by lionelgreenstreet@[EMAIL PROTECTED]
Jun 30, 2008 at 07:32 AM
In addition to my previous post, i've simulated an transimpedance
amplifier with:
1)Rf=10kohm
2)ideal current generator=0.2mA
3)OPA128 (with 1MHz as open loop bandwidth)
4)only parasitic capacitance(3pf)
IId this circuit cut off frequency is equal to
Vout=(-Rf*Iin)/(1+j(f/fc))
fc=A0/(2*pi*Rf*Ci)
with
A0=open loop gain at frequency pole
because:
Rf'=Rf/A0 for miller effect
Ci=parasitic capacitance (there is no miller effect for common mode
and differential mode parasitic capacitance)
So,i've obtained fc equal to 2,3Mhz (calculated and simulated), but i
not conviced: its strange that i have a bandwidh greater then open
loop gain...I know that open loop bandwith is obtained as voltage
amplifier (instead fc is obtained as transimpedance amplirier), but
i've some doubts on these results....what do you think?
