Diode failure in AC line

[RK NAIR]

Dear friends,

I need one help on this topic, please.
two rectifier diodes 6A4 are connected in parellel and it is connected in seires with a high speed AC motor(22000rpm) to reduce the speed with a switch arrangement.If we put switch in one direction the motor will be directly connected to the line and in other position it will connected through the diodes.

Now these diodes are failing frequently , will you able to suggest a reason and remedy for this problem.Motot input voltage:230V
max current : 5A
Diode no : 6A4 / 400V PIV/ 6A

waiting for a positive responce

RKN

 

[Nigel Goodwin]

Now these diodes are failing frequently , will you able to suggest a reason and remedy for this problem.Motot input voltage:230V
max current : 5A
Diode no : 6A4 / 400V PIV/ 6A

I would suggest 400V PIV is rather too low, use 800V diodes instead. You also don't mention and current balancing resistors?, if you're just paralleling the two diodes one may be passing all the current.

But for a start, get some decent voltage diodes - 230V RMS is 650V peak to peak!.

 

[JohnBrown]

1. As far as I know, it's not a very good idea to use diodes in parallel.
2. Maybe there's some big back-EMF spike type thing damaging your diodes?
3. Does the motor pull 5A when running? If so, what does it take at start-up or stalling speeds?
4. I realise English is not your first language, althought it's a darn sight better than that of most of the native English speakers using this forum, so please take it as a compliment when I bother to set you straight on the way to spell "response" and "parallel"

 

[RK NAIR]

Dear John,

Happy to read your response and thanks for pointed out the spelling error happened while typing.Very rare people will do this.Great.
This motor will consume only 4.5amp if the orifice is completly opened.ie fully loaded.If it is closed it will consume only 2.5-3amps. While starting it will take arround 10-13amps for a moment.
For your information , some of the diodes are not failing under the worst test conditions also.Please consider this for your analysis.
Waiting for your valuable comment.

1. As far as I know, it's not a very good idea to use diodes in parallel.
2. Maybe there's some big back-EMF spike type thing damaging your diodes?
3. Does the motor pull 5A when running? If so, what does it take at start-up or stalling speeds?
4. I realise English is not your first language, althought it's a darn sight better than that of most of the native English speakers using this forum, so please take it as a compliment when I bother to set you straight on the way to spell "response" and "parallel"

 

[pebe]

You state that the motor is AC and runs at 22000rpm. So it's not the normal squirrel cage induction motor. Is there perhaps something about the type of motor that is causing the failures? Do the diodes fail short or open circuit?

 

[Nigel Goodwin]

You state that the motor is AC and runs at 22000rpm. So it's not the normal squirrel cage induction motor. Is there perhaps something about the type of motor that is causing the failures? Do the diodes fail short or open circuit?

Everybody seems to be ignoring my previous post, where I pointed out his diodes are too low a voltage for 230V mains?.

400V PIV diodes are too low!, they are bound to fail! - use 800V ones.

 

[Oznog]

Now is that 5 amp "max current" rms or peak? If it's rms you're exceeding 6 amps at times.

As a diode heats up, its voltage drop decreases. So the warmer diode may turn on at .65V and the cold one won't turn on till .75 volts. Thus whichever diode turns on first warms up and always switches on before the other, bears a disproportionately higher amount of current, which only makes it warmer still. That's why you can't share current between diodes without resistors which creates a large enough voltage drop to make the forward voltage variations due to heat and just mfg differences inconsequential. Unfortunately, they also make some heat, reduce efficiency, and add to the output impedance of the supply.

So it's really, really a good idea to start with a diode with the appropriate rating rather than trying to parallel diodes. Just don't do it, there's almost always another way.

 

[audioguru]

Hi Oznog,
Good point. Thermal runaway favours the "best" diode of paralleled ones and thus makes it even more of a hog until its eating of too many amps kills it.

Hi Nigel,
I can't find the datasheet for the diode but perhaps it has a sharp avalanche characteristic (unlike an ordinary diode that just leaks a little more), and due to its voltage rating that is too low, creates a lot of heat.

So forward current is killing some diodes and reverse voltage the others.
What a poor choice of a diode.

 

[RK NAIR]

motor type

This is not a motor which u stated down, Its a AC High speed conduction motor with carbon brush .failed diodes are found to be in open condition.

You state that the motor is AC and runs at 22000rpm. So it's not the normal squirrel cage induction motor. Is there perhaps something about the type of motor that is causing the failures? Do the diodes fail short or open circuit?

[/b]

 

[RK NAIR]

Now these diodes are failing frequently , will you able to suggest a reason and remedy for this problem.Motot input voltage:230V
max current : 5A
Diode no : 6A4 / 400V PIV/ 6A

I would suggest 400V PIV is rather too low, use 800V diodes instead. You also don't mention and current balancing resistors?, if you're just paralleling the two diodes one may be passing all the current.

But for a start, get some decent voltage diodes - 230V RMS is 650V peak to peak!.

Hi Niegel,

As u said ,if we use 6A8 or 6A10 will solve the problem?
Please let me know the value of the current balancing diode which we have to use in series with this.

 

[RK NAIR]

Hi OZONG,

This 5Amp is an RMS reading of the ammeter when the motor is fully loaded.I dont know what will be P-P current. There are two dides in parellel so the rating will be 12Amp , am i wright ?

please let me know your views and ideas on this topic please.

Now is that 5 amp "max current" rms or peak? If it's rms you're exceeding 6 amps at times.

As a diode heats up, its voltage drop decreases. So the warmer diode may turn on at .65V and the cold one won't turn on till .75 volts. Thus whichever diode turns on first warms up and always switches on before the other, bears a disproportionately higher amount of current, which only makes it warmer still. That's why you can't share current between diodes without resistors which creates a large enough voltage drop to make the forward voltage variations due to heat and just mfg differences inconsequential. Unfortunately, they also make some heat, reduce efficiency, and add to the output impedance of the supply.

So it's really, really a good idea to start with a diode with the appropriate rating rather than trying to parallel diodes. Just don't do it, there's almost always another way.

 

[Nigel Goodwin]

As u said ,if we use 6A8 or 6A10 will solve the problem?
Please let me know the value of the current balancing diode which we have to use in series with this.

Using 800V diodes (or higher) will probably solve your problem, but it's good practice to have a resistor in series with each one to ensure equal current sharing. A 0.1 or 0.22 ohm 3W or 5W should be fine.

Paralleled diodes are commonly used (without balancing resistors) in much of the cheap far eastern domestic electronics equipment.

 

[RK NAIR]

As u said ,if we use 6A8 or 6A10 will solve the problem?
Please let me know the value of the current balancing diode which we have to use in series with this.

Using 800V diodes (or higher) will probably solve your problem, but it's good practice to have a resistor in series with each one to ensure equal current sharing. A 0.1 or 0.22 ohm 3W or 5W should be fine.

Paralleled diodes are commonly used (without balancing resistors) in much of the cheap far eastern domestic electronics equipment.

You mean .1ohm/0.22ohm low value resistor with 3/5 watt
pls confirm

 

[samcheetah]

yes thats what he means. the low value is because there shouldnt be much effect of the resistors. and the high wattage is because it will be carrying alot of current. hence power dissipation will be high.

i hope that helps

 

[Nigel Goodwin]

yes thats what he means. the low value is because there shouldnt be much effect of the resistors. and the high wattage is because it will be carrying alot of current. hence power dissipation will be high.

Yes, thank you samcheetah - I've been busy 8)