- #1
papasmurf
- 22
- 0
The gain of an internally compensated op amp may be expressed by
Equation 1) A(s) = A0 / ( 1 + s/[itex]\omega[/itex]b) where [itex]\omega[/itex]b is the 3dB frequency or "break frequency" and A0 is the dc gain.
Plugging in s=j[itex]\omega[/itex] and [itex]\omega[/itex] = 2*pi*f gives
Equation 2) A= A0 / (1 + j*f / fb) where fb is still the "break frequency"
I understand how all of that works.
What I don't understand is how you can go from Equation 2 to this:
Equation 3) A = A0 / sqrt( 1 + (f/fb)^2).
I hope all of this makes sense and someone can help me. Thank you.
Equation 1) A(s) = A0 / ( 1 + s/[itex]\omega[/itex]b) where [itex]\omega[/itex]b is the 3dB frequency or "break frequency" and A0 is the dc gain.
Plugging in s=j[itex]\omega[/itex] and [itex]\omega[/itex] = 2*pi*f gives
Equation 2) A= A0 / (1 + j*f / fb) where fb is still the "break frequency"
I understand how all of that works.
What I don't understand is how you can go from Equation 2 to this:
Equation 3) A = A0 / sqrt( 1 + (f/fb)^2).
I hope all of this makes sense and someone can help me. Thank you.