adding ressistor on the output to cure problematic slope in bode plot

[yefj]

Hello , on the third minute of the manual shown in the link at minute 3 he says that when we add a resistor our added pole is 1/(Rop+Radd)C
Then he says that when the impedance of the capacitor in the same order ar the added resistor then we have a ZERO effect and the slope gets less steep.
How does a pole can turn to zero?
What is the math behind it?
Thanks.

 

[D.A.(Tony)Stewart]

http://web.mit.edu/2.14/www/Handouts/PoleZero.pdf

When Impedance of C matches series Rs then you are near critical damping.
My RLC impedance 4D graph helps look up these values.

e.g. here L=C=1e-9 and Zo=sqrt(L/C)= 1 so adding Rs=1 make it "near" critically damped with a very low peak in frequency response and shifted slightly lower. (Q = 1⁄2) is said to be critically damped.

There are tables with direct correlation to 2nd order Q and overshoot.

Q factor - Wikiwand

In physics and engineering, the quality factor or Q factor is a dimensionless parameter that describes how underdamped an oscillator or resonator is. It is defi...

www.wikiwand.com

 

[danadakk]

If we ignore the internal z's of the OpAmp output, we have a series
R and C to ground, so

Z = R + 1/sC = (RCs+ 1) / sC = [ RC (s + 1/RC) ]/sC, hence you can see the zero
in numerator.


Regards, Dana.

 

[D.A.(Tony)Stewart]

If we ignore the internal z's of the OpAmp output, we have a series
R and C to ground, so

Z = R + 1/sC = (RCs+ 1) / sC = [ RC (s + 1/RC) ]/sC, hence you can see the zero
in numerator.


Regards, Dana.

but at unity gain frequency then internal Z of op Amp may be relevant if R approaches Zo.

e.g. Op Amp Zo may vary from 20 to 250 ohms in BJT type

So for critical damping step response, include internal Zo and choose Rtotal = 1.85 to 2 * X(f) at breakpoint, or fo resonance

It depends if you want slightly faster step response with overshoot, or no overshoot.
But recall Vin+ always has greater BW.