Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
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
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.
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
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.