

I would not drive it to that extreme.

Get one of those big processor coolers perhaps.

The came up with some thing new since I worked on PCs, heatpipes!

That might work if I can figure out how to mount it.

Fred

 

[JosephKK]

I used the LM3886 it has better than excellent rejection. It was
designed to run off an unregulated supply. the problem came in trying
to get 50 watts into 4 ohms. it requires slightly more then +/-24V at
5A. My 24 volt transformers would rise to +/-42 during no-load
conditions. That is the absolute maximum limit of the chip. I
destroyed two chips, and they have all sorts of built in protections,
driving them to full output with those transformers.

worse than 50% regulation? 42/24 = 1.75. Ridiculous. Use a better
transformer.

 

[JosephKK]

Hey Graham,

Let me see if I can address these in order:

I did the thermal calculations. I used a 3C/W heat sink, and directed
a 40mmx40mm fan at during testing because it got hot when at full
load.

The 0.1uf and the 470uf were local bypass close as possible to the
pins of the device. In the gain clones, its those caps, a bridge, and
the tranny. Absolutely minimalistic.

My uinderstanding of transformer ratings is that the regulation is
drived from the voltage rise from load to no-load, and they deliver
the stated VA when resistively loaded.

I'm unsure about what difference a bridge and cap filters would do to
the output of the PS besides lower the VA a bit due to the diode drop
and the caps ESR

Fred.

Check out the diode conduction angles in the fully loaded case,
lighter loads just reduce the conduction angle. This will also
provide ripple data.