Help to understand opamp compensation

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- idkfaidkfaidkfa
  
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posted
15 years ago

Thu, Jan 8, 2009 7:49 AM

I've found this interesting lecture

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but i don't understand open loop gain with resistor series capacitor as load.In fig. 6.15 opamp's open loop gain is represented: it is calculated (i think) at node Vout of fig 6.14, so pole is equal to 1/ (Ro+Riso)*Cl. I don't understand how zero is obtained.... Can you help me? Thanks

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- Tim Wescott
  
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posted
15 years ago

Thu, Jan 8, 2009 3:06 PM

Try replacing the amplifier inside the opamp (ahead of the 28.something resistor) with a voltage source, and find the transfer function from that source to Vout.

Do your calculations right and you'll see both the pole at fpo1 and the zero at fzo1.

Report back here...

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Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

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"Applied Control Theory for Embedded Systems" by Tim Wescott
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