consider the step response of a single pole OP, like A(s)=A0/(s/ω0+1), the output of this OP configured as closed loop (buffer with feedback factor β), will be Acl(s)=A(s)/(1+A(s)*β)avinash said:dear btrend,
why and how you have chosen bandwidth=5•fin
if the loading cap is not so large, it is not so hard to get fu=600MHz. the thing u pay to earn this fu is the large current it consume.which opamp can provide the UGB of 600MHz as you mentioned.simple folded cascode can provide upto 100-150MHz .
from ur spec. Vout is only 4*1.25=5V, and ur VCC=12V. so the Vout is not so large as compare to VCC. I think GM amp. can fit ur need. usually large output swing is refered to Vout > 0.8Vcc in my knowledge.so if I use a simple Gm amplifier, how could I obtain large output swing?
Acl(s)=[A0/(1+A0*β)]*{1/[1+s/[(1+A0*β)*ω0]}~=Acl0*[1/(1+s/(β*ωu)]
the ωu is unit gain bandwidth, and it is equal to A0*ω0. and this value is constant, and β is the feedback factor, which happen only in closed loop .in my opinion, it should be:
Acl(s)=[A0/(1+A0*β)]*{1/[1+s/[(1+A0*β)*ω0]}~=Acl0*[1/(1+s/ωu]
where ωu=(1+A0*β)*ω0.
SR limits the maximum power bandwidth, fp, defined as the frequency at which a sine wave input, at the rated output voltage, begins to exhibit distortion. So fp>30E6, output Vr=4*1.25=5V, then SR>fp*(2*pi*Vr)=942V/us, I>SR*CL=85mA.wholx said:I = Cload * dV/dt = Cload * swing voltage / (1/fu) = 90E-12 * 1.25 * 600E6 = 70mA
fu = gm/(2*pi*CL) ==> gm=2*pi*fu*CL =339mA/V
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