How to figure power dissipation of this shunt?

rg12

100 kW
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I was thinking of using an Allegro 150A current sensor (instead of a shunt) for measuring current in a circuit.
The rated resistance of the component is 100uOhm (I guess it's milliohms).
How can I know how many watts can it dissipate?
It should measure up to 120V and up to 150A so how can I know the power dissipation and potential heat considering all of these numbers?
Also, how low is 100uOhm compared to the resistance of those big metal shunts?

Also, same question about this linear step down regulator:
http://www.analog.com/media/en/technical-documentation/data-sheets/7138f.pdf
At 120V and lets say 300mA of circuit power consumption, how hot can it get? I was told that if the circuit draws 5V it would need to dissipate 115V at 0.3A which is like 34W through that tiny component.
If thats true then it seems phisically impossible and if thats the case then how come it's rated for 140V and 400mA?
 
There's an efficiency graph on the datasheet for the regulator. Looks like it's a bit under 90%, so figure 10% of what goes in gets thrown off as heat and the rest goes to the output.

To figure out dissipation of the 100uOhm load use Ohms law. u means micro, so 100 millionths of an Ohm. In reality it's probably less resistance than you'd get through an equal length of copper wire. I'd do the math but I'm tired :)
 
Punx0r said:
P = I^2R

P = 100A^2 x 0.0001 ohms

P = 10000 x 0.0001

P = 1W

Thanks alot, so that seems reasonable...
Any ideas about the current sensor?
 
Oh sorry, I mixed the question about the current sensor with the one about the linear regulator.
The linear regulator is rated 140V 400mA which means that at lets say 120V input while the circuit is consuming 5V there is 115V time let's say 300mA consumption which ends up to be 34.5W which is an insane amount of heat for that tiny regulator.
How come it's rated so high if it's not physically possible to dissipate such heat?
 
rg12 said:
Oh sorry, I mixed the question about the current sensor with the one about the linear regulator.
The linear regulator is rated 140V 400mA which means that at lets say 120V input while the circuit is consuming 5V there is 115V time let's say 300mA consumption which ends up to be 34.5W which is an insane amount of heat for that tiny regulator.
How come it's rated so high if it's not physically possible to dissipate such heat?

The regulator that you linked is not a linear regulator, although it is made by the "Linear" company so that can be confusing. It's a switching converter, so it's much more efficient than a linear type. Check out the "POWER LOSS" curves on the first page. For 5V @ 300mA output, with an input of 140V, the power loss is under 1W.
 
Addy said:
rg12 said:
Oh sorry, I mixed the question about the current sensor with the one about the linear regulator.
The linear regulator is rated 140V 400mA which means that at lets say 120V input while the circuit is consuming 5V there is 115V time let's say 300mA consumption which ends up to be 34.5W which is an insane amount of heat for that tiny regulator.
How come it's rated so high if it's not physically possible to dissipate such heat?

The regulator that you linked is not a linear regulator, although it is made by the "Linear" company so that can be confusing. It's a switching converter, so it's much more efficient than a linear type. Check out the "POWER LOSS" curves on the first page. For 5V @ 300mA output, with an input of 140V, the power loss is under 1W.

Oh thats good to know, but for some reason I have a problem that it randomly fails and the unit doesn't turn on until I change the component and it is expensive as f!
 
rg12 said:
Addy said:
The regulator that you linked is not a linear regulator, although it is made by the "Linear" company so that can be confusing. It's a switching converter, so it's much more efficient than a linear type. Check out the "POWER LOSS" curves on the first page. For 5V @ 300mA output, with an input of 140V, the power loss is under 1W.

Oh thats good to know, but for some reason I have a problem that it randomly fails and the unit doesn't turn on until I change the component and it is expensive as f!

What component are you changing? The IC chip?

If the IC is failing, maybe it's still overheating. For a tiny IC, 1W can still be quite a lot of power to dissipate, hopefully the PCB has a good layout for this.
 
Addy said:
rg12 said:
Addy said:
The regulator that you linked is not a linear regulator, although it is made by the "Linear" company so that can be confusing. It's a switching converter, so it's much more efficient than a linear type. Check out the "POWER LOSS" curves on the first page. For 5V @ 300mA output, with an input of 140V, the power loss is under 1W.

Oh thats good to know, but for some reason I have a problem that it randomly fails and the unit doesn't turn on until I change the component and it is expensive as f!

What component are you changing? The IC chip?

If the IC is failing, maybe it's still overheating. For a tiny IC, 1W can still be quite a lot of power to dissipate, hopefully the PCB has a good layout for this.

The regulator.
Just tested again, it works well with 40-50v but when trying 60v and above the unit dies and doesn't work anymore even with low voltages.
 
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