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Old 23rd August 2010, 12:21 PM   #1
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Default Feedback and output impedance

I understand how loop/global negative feedback reduces distortion, but how exactly does it reduce the output impedance? I've searched around here and on google without a definitive answer - just some general hand waving about the reduction of distortion, increase in bandwidth and reduction in output impedance.
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Old 23rd August 2010, 01:07 PM   #2
jcx is offline jcx  United States
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if you really know how output distortion is reduced by negative feedback then you already know how output impedance is reduced - feedback isn't "smart" enough to know if it is correcting a nonlinear distortion or a linear voltage drop from load current flowing in device impedance - both result in a measured error at the summing input and the error is amplified by the excess loop gain (feedback factor)

Blackmans Theorem is a keystone in feedback theory - look for it in your textbook index to determine if the book is useful in really explaining feedback

Last edited by jcx; 23rd August 2010 at 01:16 PM.
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Old 23rd August 2010, 01:13 PM   #3
PMA is offline PMA  Europe
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Very high feedback = necessity to use output coil = higher output impedance with frequency

Quite funny.

Take a look at John Curl JC1. This power amp does really have LOW OUTPUT IMPEDANCE.
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Old 23rd August 2010, 03:04 PM   #4
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The amp Zout is not a single physical impedance, but rather the effect of a sagging output level with increasing load. If, say, with a load increase of 5 amps, the output level drops 1V, we say (with a wink to Mr. Ohm) that Zout = 1/5 ohms = 200milli ohms.
If you use neg feedback, the drop of the output level is 'counteracted' by the feedback (or if you prefer, compensated). So now you only have a level drop of say 100mV with a 5A output load increase. Now Zout = 0.1/5 = 20 milliohms.

BTW That's also an easy way to measure Zout. Measure the output level drop for a given output load increase, and divide.
Be aware that it is frequency dependent and possibly level dependent.

jd
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Old 23rd August 2010, 05:39 PM   #5
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thank you janneman - very succinct and it makes perfect sense
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Old 23rd August 2010, 06:12 PM   #6
forr is offline forr  France
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Quote:
Originally Posted by kstagger View Post
I understand how loop/global negative feedback reduces distortion, but how exactly does it reduce the output impedance? I've searched around here and on google without a definitive answer - just some general hand waving about the reduction of distortion, increase in bandwidth and reduction in output impedance.
The initial purpose of negative feedback when invented was to stabilize the voltage gain, making the value of the output voltage being a multiplied value of the input voltage, and this most, in the most independant way as possible from the inside of the circuit or from the load at the output.
All the effects you describe are only consequences of this stabilisation of the gain, which is set by two resistors.
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Old 23rd August 2010, 07:17 PM   #7
DF96 is offline DF96  England
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Negative feedback only lowers output impedance if it senses the output voltage, and tries to make this a defined multiple of the input voltage. You can arrange negative feedback to sense the output current instead, and then it raises output impedance. This is how an unbypassed emitter/source/cathode resistor works. You can even have a combination of the two extremes, and use feedback to set a particular impedance.

This may be why you could not find a simple explanation of why NFB reduces output impedance. Sometimes it doesn't!
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Old 23rd August 2010, 07:20 PM   #8
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Quote:
Originally Posted by DF96 View Post
[snip] You can arrange negative feedback to sense the output current instead, and then it raises output impedance. This is how an unbypassed emitter/source/cathode resistor works. [snip]
An unbypassed cathode/emitter res is negative voltage feedback (100% of it) and decreases Zout.

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Old 23rd August 2010, 07:26 PM   #9
SY is offline SY  United States
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Quote:
Originally Posted by janneman View Post
An unbypassed cathode/emitter res is negative voltage feedback (100% of it) and decreases Zout.

jd
Really? It decreases output impedance at the plate/collector?
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Old 23rd August 2010, 07:40 PM   #10
DF96 is offline DF96  England
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I suspect janneman may be confusing emitter unbypassing with emitter following. It all depends on where you take your output from.
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