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[SOLVED] Bipolar Sample and hold?

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engrMunna

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Hi, in the figure if the current Ib is on then how come the input signal equal to the output? This is not the normal configuration of a diode bridge I have seen in a rectifier...I mean I can't see any path for the signal from Vin to Vout

 

FvM

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There's a path as long as the signal current magnitude is smaller than the switching current: |Is| < Ib/2.
You can best analyze the circuits small signal behaviour by calculating with differential diode resistances, rd = 26 mV/Id for silicone junction diodes.
 

chuckey

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When Ib flows all the diode act like a short circuit (or low impedance) when there is no current all the diodes are off and the current path from input to output is via the reverse junction of the diodes. The difficulty with the circuit is keeping voltage drops associated with the switching pulses out of the storage capacitor, so a balanced (about earth) current drive is what you want. Also when the switching pulses are there, you don't want your input signal going into the switching current source so this must be high impedance.
Frank
 

engrMunna

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Thanks...your idea of using small signal model helped a lot!
 

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