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14th August 2005, 05:18 PM  #1 
diyAudio Member
Join Date: Nov 2002
Location: sg

a question about local feedback
i was pondering bout my next amp project and it hit me that i didn`t have an idea what local feedback did to the output impedance of the stage. and if implemented in an output stage, what kind change in output impedance would i see...? is it very much like global feedback?
thanks! 
14th August 2005, 05:48 PM  #2 
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Join Date: Apr 2004
Location: Monroe Township, NJ

Local or global, voltage NFB is voltage NFB. It raises I/P impedance and lowers O/P impedance. The math stays the same.
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Eli D. 
14th August 2005, 05:53 PM  #3 
diyAudio Member
Join Date: Nov 2002
Location: sg

with 3db of feedback, how much change in the o/p Z am i expecting to see change?
thanks! 
14th August 2005, 11:50 PM  #4 
diyAudio Member

3 decibels (power) = x 2 = 6 decibelvolts.
dBV are double (duodecibels?) because doubling voltage doubles current into a resistance, so power actually goes up by a factor of four (i.e., +6dB). At any rate, linear negative feedback reduces Zo, distortion, gain and, if connected properly, increase Zin all by the same factor. This factor is the amount of feedback. Tim
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15th August 2005, 11:37 AM  #5  
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Join Date: Jan 2003
Location: Sweden

Quote:
dB for voltage = 20 * LOG(V2/V1), so if V2/V1 is 2, (double voltage) it gives 6dB dB for Power = 10 * LOG(P2/P1), so if V2/V1 is 2, (double power) it gives 3dB However 3dB for voltage and 3dB for power is the same thing V2/V1 = 1.4142 gives 3dB and an increase of 1.4142 in voltage gives an increase of 2 times in power, (P ~ V^2) and 2 times in power is also 3dB. Back to the original question, 3dB feedback is 1.4142 times in voltage so the decrease in output impedance will be 1.4142 times, (I assume that what is meant with amount feedback here is how much the gain decreases) Regards Hans 

15th August 2005, 02:19 PM  #6 
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Join Date: Nov 2002
Location: sg

thanks a million, now i`ve got an idea how my amp will turn out.

16th August 2005, 03:13 AM  #7  
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Quote:
By numbers alone, 3dB (power) = factor of two = 6dBV. I noted lower that, into a resistance, 3dB = 6dBV. Tim
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16th August 2005, 07:31 AM  #8  
diyAudio Member
Join Date: Jan 2003
Location: Sweden

Quote:
"dB for voltage = 20 * LOG(V2/V1), so if V2/V1 is 2, (double voltage) it gives 6dB dB for Power = 10 * LOG(P2/P1), so if V2/V1 is 2, (double power) it gives 3dB" This is the definition of dB. What it means is that 3dB is 2 times for Power but only SQRT(2) for voltage. An example: an amplifier with 40dB gain has power gain of 1000 times but a voltage gain of only 100 times, (assuming same input and output impedance) In this case, (what Hacknet asked about) we had a feedback of 3dB which means that the power AND voltage gain decrease by 3dB i.e. SQRT(2) ~= 1.4142 times but a power gain decrease of 2 times. Regards Hans 

16th August 2005, 07:42 AM  #9 
diyAudio Moderator Emeritus
Join Date: Jan 2004
Location: Jakarta

This thread, which I strarted last January, discusses the effectiveness of partial/local feedback for improving damping factor. There were many helpful contributions from people with a lot of experience.

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