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# Question about op amp frequency response?

Hello , i have some question about op amp frequency response.

i use non inverting like the scematic below.If i set R1=10kΩ , C1=220nF , R2=100Ω , C2=1uF . R1 and C1 will set low frequency response of opamp at 72Hz . R2 and C2 will set high frequency response of opamp at 1592Hz.

But how much it affect the signal in dB ? and how its affect the signal if i set R2=100k ,C2=1n (gain~11)?

Can you give me the formula please ?

Thanks very much!

Sorry for my bad English!

## Discussions

The formula for the transfer function for a non-inverting amplifier, like the one shown is just:

(Vout/Vin) = 1 + (X2/X1)

and for this circuit in particular,

X1 = R1 + (1/(s*C1)) = (1+s*R1*C1)/(s*C1)

X2 = R2 parallel (1/(s*C2)) = R2/(s*R2*C2+1)

where, as usual, s = j*ω

When I substitute those expressions for X1 and X2, I get this big ugly transfer function, with

numerator= s*s*R1*C1*R2*C2 + s*R1*C1 + s*R2*C2 + s*R2*C1 + 1and

denominator= (s*R2*C2 + 1 )*(s*R1*C1 +1 )Assuming I did the math right, there are two poles at s=-1/(R1*C1), -1/(R2*C2), and two zeros somewhere...

Also I think it has gain=1 at DC (at ω=0), but you probably could have guessed that just from looking at the circuit and imagining the capacitors as open circuits.

Anyway, from here I think you can substitute in whatever your actual Rs and Cs are, and feed this transfer function to your favorite number crunching program, like MATLAB, or Octave, and do a Bode plot to see what your frequency response is doing.

The corners are at 1/r1c1 and 1/r2c2 and both will both be changing at 20dB/decade, if that's what you mean. Since its non-inverting, the minimum gain is always 1.