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13th March 2003, 09:44 PM  #1 
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Join Date: Feb 2001
Location: Mars

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? 
13th March 2003, 09:48 PM  #2 
diyAudio Moderator

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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13th March 2003, 10:37 PM  #3 
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Join Date: Feb 2001
Location: Mars

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? 
13th March 2003, 10:45 PM  #4 
diyAudio Moderator

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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The more you pay for it, the less inclined you are to doubt it. George Smiley 
13th March 2003, 10:56 PM  #5 
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Join Date: Feb 2001
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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 ? 
13th March 2003, 11:10 PM  #6 
diyAudio Moderator

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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The more you pay for it, the less inclined you are to doubt it. George Smiley 
14th March 2003, 02:30 AM  #7 
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Join Date: Feb 2001
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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? 
14th March 2003, 02:44 AM  #8 
diyAudio Moderator

Yes, that's the trick. Hardly a "current source," but it will mimic the frequency response changes of a high outputZ amp, like a typical tube amp.
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The more you pay for it, the less inclined you are to doubt it. George Smiley 
14th March 2003, 02:50 AM  #9  
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Location: AveiroPortugal

The Carver...
Quote:
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14th March 2003, 06:37 AM  #10 
diyAudio Moderator

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 realworld speakers, those that have nonconstant 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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The more you pay for it, the less inclined you are to doubt it. George Smiley 
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