50 Amps?

Thread Starter

twister007

Joined Feb 29, 2012
81
If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
 

MrSalts

Joined Apr 2, 2020
2,767
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.
 

WBahn

Joined Mar 31, 2012
29,976
Could you link to a data sheet for that transistor. A Google search didn't come up with any results for me.

Without the data sheet, there's no way to know (other than trial-by-fire) what it can and can't take.

For more general answer, are there transistors that can handle those kinds of currents? The answer is most definitely yes. Depending on how you operate them, the biggest factor is usually how good your heat sinking is.
 

dl324

Joined Mar 30, 2015
16,839
If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
It depends. This device came up when I searched for the part number you provided:
1672702830733.png

Under the right conditions, it will handle 50A continuous current.
 

sagor

Joined Mar 10, 2019
903
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.
 

Beau Schwabe

Joined Nov 7, 2019
155
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.
 

dcbingaman

Joined Jun 30, 2021
1,065
Not without a large heatsink:

1672705619136.png
Dissipation max is 50A^2*0.022ohms=55 Watts
No heatsink: 62.5C/W*55W=3,437C (Forget it!)
 
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dcbingaman

Joined Jun 30, 2021
1,065
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.
Probably because the forward resistance of that thing according to the datasheet is around 1mOhm. That is 30^2*0.001 =
0.9W. The FET is much higher forward resistance so not a good comparison.
 

Thread Starter

twister007

Joined Feb 29, 2012
81
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.
[
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.
Tthanks. How did you arrive at that temperature and amps?
 

dcbingaman

Joined Jun 30, 2021
1,065
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=
60W at 100C, if I did that correctly. But again that has to assume a heatsink because without it you have 62.5C per watt.
 

strantor

Joined Oct 3, 2010
6,782
If I attach welding cables to the skinny legs of this FPQ50N0 Transistor, and bias it on, will it really pass 50 amps?
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.
 

dcbingaman

Joined Jun 30, 2021
1,065
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.
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:
 

Attachments

WBahn

Joined Mar 31, 2012
29,976
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.
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.

The max current rating and the max power rating (and derating) are two different, but somewhat related things. There are several factors that place limits on a parameter and the one that matters is the one that places the most stringent limit.

One limit is Junction Temperature, Tj. It is stated as having a max temperature of 175°C. If the case temperature is held at 25°C, that that means that the max temp difference between the junction and the case is 150°C. At 1.24 °C/W, that means that the device is limited to 121 W (call it the 120 W in the data sheet).

As for the 3.75 W given for an ambient temperature of 25 °C, I'm not quite sure what the assumptions are there -- that asterisk might tell us. If we assume no heat sink, then the thermal resistance is 62.5 °C/W (unless the case in SamR's datasheet is different than the other datasheet above) and 150 °C rise would occur at just 2.4 W.

If the on-resistance is at the max shown of 0.022 Ω, then the current would have to be ~74 A to get to 120 W. However, that resistance is spec'ed at a current of 25 A and a JUNCTION temperature of 25 °C.

As the junction temperature increases, so does the on-resistance.

1672713680158.png
You can see that it will me more than double at a junction temperature of 175 °C.

To get 120 W at 50 A, the on-resistance would need to be 48 mΩ, which is a factor of 2.18 times the max on-resistance specified at 25 °C, which agrees very closely with the chart above.
 

strantor

Joined Oct 3, 2010
6,782
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:
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 did simplify things a bit; There is more to it than just the resistance of the legs. If you do the math on the cross sectional area of the legs of a TO-220 package it will probably indicate that 50A or 75A should not cause enough temperature rise to melt the legs. But all of the heat causing melting of the legs isn't coming just from the legs. The heat from inside the IC propagates out and down the legs. So a TO-220 package transistor having the theoretical infinite heatsink they base the datasheet values on, probably wouldn't melt off at the legs like I described.

But that heatsink doesn't exist, a lot of heat is generated in the IC, and that heat travels to the legs where it adds to the heat that the legs are also generating, and as the legs heat up their resistance goes up, causing them to drop more voltage and dissipate more power, generate more heat, and so forth. Thermal runaway, melty legs.

Since your hall sensors aren't doing much (as opposed to a transistor) the IC part isn't going to dissipate near as much power, heat up near as much, or contribute much if any heat to the package or legs.
 
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Thread Starter

twister007

Joined Feb 29, 2012
81
I think he got that from this graph on the datasheet:
View attachment 284355
Thanks. I see said the blind man. Ha! I should learn to read data sheets better!
If I have 6 transistors heat sinked to the sides of
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?
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.
 
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