Ken123
Joined: 06 Oct 2006
Posts: 2
Location: Singapore
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 Posted:
Wed Dec 13, 2006 6:57 am Post subject:
switched capacitor filter |
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i have some questions regarding basic operation of switched capacitor filter. I wonder whether anyone is kind enough to help me. i attached a picture of inverting and non-inverting switched capacitor filter.
1) In picture (a) it is non-inverting switched capacitor filter.
My understanding of its' operation is :
Phase 1 (Vi is +ve)
- C1 is charged up by signal Vi. Left side of C1 is +ve.
Phase 2 (Vi is +ve)
- C1 discharge and current flows from right to left of C1 and C2 because left side of C1 is +ve. Thus Vo is +ve, and it is non-inverting.
Phase 1 (Vi is -ve)
- C1 is charged up by signal Vi.
Phase 2 (Vi is -ve)
- C1 discharge and current flows from left to right of C1 and C2. Thus Vo is -ve, and it is non-inverting.
Is the above understanding correct?
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2) In picture (b) it is inverting switched capacitor filter.
My understanding of its' operation is :
Phase 1 (Vi is +ve)
- It will charged up C1, left side will be +ve and current flows from left to right. This current will flows through C2 from left to right causing the output Vo to be -ve. Thus Vo is -ve and it is inverting.
But i don't understand phase 2. What will happen? C1 discharge by short circuit across it?
Phase 1 (Vi is -ve)
- It will charged up C1, left side will be -ve and current flows from right to left. Vo is +ve.
Again i don't understand phase 2.
help explaning to me ?
Thank You
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neon
Joined: 25 Feb 2006
Posts: 579
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Wed Jan 17, 2007 8:18 pm Post subject:
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| change the source to a DCvolts then set up a table logical table as s1,s2,s3,s4 the close and open each S's as 0,1. Basicaly you will charge and discharge the first cap whatever hapens on c1 will transfer to the output a opposite polariry. because of inversion. Change the source polarity will reverse output polarity. s1 and s2 cannot be closed at the same time obviously you will short the source. but for paper analisys it is a valid premise. |
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