/diy/ - Do It Yourself

>>2403822
Calculate the transfer function for something quantitative:
>X_C = 1/(s*C)
>X_bottom = XC2 || R2 = XC2*R2/(XC2+R2) = R2/(1+sR2C2)
>X_top = XC1 + R1 = R1 + 1/sC1
>Vout/Vin = X_bottom / (X_top + X_bottom)
>Vout/Vin = R2/(1+sR2C2) / (R1 + 1/sC1 + R2/(1+sR2C2))
>etc.
>Vout/Vin = sR2C1 / (sR1C1 + R1R2C1C2s^2 + 1 + sR2C2 + sR2C1)
Each term in that equation (except for 1) represents the complex angular speed or frequency "s" multiplied by some effective frequency.

If the resistances and capacitances are all equal (R1=R2=R, C1=C2=C) then you get a simpler equation:
>Vout/Vin = sRC / (3sRC + (sRC)^2 + 1) = 1/(3s + sRC + 1/(sRC))
Here the resulting graph has a somewhat rounded top, a passband gain somewhat below 1 at 0.33 or -9.5dB, and the standard ±20dB/decade rolloff on either side.

More likely though, if you'll be having a substantial flat area in the middle of the band-pass, you can make it such that two values are much larger than two others. Say the passband gain, we want to be as high as possible, so R1 will be very small and R2 will be very large. Then for similar reasons we make C1 much larger than C2. This way we get a simple graph where the passband goes from R1C2 to R2C1, indicating that those two parts in the transfer function refer to the edges of the passband. This is a little easier to see if you express it as:
>Vout/Vin = 1 / (R1/R2 + sR1C2 + 1/(sR2C1) + C2/C1 + 1)
The only terms are either the corner frequencies or simple dimensionless ratios. These ratios are much larger than 1, so we can simplify:
>Vout/Vin = 1 / (sR1C2 + 1/(sR2C1) + 1)
When messing about with transfer functions, you want the terms to actually mean something.

The other extreme would be making R2 and C1 much larger than their counterparts, this gives a triangular peak with a "passband" gain much lower than 1. As seen in the lower equation, the C2/C1 and R1/R2 terms are much larger than 1 and pull down the maximum value:
>Vout/Vin = 1 / (R1/R2 + sR1C2 + 1/(sR2C1) + C2/C1 + 1)