If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
It really depends how much resistance is in the load that you are welding, the resistance of the cable and the voltage you are supplying. Also, it could for a while, depending how much heat you are able to extract from the MOSFET to keep it cool.If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
Probably because the forward resistance of that thing according to the datasheet is around 1mOhm. That is 30^2*0.001 =I thought the same thing with this 50 Amp current sensor in a SOIC 8 package ... MLX91221KDC-ABR-050-SP
Were pushing 30 Amps through this thing at 240VAC (7200 Watts) and it's not even warm.
I thought the same thing with this 50 Amp current sensor in a SOIC 8 package ... MLX91221KDC-ABR-050-SP
Were pushing 30 Amps through this thing at 240VAC (7200 Watts) and it's not even warm.
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Tthanks. How did you arrive at that temperature and amps?That 50A is an "Absolute Maximum", which can only be met under ideal, and at time theoretical, conditions. That means keeping the device junction temperature below the maximum threshold. If you can do that, then it will pass 50A. Otherwise, it will overheat and destroy itself.
Notice that at 100 deg C, the current capability is down to 35.4A already.
I think that assumes a 100% efficient heat sink attached to it because the thermal data is 1.24C per watt which is the same as the number you have here being 0.8W/C = 1.25C/W but that is the junction to case, not junction to ambient.Yes, but...
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Well you might have a point. I think I am misunderstanding that data with 'derating about 25C I think it means you can run 120W but if above you must take the 120W and subtract (T-25C)*0.8W per c, thus at 100C that would be 120W - (100C-25C)*0.8W/C=Yes, but...
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That is strange, the datasheet I have for that part does not have the first entry you are showing (3.75W). What is the * on that?Yes, but...
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Let me guess... you're looking at these tiny toothpick legs of the TO-220 package and comparing them to the gauge of wire that would typically be prescribed by NEC or IEC or other regulatory bodies for 50A?If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
I see what you are saying. I wonder why Hall effect current sensors like the following state they can handle such high currents? The 16pin device uses 4 pins for in and 4 more for out and it is in a SOIC package?Let me guess... you're looking at these tiny toothpick legs of the TO-220 package and comparing them to the gauge of wire that would typically be prescribed by NEC or IEC or other regulatory bodies for 50A?
Well, yeah, you're right. Sorta.
Keep in mind that the reason why wire "must" be so thick to carry such current is because only very little temperature rise is allowed inside walls made of flammable materials. And because the insulating materials on wires be rapidly degraded with high temperature. When you're talking about components on a PCB, much higher temperature rise is acceptable. So we can get away with using much smaller current paths.
But not that small. 50A through the tiny legs of a TO-220 package is ridiculous. Take the values you read in datasheets with a grain of salt. They refer to the semiconductor inside the package, and not the package itself. Some packages will carry high current better than others, and even if they have the same semiconductor inside. TO-220 is not the package to choose for high current. I have read that the some tests of the TO-220 package revealed that the legs will melt between 63 and 75A. That means glowing red, then glowing white, then turning to liquid, and that is the actual legs. Not the solder securing the legs. That solder will melt way before the legs themselves melt. Probably well below 50A.
Your suspicion is justified. Don't try to pass 50A through those tiny legs. Get a bigger transistor.
Notice that the ratings are for a CASE temperature, Tc, of 25°C. So this requires that you are use whatever heatsinking and heat transport mechanism is needed in order to keep the case at 25°C.I think that assumes a 100% efficient heat sink attached to it because the thermal data is 1.24C per watt which is the same as the number you have here being 0.8W/C = 1.25C/W but that is the junction to case, not junction to ambient.
Yeah sharing the current across 4 parallel legs instead of just one probably helps (maybe by a factor of 4, not sure of the cross sectional area).I see what you are saying. I wonder why Hall effect current sensors like the following state they can handle such high currents? The 16pin device uses 4 pins for in and 4 more for out and it is in a SOIC package?
MLX91221 Integrated Current Sensor IC:
Thanks. I see said the blind man. Ha! I should learn to read data sheets better!I think he got that from this graph on the datasheet:
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If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50
If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?Tthanks everyone for your replys! I now have a much greater understanding of the data sheets and that transistor.If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?