Help to understand opamp compensation

I've found this interesting lecture
www.analogzone.com/acqt0704.pdf
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

 

[Tim Wescott]

On Wed, 07 Jan 2009 23:49:15 -0800, idkfaidkfaidkfa wrote:

I've found this interesting lecture
www.analogzone.com/acqt0704.pdf
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

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