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Old 13th March 2003, 09:44 PM   #1
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Default Damping factor

Anyone dispute this ?

Damping Factor =Zload/Zamp

Zload = speaker impedance
Zamp = output impedance of amplifier

Is the output impedance of the class ab amplifier a
constant, ie restistive measure or is
there any AC components which would
cause the output impedance to be a variable?

On a typical class ab amplifier, if you were
to trace the path that contribute to output impedance,
what components are those? ie, speaker terminal --> output
coil --> output stage transistor w/associated series resistor + what else?
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Old 13th March 2003, 09:48 PM   #2
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Damping factor as a specification is defined as 8 ohm divided by the source Z of the amp, normally at 1 kHz. It's one of the most useless specs out there.

Tracing the source Z is a little tricky, since feedback enters the picture.
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Old 13th March 2003, 10:37 PM   #3
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You are saying that it's frequency dependent? ie,
1khz testing.

Is it safe to say then;

If I have a high current output stage with
an array of paralled transistors *and* a high
current power supply, this should give me a low
output impedance even at low frequencies such
as 20hz? as opposed to a weak power supply
and no paralled transistors of the same ratings.

if above true, then can I say this;

If I replaced the high current power supply
with a low current power supply and do
the same 20hz test, then the current limiting
effects will create a condition of higher output impedance?
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Old 13th March 2003, 10:45 PM   #4
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Maybe, maybe not. A high current amp may or may not have a low source impedance. Consider, for example, a power current source rated at 10 amps. That will have a much higher source impedance (read, lower damping factor) than a weedy little LM380.

Normally, damping factor does vary with frequency; with most SS amps of modern topology (i.e., no output coupling cap), the source Z tends to go up at *high* frequencies, since feedback is reduced due to falling open loop gain

Don't confuse source impedance with maximum current capability. Also, don't confuse "damping" with "damping factor." The latter is a strictly defined commercial spec measurement, performed within the amp's ratings.
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Old 13th March 2003, 10:56 PM   #5
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--Consider, for example, a power current source rated at 10
--amps. That will have a much higher source impedance (read,
--lower damping factor) than a weedy little LM380.

A class ab design that can deliver 10 amperes into
the load *may* have much higher output impedance
than the weedy 2.5w LM380 ?

Is the frequency the variable that determines this ?
ie, people use film capacitors to bypass electrolytics
because at high frequencies the electrolytics are weak?

Is this a similar analogy ?
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Old 13th March 2003, 11:10 PM   #6
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I don't get the analogy.

Take that class AB amp you posit. Let's assume that its rails are at some high voltage. And that its source impedance is close to zero, say 0.01 ohm. So this amp has an extremely high damping factor, about 800. Now put an 8 ohm resistor in series with the output and call this a new amp. Your damping factor of the new amp is now about 1, yet the amp can still drive 10 amps through a load. So it can swing more voltage and it can drive more current than that LM380, yet its damping factor is lower because of the 8 ohm source impedance.

Starting to see why I called DF a relatively useless spec? To understand that uselessness even further, look at the frequency response of a speaker load driven by an amp with a DF of 800 versus the same response with a DF of 10. Damned little difference, probably inaudible.
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Old 14th March 2003, 02:30 AM   #7
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Good explaination.

/what if ........ LOLl

Taking your example of 0.01 ohms
output impedance, high voltage source
driving 10 amperes thru an 8 ohm load
(which represents a simple load, not
exactly a speaker).

DF = 800
Output Power = 10A^2 * 8 ohms
=800w
Vload = 10A* 8 ohms = 80v

True/false;

If all those variable remain constant
and I switch in another 8 ohm
series resistor on the output stage
to increase the output impedance
of the amplifier (lower DF to 1),
will this not decrease the 10 amperes
flowing thru the load resistor from 10 amperes down to 5 amperes (16 ohm
load total).

DF = 1
Output Power = 5A^2 * 8 ohms
=200w
Vload = 5A* 8 ohms = 40v

Did Bob Carver place a series
resistor on his amp output
to convert it to a current source
trying to emulate tube amp?
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Old 14th March 2003, 02:44 AM   #8
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Yes, that's the trick. Hardly a "current source," but it will mimic the frequency response changes of a high output-Z amp, like a typical tube amp.
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Old 14th March 2003, 02:50 AM   #9
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Default The Carver...

Quote:
Yes, that's the trick. Hardly a "current source," but it will mimic the frequency response changes of a high output-Z amp, like a typical tube amp
yes if memory serves is a 1 Ohm resitor in series...hardly a current souce as you say!
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Old 14th March 2003, 06:37 AM   #10
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Sorry, thylantyr, I forgot to answer your question. Yes, true. That's why high source Z leads to measurable and audible frequency response differences with real-world speakers, those that have non-constant Z over the audio band.

Now, if you connect the feedback loop at the load end of the 8 ohm resistor you added, the source Z becomes 8 divided by the feedback factor. If you have a moderate 20 dB of feedback, that source Z drops to only 0.8 ohm.
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