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We have the following circuit

enter image description here

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?

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    \$\begingroup\$ your secondary SLEWRATE is 2,000 volts per second. To swing 0.8 volts will need 0.8/2000 = 0.0004 seconds approximately. \$\endgroup\$
    – analogsystemsrf
    Jun 27, 2019 at 15:36

2 Answers 2

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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.

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  • \$\begingroup\$ Mr Sinno... In real world , we say Vf=0.8V @ If= (Vs-0.8)/R which is more than the vanish point or near 0, so the time delay to reach this voltage at some phase is actually a tad larger. Vf=0.8 is 1A for a typ. 1N4005 power diode. but for now neglect this. \$\endgroup\$
    – Tony Stewart EE75
    Jun 27, 2019 at 15:59
  • \$\begingroup\$ Mr Roughbeast. Now for an expert question. ( for fun) What transformer parameters (L=?, coupling coefficient=?) with at least 1 sig.fig) are necessary to achieve Vf=0.8A say for 1kOhm load and what must be the PIV rating.? (min) \$\endgroup\$
    – Tony Stewart EE75
    Jun 27, 2019 at 16:20
  • \$\begingroup\$ For Uber Analog EE's, what is the diode bulk resistance, Rs for that PIV diode to conduct enough current to reach 0.8V ? to at least 1 sig.fig. \$\endgroup\$
    – Tony Stewart EE75
    Jun 27, 2019 at 16:32
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    \$\begingroup\$ Uber Answer: No solution as Diodes do not have a Peak Inverse Voltage rating of >=10kV with a forward Vf=0.8. It would have to be at least 10 diodes in 1 thus Vf >8V so not possible to have Vf=0.8V with Vr=10kV=PIV rating \$\endgroup\$
    – Tony Stewart EE75
    Jun 27, 2019 at 16:46
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\$10sin(\omega t)=0.8\$

\$sin(\omega t)=0.08\$

\$arcsin(0.08)=\omega t=0.080086\:rad\$

\$\therefore \:t=0.080086/200=0.40043\: ms\:\approx 0.4 \:ms\$

Note: \$sin(\theta)\approx\theta\:\$ for small values of \$\theta\$, where \$\theta\$ is in radians. So we could get to the approximate answer without doing the \$arcsin\$.

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    \$\begingroup\$ good use of Trigonometry and Mathjax +1 Care to try my Uber EE challenge or the other? \$\endgroup\$
    – Tony Stewart EE75
    Jun 27, 2019 at 16:34

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