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Is this correct sallen-key bandpass filter
- Thread starter tim_3491
- Start date
Assuming that the circuit works correctly, V+ = V-. Because of the two 10k's, V- = Vout/2. Therefore, V+ = Vout/2. Also, since I5 = -(V+/20k) = -Vout/40k. Then I4 = -I5, so I4 = Vout/40k = (Vx - Vout/2)*jwC. Solve for Vx, and you will have Vx in terms of Vout. The rest you seem to have well in hand.
While I can't answer your question, any help on this particular circuit would be much appreciated 
. Though one thing is, I looked at your working and I found a mistake:
You've multiplied by 1/jwC (fair enough) but then when you minus V- on both sides, you have suddenly brought it so it is over 2R (20k) even though I needs to be multiplied by 2R before you can do this (adding fractions involves the fractions having the same denominator).
Other than that though, you are going about this question the same way as me really. I've analysed KCL at V+, V- and Vx (as you call it) and I keep getting answers but they are wrong! (They produce complex numbers or a gain that is too high). If you actually find the answer, do share
EDIT: Also V+ and V- are not necessarily 0V - they are simply the same
. Though one thing is, I looked at your working and I found a mistake:You've multiplied by 1/jwC (fair enough) but then when you minus V- on both sides, you have suddenly brought it so it is over 2R (20k) even though I needs to be multiplied by 2R before you can do this (adding fractions involves the fractions having the same denominator).
Other than that though, you are going about this question the same way as me really. I've analysed KCL at V+, V- and Vx (as you call it) and I keep getting answers but they are wrong! (They produce complex numbers or a gain that is too high). If you actually find the answer, do share
EDIT: Also V+ and V- are not necessarily 0V - they are simply the same
Last edited:
The Electrician
- Joined Oct 9, 2007
- 2,986
