get switch cap current

How can I get a current I=(Vref-Vin)/R, R is related to C*f from a switched capacitor network? Anyone can help? Thanks.

hi shanmei,

you may be misinterpreting what i'd call this an equivalent dc current for a cap being refreshed at frequency "f" for a continuous *true* DC current. see if i can explain,

Imagine a pulse train from Vref (low voltage) to Vin (high voltage) at frequency "f" with duty cycle of 50%. A square wave is at frequency, f,has a period of T = 1/f. In other words, this voltage is at Vin voltage for 50% of T and at Vref for the other 50% of T.

Take a capacitor with terminal A and B and hold terminal B at Vref. On terminal A, apply the aforementioned pulse train. For the first half of T, when terminal A is at Vin, the charge stored on the cap is Q=C*(Vin-Vref). Then the second half of T, the charge stored on the capacitor is Q=C*(Vref-Vref)=0 Coulomb.

Now imagine the pulse trains is created by two independent sources: (1) Vin and (2) Vref and a magical, ideal, switch connecting either source to the terminal A.

Per each period, T, how much of Q is lost by Vin? Answer: C*(Vin-Vref). Since for the second half, the capacitor is drained of its charges by having potential of 0V across its two terminals, every cycle, Vin must provide same amount of charge Q=C*(Vin-Vref).

Units of current, I, is Coulomb/second. From charge perspectives, Vin looses C*(Vin-Vref) ever "T" time. So the current Vin is providing is I=C*(Vin-Vref)/T. You know T = 1/f. Thus, the Vin source has an equivalent current draw of C*f*(Vin-Vref).

If you wanted to treat this packets of charges being supplied over time as a continuous current for calculations, then one can say the source Vin has equivalent resistance of "1/(C*f)" only then you can say I=(Vref-Vin)/R.

I didn't realize how difficult it'd be to convey this notion without a diagram. I am too lazy to draw a picture and do an upload. hope this helps.

-gujaratibhai

thank you for your detailed explanation,gujaratibhai.