Class "D" power amp conundrum

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I just ordered a kit for a switching power amp from www.41hz.com, and
have run up against a conundrum. Suppose, for the sake of argument, you
have two chip amps, both running on the same regulated power supplies,
delivering the same output signal to the same load. One is Class AB,
with roughly 60% efficiency. The other is a switching amp with 90%
efficiency. But...

1. Both amps feed the same voltage & current waveform to the load.

2. Both amps draw the same current waveform from the power supplies.

One amp dissipates more heat than the other, but it doesn't seem like
the difference in heat is coming from anywhere else in the system. I
have to assume that my second assumption is faulty, but I can't figure
out how.
 
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On 7 Aug 2005 19:07:38 -0700, elephantcelebes@yahoo.com wrote:

>I just ordered a kit for a switching power amp from www.41hz.com, and
>have run up against a conundrum. Suppose, for the sake of argument, you
>have two chip amps, both running on the same regulated power supplies,
>delivering the same output signal to the same load. One is Class AB,
>with roughly 60% efficiency. The other is a switching amp with 90%
>efficiency. But...
>
>1. Both amps feed the same voltage & current waveform to the load.
>
>2. Both amps draw the same current waveform from the power supplies.
>
>One amp dissipates more heat than the other, but it doesn't seem like
>the difference in heat is coming from anywhere else in the system. I
>have to assume that my second assumption is faulty, but I can't figure
>out how.

The linear amplifier draws (at each instant in time) approximately
the same current from the power supply that it delivers to the load,
as the current is passed in series from the power supply through the
output transistors and to the load. This covers half of your
assumption 2.
The switching amplifier draws LESS current from the power supply
than it delivers to the load (ignoring inefficiencies, and it's
probably true even with 90 percent efficiency). It acts like a
transformer, much as any switching power supply does. Pulling 10
amperes from a SMPS's 5 volt line doesn't result in anywhere near 10
amperes draw from the 120 volt power line input to the SMPS. That's 50
watts, and with reasonable efficiency less than 1 ampere will be drawn
from the power line.

-----
http://www.mindspring.com/~benbradley
 
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In <1123466858.787501.210490@z14g2000cwz.googlegroups.com>, on 08/07/05

at 07:07 PM, elephantcelebes@yahoo.com said:


>I just ordered a kit for a switching power amp from www.41hz.com, and
>have run up against a conundrum. Suppose, for the sake of argument,
>you have two chip amps, both running on the same regulated power
>supplies, delivering the same output signal to the same load. One is
>Class AB, with roughly 60% efficiency. The other is a switching amp
>with 90% efficiency. But...

>1. Both amps feed the same voltage & current waveform to the load.

>2. Both amps draw the same current waveform from the power supplies.

>One amp dissipates more heat than the other, but it doesn't seem like
>the difference in heat is coming from anywhere else in the system. I
>have to assume that my second assumption is faulty, but I can't figure
>out how.

With an output filter on the Class-D amplifier, 1. could be mostly
true.

If 1. & 2. are true then the amplifier efficiencies are incorrectly
stated.

If you've discovered a simple electronic process that creates or
destroys energy, at least try to patent it. In the past the Patent
office has taken a dim view of such claims, but you could get lucky --
especially if you claim it is a software process.

When the output transistors are switched ON, the amplifier draws
current from the power supply. In the Class-AB amplifier, output
transistor current is always being drawn. In the Class-D amplifier,
output transistor current flows intermittently.

When a transistor is full on or full off, very little waste heat is
generated. When a transistor is partially conducting, quite a lot of
waste heat can be generated. Calculate a few waste heat points by
multiplying the voltage drop across the transistor by the current
flowing through it. When the transistor is full on, the voltage drop is
small. When the transistor is off, current is zero. If the transistor
is half on, waste heat is high. In the Class-D amplifier, the
transistor is full on or off and little heat is generated. (mostly
while the transistor is in transition between on and off)

In the Class-D amplifier a post filter integrates the pulses and the
resulting output can be identical to the Class-AB situation.

Measuring the voltages and currents can be difficult because some
metering systems respond poorly to the Class-D pulses.

---

If you haven't already considered measurement issues, think about the
power associated with a simple sine wave current. Measuring the peak
value is easy enough, but it is not very useful because the dwell time
at the peak is brief. The next obvious thought is "average" over one
cycle. Unfortunately, the average value of a sine wave is zero. While
"average" is a good concept, but we have a computational difficulty.

If you haven't been through all of this, find a good text and do some
reading.

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wordgame:123(abc):<14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13> (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
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Don't worry about me inventing a perpetual motion machine. My
three-year-old boy is doing that job just fine. I was mentioning the
conservation law to drive home that one of my two assumptions is
incorrect, but I could not figure out which one. Ben Bradley answered
the question quite succinctly.

I'm not sure which question you are answering. It is not a
computational difficulty to calculate an average. Is there something
that Ben missed in his answer to my question?
 
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Sorry about being snippy. You are correct that I did not indicate my
level of knowledge, but the "go read a textbook" response doesn't
indicate your level of knowledge either ;-)
 
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In <1124495805.919149.122360@f14g2000cwb.googlegroups.com>, on 08/19/05

at 04:56 PM, elephantcelebes@yahoo.com said:

>Sorry about being snippy. You are correct that I did not indicate my
>level of knowledge, but the "go read a textbook" response doesn't
>indicate your level of knowledge either ;-)

No, I threw that in because I was not prepared to go through a complete
discussion of "RMS" vs "average" vs "peak" power.

-----------------------------------------------------------
spam: uce@ftc.gov
wordgame:123(abc):<14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13> (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------
 
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Understood. And I should have indicated my level of knowledge, like you
suggested. I am a physicist, and have done extensive work with
electronics. Had I thought about things a bit more, I would not have
gotten into my quandary. Chalk it up to intellectual laziness.

The kit is built... I hook it up to a power supply tomorrow.
 

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