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Overloading an output , blows transistors why?

duke37

Jan 9, 2011
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Ohms law.
V = I * R or I = V/R or I = V/R
Things can get complicated if values change, you may then need to resort to calculus but this is rarely necessary.

Consider a power supply rated at 10V and 500mA. This means that the supply will give out 10V and will be happy providing up to 500mA output.
Connect a 100R resistor across the output, the current will be 100mA and the supply will hardly notice it.
Connect another 100R across the output, the current will now be 200mA and the supply is beginning to work.
Connect three more 100R resistors, the current is now 500mA and the supply is fully loaded. It will probably get warm from the resistance in its innards but should not break into a sweat.
Connect some more resistors and the supply will be overloaded, it will try to provide enough current to maintain 10V but may not manage it. The excess current will heat the supply and it may fail. A fuse is intended to melt if the current is too high so limiting the heat in more expensive components. Transitors make fast fuses!

If you wish your circuitry to last, do not ask for more than the rated current or supply more than the rated voltage
 

Danny Daviss

Mar 16, 2013
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Connect three more 100R resistors, the current is now 500mA and the supply is fully loaded.

Yes I understand, because when you connect resistors in parallel the resistances goes down so the current increases

Connect some more resistors and the supply will be overloaded, it will try to provide enough current to maintain 10V but may not manage it

The Supply Voltage will drop when overloaded?

The output transistors are "fulling open" when the power supply is overloaded?
 

Danny Daviss

Mar 16, 2013
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There is a formula my manage did, he took the output AC voltage which was 120 volt AC and did a formula to get the DC power supply voltage

How do u get formula?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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The Supply Voltage will drop when overloaded?

it might, but then again, it might not. If it fails, the voltage may go higher, or lower.

The output transistors are "fulling open" when the power supply is overloaded?

Maybe, but probably not.

A full load is essentially the point where something reaches a design limit.

It may be that you've hit the maximum current for the transistor, or the maximum dissipation, or a thermal limit is reached due to the amount of heatsinking provided.

It also means you're likely to be near other limits as well. There is typically no point in designing a circuit to handle vastly more than the maximum load, so whilst there may me one component which actually dictates what the load limit is, others will not be far behind.

A good designer may allow some leeway, but that can be eaten up by (for example) dust on a heatsink, or a faulty fan, or changes in mains voltage, or the aging of components.

Just like the overloaded car, one cannot typically say exactly what will fail, when it will fail, or how it will fail. One can be sure though, that you risk failure.
 

Danny Daviss

Mar 16, 2013
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The IGBT are rated at 150Amps , so the current rating is going higher than 150 amps because it keeps blowing the IGBT transistors gates

120AC , 150 amps = what DC supply voltage? what formula do u use to get the DC supply voltage?

it might, but then again, it might not. If it fails, the voltage may go higher, or lower.

So the DC supply voltage can go higher than its rated for?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Why do you think the current is exceeding 150 amps? This isn't the only reason IGBT's fail.

And I can't tell you how your power supply fails. Is it failing?

At 150A, my guess is that the device fails when it is being switched off. Perhaps you are switching it off too slowly, and perhaps you are switching it off too fast.

Since you're not prepared to show us the circuit, that's about as far as we can go.

I think it's about time to close all of your threads.
 
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