Inrush current in Diodes

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mivey

Senior Member
Do you mean through the diode? A diode has a peak current rating. There is a slight voltage drop in the diode. Through current & voltage drop = heat. Exceed the inrush rating of the diode and you can burn it up.
 

Hameedulla-Ekhlas

Senior Member
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I have attached the lecture file and it shows the circuit.

Look at point number five it says regarding to surge or inrush current.

In the output there is a capacitor. It says that when we close the switch at positive zero the voltage across the capacitor is zero and in the input there is full current. This current it says is a inrush or surge current and diode should withstand this large current.

I dont know is it normal current when switch closes or surge current. Inrush or surge current is much higher than normal current.

I have not heard regarding this before. If you have any farther information or other please I want to know.
 

mikeames

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Germantown MD
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ELI the ICE MAN. In this circuit current leads coltage. When the switch is close the cap starts to charge droping the full voltage across the cap. As the cap charges the current reduces and voltage increases accross the cap? Is that what you were getting at ? The diode itself does not have an inrush but a voltage drop of .7 volts for si type and .3v for germanium type. In the circuit above you would need a diode that could take the inrush of the cap.
 
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charlie b

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But I would not describe that as a surge or inrush current. I would use those terms for a motor or transformer, something that has a winding that permits very high initial currents but that will "eventually" (i.e., within a fraction of one second) create a counter-emf that reduces the current to "normal" values. When you apply a very small voltage to a diode, smaller than the "forward voltage - on the order of 0.3 to 0.7 volts, no current will flow. Apply more voltage than that, and current will flow. It will continue to drop the same small voltage, but otherwise acts as if it were a wire with no resistance. There is no surge or inrush. Current is either zero, or it is a value that is determined by the rest of the components in the circuit.
 

charlie b

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Look at point number five it says regarding to surge or inrush current.
That inrush is a result of the capacitor charging, and is not related to the diode. What they are saying is that you must figure out how large that inrush can be, and make sure you select a diode that can survive that much current.

 

Hameedulla-Ekhlas

Senior Member
Location
AFG
But I would not describe that as a surge or inrush current. I would use those terms for a motor or transformer, something that has a winding that permits very high initial currents but that will "eventually" (i.e., within a fraction of one second) create a counter-emf that reduces the current to "normal" values. When you apply a very small voltage to a diode, smaller than the "forward voltage - on the order of 0.3 to 0.7 volts, no current will flow. Apply more voltage than that, and current will flow. It will continue to drop the same small voltage, but otherwise acts as if it were a wire with no resistance. There is no surge or inrush. Current is either zero, or it is a value that is determined by the rest of the components in the circuit.

The picture that I have attached and if you look at it. It show at the end that written "short time surge energy that the diode can withstand"

It looks like motor or transformer inrush current because it is for very short time.
 

mivey

Senior Member
I have attached the lecture file and it shows the circuit.

Look at point number five it says regarding to surge or inrush current.

In the output there is a capacitor. It says that when we close the switch at positive zero the voltage across the capacitor is zero and in the input there is full current. This current it says is a inrush or surge current and diode should withstand this large current.

I dont know is it normal current when switch closes or surge current. Inrush or surge current is much higher than normal current.

I have not heard regarding this before. If you have any farther information or other please I want to know.
What they are telling you is that the diode has a maximum amount of peak current that it can handle before it gets damaged. When an inrush current flows through the diode, it can cause it to heat up.

In some circuits you will see a series resister in line with the diodes. This helps reduce the amount of inrush current seen by the diode.
 

mivey

Senior Member
That inrush is a result of the capacitor charging, and is not related to the diode. What they are saying is that you must figure out how large that inrush can be, and make sure you select a diode that can survive that much current.
There you go. H-E, think of it like making sure a conductor is big enough to handle the peak load. The conductor does not create the load or current and neither does the diode. They are just conduits, so to speak.


Add: LEDs are a different story, but that is not the type diode under discussion in the lecture.
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
What they are telling you is that the diode has a maximum amount of peak current that it can handle before it gets damaged. When an inrush current flows through the diode, it can cause it to heat up.

In some circuits you will see a series resister in line with the diodes. This helps reduce the amount of inrush current seen by the diode.

It means when there is only a diode and a capacitor due to inrush current which is caused by a capacitor, the diode should withstand that.

And it also mentioned that in this positive zero which inrush current occurs the voltage across capacitor is zero. Now since there is a zero voltage across the capacitor, does it make any affect on inrush current or inrush current is due to capacitor since there is a zero voltage.
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
There you go. H-E, think of it like making sure a conductor is big enough to handle the peak load. The conductor does not create the load or current and neither does the diode. They are just conduits, so to speak.


Add: LEDs are a different story, but that is not the type diode under discussion in the lecture.

Yes mivey, Capacitor starts charging after positive zero time. I only want to know since we close the switch and in positive zero the voltage is also zero across the capacitor so in this positive zero time why inrush current occurs and why it is not a normal current.
 

mivey

Senior Member
Yes mivey, Capacitor starts charging after positive zero time. I only want to know since we close the switch and in positive zero the voltage is also zero across the capacitor so in this positive zero time why inrush current occurs and why it is not a normal current.
Recall from school that the capacitor charge is an exponential function where the current increases sharply at first then gradually levels out as the capacitor gets charged.

Normally charge current is V/R*e^(-t/RC). Here we have practically zero "R" so the voltage is applied in a huge step across the capacitor which results in a very steep charging current curve.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Yes mivey, Capacitor starts charging after positive zero time. I only want to know since we close the switch and in positive zero the voltage is also zero across the capacitor so in this positive zero time why inrush current occurs and why it is not a normal current.

But are you sure the switch will close at T=0, when the input voltage is zero? You can only know this if you have an electronic switch that is synchnorized to the input voltage.

If it is a manual switch, it may be closed when the input voltage is at a peak. If this is the case, the full input voltage will be applied across the small wire and capacitor resistances. This, and the circuit inductance are the only things limiting current flow. At the instant the switch is closed, the diode has to withstand what is basically a short circuit.

As Mivey mentioned, many designs use small resistances to limit inrush, and to limit the heat the diode has to withstand (the I^2*t ).
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
Recall from school that the capacitor charge is an exponential function where the current increases sharply at first then gradually levels out as the capacitor gets charged.

Normally charge current is V/R*e^(-t/RC). Here we have practically zero "R" so the voltage is applied in a huge step across the capacitor which results in a very steep charging current curve.

Good point
 

dbuckley

Senior Member
Its actually the only important point.

When the switch is closed, the cap is almost a short circuit and the charging current will be limited by the impededence of the source and the series impedence elements. In any real-world half wave rectifier scenario, as you have drawn, it is just the first half-cycle where this is a problem, as by the time the first half-cycle reaches its peak voltage then the cap is fully charged, and then it gets recharged on subsequent half-cycles, but not from zero volts, its just a top-up, so the current flow is smaller.

Diodes are semiconductor devices, so they killer is heat, and the surge rating is based on the amount of heating the diode can withstand before failing. The issue is getting the heat away from the junction, so hot spots do not build up, which is where the failure occurs.
 

Besoeker

Senior Member
Location
UK
Greeting all,


Can any body explain me the surge or inrush current in diodes.
Diodes have a single cycle surge current rating usually denoted by Itsm or Ifsm and a related i^2t value.
These are a measure of the maximum surge current the diode can take and survive. The i^2t value is mostly used for selecting fuses that will protect the diode, with the fuse having a lower total i^2t let through than the diode.

Typically the Itsm is about 10-15 times the maximum average forward current Ifavm.

But note that all of those values depend on junction operating temperature.
If you go to a manufacturer's web site you can get full data sheets. Try for example, Infineon. The D188S is one that I know is on there and that may give you an idea of what some of the parameters are.
 
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