Gain-Bandwidth product=constant

eeweiboy

在回授系統中,為何說 Gain-Bandwidth product=constant ??
有高手願意解答ㄇ

xcoder

这是single pole时的情况
let an amplifer:　A(s)= A/(1+s/p1)
the feedback factor is : beta
then the close loop gain of the feedback system is : A(s)/[1+ A(s)*beta]=A/(1+s/p1+A*beta)
  =>s=-(1+A*beta)p1 , that is the dominate pole of the feedback system, p_fb=(1+A*beta)p1
if A*beta >> 1, then p_fb=A*beta*p1, => p_fb*(1/beta)=A*p1
here:  
1/beta is the close loop dc gain of the feedback system, so any (single pole ) feedback system, the gain * the bandwith is a constant which equals to the dc open loop gain of the amplifier * the bandwidth of the amplifier, the gain bandwidth of the amplifier with any feedback factor is not changed!

hope helpful

teddybear

miller compensation makes p1>>p2,
so in such 2 stage op, you can consider it as a simple pole system
then above formula is right

tq01qaz

仅仅是单极点近似才成立