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1st April 2003, 01:40 AM  #1 
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Join Date: Jun 2002
Location: Melbourne, Australia

Opamp phase shift  open and closed loop.
As we all know, any opamp has a certain amount of phase shift that gets greater with frequency. Manufacturers publish these figures for open loop. What happens to the phsae shift when the loop is closed though? Obviously the phase shift still exists inside the feedback loop, but what if you measure it outside the loop? Does it decrease in proportion to gain reduction like distortion does? Does it stay the same?
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1st April 2003, 06:27 AM  #2 
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If you have a flat response and you are not near the "drop" the phase shift is zero, but inside the amp and the feedback loop the phaseshift is probably 90 degrees up to 120140 degrees at  3 dB.
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1st April 2003, 07:08 AM  #3  
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Location: Sacramento, CA

Quote:
That's why group delay is really the only meaningful figure of merit when it comes to accurate reproduction. se 

1st April 2003, 04:59 PM  #4  
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Join Date: Nov 2002
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Quote:
But is it the only meaningful figure? Dropping open loop gain with frequency results in reducing distortion reduction with frequency. 

1st April 2003, 06:24 PM  #5 
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Hi Circlotron and others,
Open and closed loop phase shifts are related. Then, both are related to the opamp’s bandwidth. The higher the bandwidth, the lower the phase shifts at given frequency. Also, opamp set to lower gain has higher bandwidth and hence lower phase shift. Here are the numbers. With gain of 20, LM6172’s (100MHz at unity gain opamp) phase shift at 20kHz is less than 0.2deg, and group delay (difference between 1kHz and 20kHz) is notably less than 1picosecond. In the same circuit OP27 (8MHz at unity gain) has more than 3degs phase shift and difference in group delay is more than 1ns. The same OP27 when works with gain of 4 has phase shift less than 0.5deg and group delay of a few picoseconds. If you check these numbers calculating these shifts from corresponding group delays you’ll se, as Steve mentioned, the bulk of these phase shift numbers are caused by the propagation delay and not by the nonlinear groupdelay. But again, propagation delay is in relation with the nonlinear group delay. Pedja 
1st April 2003, 06:47 PM  #6  
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Quote:
Quote:
se 

1st April 2003, 07:26 PM  #7  
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Join Date: Sep 2002
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The phase shift for a given frequency and a given delay is a simple calculation of the time delay times the frequency times 360 degrees. For example, with a constant delay of 12.5 microseconds, the phase at 20 Hz will be: 0.0000125 x 20 x 360 = 0.09 degrees. At 200Hz it will be: 0.0000125 x 200 x 360 = 0.9 degrees. At 2kHz it will be: 0.0000125 x 2,000 x 360 = 9.0 degrees. And finally at 20 kHz, it will be: 0.0000125 x 20,000 x 360 = 90 degrees. Yet in spite of a 90 degree phase shift at 20 kHz, all frequencies will arrive at the same time, so there's no smearing or phase distortion, etc. se 

23rd April 2003, 06:58 AM  #8 
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Location: Sunnyvale, CA

From a theory point of view, in a unity feedback system (buffer) the frequency response of the closed loop system is G/(G+1) where G is the complex open loop response. I made a plot showing the relationship for an arbitrary system with a delay.
The simple way of looking at it is that closing the loop reduces phase lag in the 0dB gain neighborhood, but not by much. 
23rd April 2003, 07:06 AM  #9 
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Join Date: Oct 2000
Location: Sunnyvale, CA

If anyone cares, here it the spreadsheet I threw together to make that plot.
http://members.cox.net/quotes/closed%20open.xls Also, you can see some wierdish mechanical systems measured by a program I make here using the same principles: http://support.motioneng.com/Utiliti...e/bode_07.html http://support.motioneng.com/Utiliti...e/bode_08.html 
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