Help to understand opamp compensation

Jan 08, 2009 1 Replies
I

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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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...

Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
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