Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
thats all the circuit is, a high-pass rectifier. I am learning about thevenins therom and think that R1 and R2 are seen in parallel but would like clarification on that please
how would the current in the circuit be calculated?
the way I see it, which could be wrong in parts, is
if F3db = 491 hz
a) Rt = R1//R2 = 690 ohms
b) XC = 1/(2PI*F*C) = 690 ohms
therefore
c) Z=SQRT(XCsqr + Rtsqr) = 976 ohms
d) I=V/Z = 5/976 = 5.1 mA
this isnt the same result I get in the simulator. How does the .6V drop across the diode affect the current? ie when V>.6v the diode conducts but surely the above calculations assume the diode is conducting all the time? or at least not in the circuit?
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.