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confused about full bridge rectifier

AdhiWay

Mar 15, 2022
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Hi everyone, i have another problem, and i appreciate your help in my previous post it was really useful, so thank you all,

i have a problem about full bridge rectifier circuit, i builded this circuit several time before and alwayse i faced problem that the diodes begins the burn out when i use 20 or 30 volt power supply without any resistor, so i always ending by giving up,
i use diodes with the following serial number 1N4007, so according to my research these diodes works with hight voltage, so i am really confused i dont know what is the problem,
thank you all for any help, i really appreciate it.

bridge1.jpg bridge1.jpg
 
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Bluejets

Oct 5, 2014
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Assuming 30 volt is the AC input, circuit layout looks correct.
1N4007 diode is 1Amp maximum and will withstand 50v peak voltage.
You are getting very close to your peak rating.(approx 42V)
Is there some reason you need 30V AC ...???
You have no detail on the current that the coil draws.
If that capacitor is as I read it 10,000uf its way too large.
Rule of thumb is around 1,000uF per amp load.

As a side note, get used to drawing a bridge rectifier as a diamond shape, much easier to read.
You can look that up on Google.
 

crutschow

May 7, 2021
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1N4007 diode is 1Amp maximum and will withstand 50v peak voltage.
The 1N4007 is a 1000V diode.
The 1N4001 is the 50V version.

Are you grounding the bridge input supply to the bridge output DC ground?
If so, that will cause a dead short across a couple of the diodes and zap them.
The bridge input must be completely isolated from the bridge output.
 

AdhiWay

Mar 15, 2022
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Assuming 30 volt is the AC input, circuit layout looks correct.
1N4007 diode is 1Amp maximum and will withstand 50v peak voltage.
You are getting very close to your peak rating.(approx 42V)
Is there some reason you need 30V AC ...???
You have no detail on the current that the coil draws.
If that capacitor is as I read it 10,000uf its way too large.
Rule of thumb is around 1,000uF per amp load.

As a side note, get used to drawing a bridge rectifier as a diamond shape, much easier to read.
You can look that up on Google.


thank you for replying,
yes, its 30v AC,
the capacitor is 1000 uf,
and the coil is a electromagnetic of an old 220v / 380v induction motor i have, it has a electromagnetic that moves a disk to connect it to the main rotor,
the circuit is working fine i teste it with the motor, the problem is the diodes started burning out imidiately, and its not the first time that its happen to me, i always face this problem when i make bridge rectifier using 20 or 30 volt AC to DC, i dont have this problem in lower voltage,
thanks again for your help.
 

AdhiWay

Mar 15, 2022
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The 1N4007 is a 1000V diode.
The 1N4001 is the 50V version.

Are you grounding the bridge input supply to the bridge output DC ground?
If so, that will cause a dead short across a couple of the diodes and zap them.
The bridge input must be completely isolated from the bridge output.

thanks for repying,
no, the bridge output is connected to the coil
 

Audioguru

Sep 24, 2016
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Of course the diodes will burn up when the DC current in the coil is too high.
I cropped the schematic and turned it around.
 

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AdhiWay

Mar 15, 2022
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Of course the diodes will burn up when the DC current in the coil is too high.
I cropped the schematic and turned it around.

but they say that those diodes 1N4007 can sustain a peak repetitive reverse voltage of 1000 volts, this is what confuse me
 

Kiwi

Jan 28, 2013
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What current does the coil draw?

What is the resistance of the coil?

1N_4007 diodes are 1000v 1AMP. They will burn out if you try to put more than 1Amp through them.
 

Harald Kapp

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but they say that those diodes 1N4007 can sustain a peak repetitive reverse voltage of 1000 volts, this is what confuse me
It is not the voltage that destroys the diodes, it is the current, that is why you have been sked for the current drawn by the coil.
the coil is a electromagnetic of an old 220v / 380v induction motor
When used with AC the impedance of the coil is much higher than with DC. Therefore when used on AC the current will be low -Y no problem.
When used on DC the impedance of the coil is determined by the resistance only (which you can measure with an Ohmmeter). Accordingly current will be much higher: I = V /R. Measure R and do the math. The current is most likely way above 1 A and therefore destroys the diodes.
 

AdhiWay

Mar 15, 2022
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It is not the voltage that destroys the diodes, it is the current, that is why you have been sked for the current drawn by the coil.

When used with AC the impedance of the coil is much higher than with DC. Therefore when used on AC the current will be low -Y no problem.
When used on DC the impedance of the coil is determined by the resistance only (which you can measure with an Ohmmeter). Accordingly current will be much higher: I = V /R. Measure R and do the math. The current is most likely way above 1 A and therefore destroys the diodes.

thank you for help,
i don't know how much current is going through the coil also i don't know the impedance of the coil, i still don't know exactly how to do calculation about it, i have two shitty multimeters, it doesn't have option to measure resistance of coils, or maybe i still ignorant about that,
i think measuring a resistor is different than measuring a coil resistance, isnt?, i never measured a coil resistance,

the main power supply is a 220v in / 24v out with no other informations about it like amperes......., also there is no informations about the coil,
and the coil is not small and i am kind of sure that the coil work with around 20v up to 50v DC or even more, but i am sure not less than 20v,
and the magnetic field must be strong because it tilts a solid cylinder which its kind of hard to do it even with hands,
and there is two of these magnetic coils with two cylinders and they do the seem job,
i will add more picture of the motor and the coils for more details,
thanks again.
1.jpg 1.jpg 1.jpg transform.jpg transform.jpg 2.jpg 5.jpg 6.jpg 7.jpg 8.jpg 9.jpg 10.jpg 11.jpg 12.jpg 13.jpg
 

Bluejets

Oct 5, 2014
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Label shows tappings 1 and 5 for 240v ac input.
If you are inputting 220v as you say, then you need to change the tappings to suit otherwise output will be different from your expected 24.
Possibly not a big deal but just noted.

Coil resistance just as any other resistance can be measured with a basic multimeter, no fancy gear required.
That would be for dc operation but as mentioned earlier, cannot measure the impedance for ac that way.
You are trying to operate the "coil" (no idea which one ...or two...or whatever) on dc ....why???
If they were built for ac the laminations and core will have normally a different structure.
Actually it looks to me like some form of clutch.

What are you really trying to do ...??
 

crutschow

May 7, 2021
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As BJ noted, resistance is resistance.
You should be able to measure the coil resistance with either of your shitty multimeters.
 

Harald Kapp

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it doesn't have option to measure resistance of coils, or maybe i still ignorant about that,
There is no special "resistance of coils" setting. Use the normal Ohms setting for measuring the resistance.

Why do you want to operate the coil from DC when it is designed for AC operation?
 

Audioguru

Sep 24, 2016
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The resistance of the coil is probably very low since it is simply a piece of wire. Then you can calculate the very high DC current.
You also need to measure the resistance of the meter leads so that it can be subtracted from the resistance measurement of the coil.
 

AdhiWay

Mar 15, 2022
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There is no special "resistance of coils" setting. Use the normal Ohms setting for measuring the resistance.

Why do you want to operate the coil from DC when it is designed for AC operation?

Label shows tappings 1 and 5 for 240v ac input.
If you are inputting 220v as you say, then you need to change the tappings to suit otherwise output will be different from your expected 24.
Possibly not a big deal but just noted.

Coil resistance just as any other resistance can be measured with a basic multimeter, no fancy gear required.
That would be for dc operation but as mentioned earlier, cannot measure the impedance for ac that way.
You are trying to operate the "coil" (no idea which one ...or two...or whatever) on dc ....why???
If they were built for ac the laminations and core will have normally a different structure.
Actually it looks to me like some form of clutch.

What are you really trying to do ...??

As BJ noted, resistance is resistance.
You should be able to measure the coil resistance with either of your shitty multimeters.

The resistance of the coil is probably very low since it is simply a piece of wire. Then you can calculate the very high DC current.
You also need to measure the resistance of the meter leads so that it can be subtracted from the resistance measurement of the coil.


i did tested it directly with the AC output of the transformer 24v, and the result is weak magnet it does not pull the metal cylinder, also the cylinder start vibrating because of the AC alternate i think,

and im not sure on how measure the resistance of coils, so i realy apreciate to hear your advice,
i did some measure with the multimeter by connecting resistor in series with the coil and the result is no diffrence shown in multimiter is like there is no coil connected, or i dont know the correct way,

also i posted another post related to this topic, im thinking of another way to solve this problem, so i realy apreciate if you can give a look to it, thank you
https://www.electronicspoint.com/fo...ge-rectifier-overheating.296952/#post-1827080
 
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