op amp and iits unity gain

[Gagan_SJSU]

Hi friends

I have a question to ask....if my op amp has a DC gain of 10000 and one pole at 1khz...what would be my frequency where the gain would cross unity..,also please explain me the method ..

thanks
Sunny

 

[pavel_adameyko]

Hi, Gagan_SJSU.

Transfer function of one pole circuit can be depicted as G(f)=A/(1+j*f*q), where f - frequency, j - square root of -1, q - some coefficient. If f=0 (dc) then G(f)=A= 1e+4 in your circuit. -3dB frequency is when G(f) drops square root of 2 times below its dc value. That occures when
1=| j*f*q | so q=1/f = 1e-3 in your circuit. At high f 1<<| j*f*q | so G(f) ≈A/(j*f*q). If G=1 then
f*q=A, so unity gain frequency equal to A/q=1e+7=10M for your opamp

Regards.

 

[A_Galaxy]

well, I am not sure if i understand your question correctly: but I'll try to explain it in a coherent way.

Every pole in a circuit attenuates the signal 20 dB/decade. so if you have 2 poles
for ex: pole 1= 100Mhz , Pole 2= 500Mhz. Your signal starts to attenuate at 20 dB/decade at 100Mhz and once it hits 500Mhz, it starts to go down at 40 dB/decade (20dB + 20dB = 40 dB/decade)

so in your case you have only 1 pole: so once your signal hits 1Khz it starts to role down at 20 dB/decade. so if it starts at 10000 gain; lets convert your gain to dB (decible) 10000 volts gain = 20*log (10000) = 20*4 = 80dB is your gain.

so if the signal goes down 20dB/decade then It will take 4 decades (4* 20dB = 80dB) to go to Zero.
So 4 decades after 1Khz is 10000Khz = 10Mhz. so your unity gain frequency is 10Mhz. i'm kind of sleepy, but i think the math is correct. I hope it made sense.