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anonymous
 3 years ago
Can anyone give a hand with H3P4: DIODE LIMITER, trying to find the current through Diode 1 and 2 in this circuit 
https://6002x.mitx.mit.edu/static/circuits/diodelimiter.gif
where R = 5.6k
V1 = 3.25v
V2 = 3v
Peak voltage of AC source is 7.5V
Any help on how to do the thevenin method would be great
anonymous
 3 years ago
Can anyone give a hand with H3P4: DIODE LIMITER, trying to find the current through Diode 1 and 2 in this circuit  https://6002x.mitx.mit.edu/static/circuits/diodelimiter.gif where R = 5.6k V1 = 3.25v V2 = 3v Peak voltage of AC source is 7.5V Any help on how to do the thevenin method would be great

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think the thevenin method is not required to solve the problem, as we are taking the diodes to be ideal. Just applying nodal analysis would give the answer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, dany's comment is right but someone can ask you to apply Thevenin simplification anyway. I'll try to explain what I do by words (at this moment, I have not any media to draw the circuit). Vs and the two R can be thought as forming a circuit connected to the D1V1 and D2V2 "branches". So VsRR is simplified by Thevenin as Vs/2 with a R/2 in series and connected to the two mentioned branches. If you now draw the new circuit, the analysis is strongly simplified: you have not any node! during the two positive and negative waves of Vs.  During positive values of Vs (over the V1 value), D1 is ON.  During negative values of Vs (over the V2 value), D2 is ON. Currents in (ideal) diodes are:  D1: 2x(VsV1)/R  D2: 2x(VsV2)/R. Obviously, only when Vs>V1 and Vs>V2, respectively. 0A when Vs value is between the V2 and v1 values.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Diode limiter act like a clipper:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Part 1 D1 and D2 will clip the voltage. So, the voltage at the "input" of D1 cannot be higher than V1, and the voltage at the “output” of D2 cannot be lower than V2. Therefore, the maximum positive voltage that can appear at v is 2.25 Part 2 Similarly, the maximum negative voltage that can appear at v is ‐2.5V. Part 3 Maximum current through D1: When the AC voltage source is at +8V and v = 2.25V. Current through D1 is the current from the AC source minus the current going through resistor R: (VS – v) / R – v/R = 0.00043. Part 4 Maximum current through D2: v = ‐2.5 and AC voltage source is at ‐8V. Current through D2 plus current through the resistor R is the current flowing into the AC source. Thus, current through D2 is current through the AC source minus current through the resistor R: (v – Vs ) / R + v/R = 0.00037.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in japanies style:)hahahahahha
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