Diode conduction angle calculation

Ian0

Joined Aug 7, 2020
10,244
1uH is a dead short at 100Hz, so it will depend only on the forward voltage of the diode, which will be rather a lot at 15000 Amps.
 

WBahn

Joined Mar 31, 2012
30,290
What is YOUR understanding of what conduction angle is?

Based on that, sketch out what you expect the waveform to look like and where the relevant points are with regards to finding the conduction angle.

Then, discuss how you might go about finding them.

What diode model are you using in your analysis?

As Ian0 has already pointed out, the component values you are using are absurd, but if that's what was given in the problem, you can only shake your head and move on.
 

crutschow

Joined Mar 14, 2008
34,827
As you can see from the simulation, the circuit parameters are completely unrealistic, with a peak current of about 7600 amps.

Are you sure the inductance is not in mH, as 1mH gives a more realistic peak current of 27.3A

1661607666695.png
 

MrAl

Joined Jun 17, 2014
11,688
Dear Team,

May I know how to calculate the conduction angle of the diode in the below circuit.
View attachment 274823
Hello,

Do you mean the conduction angle when the circuit first starts up?
That would probably be the first cycle or even first half cycle.
Or do you mean after steady state has been reached?
There may or may not be a difference.
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
323
I will restate me question and circuit diagram.

In the circuit given below Inductor used is ideal inductor, What is the conduction angle of diode

1661954296788.png
 

WBahn

Joined Mar 31, 2012
30,290
As the source tries to reverse bias the diode, the inductor will produce whatever voltage is required to keep the current flowing. In doing so, energy will be drawn from the inductor and dumped into the supply and dissipated by the diode.
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
323
As the source tries to reverse bias the diode, the inductor will produce whatever voltage is required to keep the current flowing. In doing so, energy will be drawn from the inductor and dumped into the supply and dissipated by the diode.
Thank you..
I can say that the conduction angle is 360 degree.Please correct me if I am wrong
 

WBahn

Joined Mar 31, 2012
30,290
Thank you..
I can say that the conduction angle is 360 degree.Please correct me if I am wrong
No. Imagine a very low frequency waveform such that all of the energy in the inductor is dissipated while the supply is on the negative half cycle. Why (and how) would the diode still be conducting?
 

MrAl

Joined Jun 17, 2014
11,688
I will restate me question and circuit diagram.

In the circuit given below Inductor used is ideal inductor, What is the conduction angle of diode

View attachment 275245

Ok so is the diode "ideal" too? That would mean 0v voltage drop for ANY current level no matter how high, possibly even an infinite current but we dont need to know that.
This would also mean the diode would never dissipate ANY energy.
 

MrAl

Joined Jun 17, 2014
11,688
No. Imagine a very low frequency waveform such that all of the energy in the inductor is dissipated while the supply is on the negative half cycle. Why (and how) would the diode still be conducting?
Hello again,

That is a practical view that would take place anywhere on earth i believe. However, with an ideal diode there is no place for the energy to go except back into the source, and indeed this kind of problem is usually handled by the volt seconds approach. Volt seconds in, same volt seconds out.

So in the ideal case it must be 360 degree right?

Normally there is some voltage drop though then we have to integrate over the portion that is above the diode voltage but with no voltage drop it seems we have to integrate over the entire half cycle, then see it go back into the source in the second half cycle.
 

MrAl

Joined Jun 17, 2014
11,688
Thank you..
I can say that the conduction angle is 360 degree.Please correct me if I am wrong
Hello,

I have to agree with you but only if the diode is ideal also. Even the slightest resistance will eat up some energy and thus the cycle with be very slightly less than 360 degrees.

All you have to do is use an ideal diode with a small resistance like 0.1 Ohms, then let the resistance start to increase. as it gets down to maybe around 1mOhm you will start to see the conduction angle start to increase, and as you go down more the conduction angle will again increase until you reach the limit of 360 degrees.

This is one of the reasons for inductor input filters in rectifier circuits. It keeps the diodes conducting at least for some longer time period which acts to reduce the peak current in the diodes.
 
Last edited:

WBahn

Joined Mar 31, 2012
30,290
Hello again,

That is a practical view that would take place anywhere on earth i believe. However, with an ideal diode there is no place for the energy to go except back into the source, and indeed this kind of problem is usually handled by the volt seconds approach. Volt seconds in, same volt seconds out.

So in the ideal case it must be 360 degree right?

Normally there is some voltage drop though then we have to integrate over the portion that is above the diode voltage but with no voltage drop it seems we have to integrate over the entire half cycle, then see it go back into the source in the second half cycle.
I'll admit that I had in mind the "ideal diode" in which which forward conducting it has a fixed voltage drop across it. But if the model is a true ideal diode with a forward drop of 0 V.

But I'm not positive that it must work out to a 360° conduction angle. It might -- I can see some handwavy arguments to support that. But since the current can't alternate directions it can't be sinusoidal, but the voltage of the supply it. So the waveform is not going to be the nice shifted-sinusoid whose symmetry ensures energy shuttling back and forth between supply and coil. Without the diode, each half cycle of the voltage waveform has an energy-in and energy-out portion in which the current direction is positive for one and negative for the other. Specifically, at the start of positive half of the voltage waveform, the current is at max in the negative direction, something that cannot happen here. I could imagine that it works out to a non-symmetric waveform in which energy is pumped into the the inductor during the positive half-cycle on one energy profile that is pumping energy over the entire half-cycle and then a very different energy profile on the negative half in which that energy is, perhaps, drawn out in less than half the cycle or, perhaps, not all of it is drawn out and there is a net build-up of energy in the inductor from one cycle to the next. Or it may work out that if the voltage waveform has odd symmetry that the energy balance is such that the energy pumped in on the positive voltage waveform is exactly taken out on the negative portion.

It shouldn't be too hard to run the math, but I don't know when I will have time to do it.
 

MrAl

Joined Jun 17, 2014
11,688
I'll admit that I had in mind the "ideal diode" in which which forward conducting it has a fixed voltage drop across it. But if the model is a true ideal diode with a forward drop of 0 V.

But I'm not positive that it must work out to a 360° conduction angle. It might -- I can see some handwavy arguments to support that. But since the current can't alternate directions it can't be sinusoidal, but the voltage of the supply it. So the waveform is not going to be the nice shifted-sinusoid whose symmetry ensures energy shuttling back and forth between supply and coil. Without the diode, each half cycle of the voltage waveform has an energy-in and energy-out portion in which the current direction is positive for one and negative for the other. Specifically, at the start of positive half of the voltage waveform, the current is at max in the negative direction, something that cannot happen here. I could imagine that it works out to a non-symmetric waveform in which energy is pumped into the the inductor during the positive half-cycle on one energy profile that is pumping energy over the entire half-cycle and then a very different energy profile on the negative half in which that energy is, perhaps, drawn out in less than half the cycle or, perhaps, not all of it is drawn out and there is a net build-up of energy in the inductor from one cycle to the next. Or it may work out that if the voltage waveform has odd symmetry that the energy balance is such that the energy pumped in on the positive voltage waveform is exactly taken out on the negative portion.

It shouldn't be too hard to run the math, but I don't know when I will have time to do it.
So the question is, does the inductor lose energy at some point.

The answer is, i believe, yes, but only at the very end of the voltage cycle which would be at 360 degrees.

The reason i believe this is because the inductor will be gaining energy right up until the end of the first half cycle because the voltage is always positive. This means the volt seconds 'into' the inductor will be whatever is dictated by the area of the voltage sine above zero. That would mean the current would increase the whole time during the first have cycle and would mean the current is entirely above zero.
Now at the start of the second half cycle, the voltage goes negative, therefore the inductor acting like a current source now will start to dump current into the source, and the diode still conducts because the voltage across the inductor would go more negative if there was a very very small drop across the diode, so we could say an infinitesimally small voltage drop and the inductor voltage would go more negative than the voltage source and thus conduct into the source.
That said, since the volt seconds is directly related to the area above zero for the first half cycle and the negative half cycle has exactly the same area as the first half, and the inductor accumulates volt seconds during the first half and loses volt seconds during the second half cycle, the inductor must lose all of the volt seconds in the second half cycle.
This would result in a current waveform that is entirely above zero, never going negative. The form would look like:
I(t)=Ipk/2-Ipk*cos(w*t)/2
where Ipk is the peak of the inductor current.
That would be sinusoidal but with an offset of exactly 1/2 of the maximum current. That would mean it could be sinusoidal but always be above zero except for that very very short time it returns to zero an instant before the first half cycle starts again.

One thing is clear i think and that is that the inductor must be either gaining energy, losing energy, or not have any energy. Since i dont believe that it can be void of energy for any time, it must always be conducting unless we want to think of the zero part of the current wave to be non conducting for an infinitesimally short time.
 
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