We have the following circuit
V2 = Vs/a;
a = 10;
Vs = 100sin(wt);
w = 200 rad/s;
Each of the diodes is
modeled by a fixed 0.8 V drop when conducting, and by an open circuit when OFF. Find the
first time instant (in ms) when diode D1 starts conducting.
I know that for a diode to be conducting there vL should have a voltage equal or higher than V2.
By KVL I found that vL = 9.2 V and put 10sin(wt) = 9.2, and solved it to get t = 0.335 (ms).
The answer however is 0.4 (ms).
What did I do wrong and how should I proceed?
Best Answer
At the instant in which the diode starts conducting, its current will be vanishingly small. So vL will be zero. Then v2 will be 0.8 volts.
Assume that Vs = 2 v2, since the transformer is overall 1:1, and solve for v2 = 0.8.