# Why does diode keep conducting even after voltage across it is negative

consider single phase two pulse semiconverter asymmetric configuration.

simulate this circuit – Schematic created using CircuitLab

Note the high value of inductance L1.

Let a be trigger angle of SCR, which is greater than t = arcsin(E/Vs). Now the SCR will work from a to pi-t and maybe beyond if current does not fall below holding current. But would not the diode D2 turn off after pi-t because voltage falls below zero?

Why does using a high value of inductance allows current to flow in D2 beyond pi-t and upto pi+a ? Does the emf E play no role in switching off the diode?

What actually happens after pi-t to pi+a? I assume that D3 cannot start conducting before pi+t and SCR4 cannot start conducting before pi+a. So where does the inductor current go?

• Why do you think it behaves this way? – jwh20 Jan 12 at 10:39
• If you can imagine it, you might simulate it. Add drivers. tinyurl.com/y3nhotco – Tony Stewart EE75 Jan 12 at 10:47
• ANother thing to consider is the diode's "reverse recovery time". The 1N4148 is pretty fast so that's probably not an issue here; if you were to replace it with 1N4001, that would be a different matter. – user_1818839 Jan 12 at 13:58
• I have to wonder whether 100 kilo-Henries is a realistic value for a high-power inductor. Certainly not for a small one. – user253751 Jan 12 at 18:42

The inductor's stored magnetic energy will try and keep current flowing through both itself and D2 by generating sufficient back-emf to forward bias D2. That's pretty much what an inductor will do in this situation.

Why does diode keep conducting even after voltage across it is negative

You might think that it's negative but, until the inductor has pretty-much got rid of most of its stored energy, the diode will be kept forward biased (like it or not).

• I feel this is the correct reason! Thank you! Would you mind checking my proposed output voltage across load: imgur.com/a/Efr8rjP output voltage is shown in green – jeea Jan 12 at 14:30
• I'm not seeing a relation to the schematic in your question. The best way is to simulate the waveform and see the effect of the inductor. – Andy aka Jan 12 at 14:36

From KVL, $$R I + L \frac{dI}{dt} + E = V_s \sin(\omega t) \tag{1}$$ for $$\a/\omega \le t \le (\pi + a)/\omega\$$ and the general solution of eq. 1 is $$I(t) = I_0 e^{-\frac{R}{L}t} + \frac{V_s}{R^2 + (\omega L)^2} (R \sin (\omega t) - \omega L \cos (\omega t)) - \frac{E}{R}. \tag{2}$$ At steady state, $$\I\$$ at the beginning and end of cycles are same, i.e. cyclic condition; $$I\left(\frac{a}{\omega}\right) = I\left(\frac{\pi + a}{\omega}\right) \tag{3}.$$ By solving eq. 3, we can get $$\I_0\$$ and the particular solution of eq. 1.

If $$\I(t) > 0\$$ for any $$\t\$$, the circuit is working in Continuous Conduction Mode (CCM) and all diodes are conducting current all the time. Otherwise, Discontinuous Conduction Mode (DCM) and diodes block the current for a period in a cycle. For DCM, eq. 3 is not applicable and initial condition $$\I(a/\omega)=0\$$ is used instead.

It is a typical case which can be found in a lot of power electronics books.