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Temperature measurement in LTspice

ericgibbs

Well-Known Member
Most Helpful Member
#2
hi,
Press the 'S' key on keyboard.
Enter this line into text box .step temp 0 50 10

This will step temp from 0C thru 50C in 10 C steps

OK.?
E

To measure the wattage dissipation of a component , place the cursor on the components and press and hold down the ALT key, a thermometer cursor should appear
Read the Watts on the bottom screen screen status text
 

alec_t

Well-Known Member
Most Helpful Member
#5
Write .temp 100
 

alec_t

Well-Known Member
Most Helpful Member
#9
Why would you expect a large difference?
 

alec_t

Well-Known Member
Most Helpful Member
#11
Depends what the component is and how it is used. An NTC thermistor, for example, might dissipate less power when hot. Semiconductor junctions drop less voltage when hot.
 

kubeek

Well-Known Member
#13
You still don´t say why do you expect the losses to change? For example mosfets will have higher Rdson, but diodes will have lower forward drop. So it really depends on the circuit in question.
 

ronsimpson

Well-Known Member
Most Helpful Member
#14
So it really depends on the circuit in question.
A resistor will have the same loss.
-----------------------
In the real world we put the power supply in a hot room and see if it survives.
In SPICE we can not have a hot room. We can only force the inside of all parts to be at 25C or 100C or .......
 

crutschow

Well-Known Member
Most Helpful Member
#15
I think y0u are confusing heat loss with a device's power dissipation.
The power dissipation is determined by the power loss in the device and does not vary with ambient temperature.
The temperature of the device increases until the power dissipation to its surroundings equals the internal power dissipation of the device, as determined by the thermal resistance to ambient.

Spice does not determine that temperature rise.
You can determine the rise by measuring the device wattage (dissipation) in Spice, and then calculating the rise from the device's thermal resistance to ambient, which is determined by the device's package, how its mounted on a circuit board, and whether it has a heatsink.
 

ronsimpson

Well-Known Member
Most Helpful Member
#17
just want to see power dissipation (w losses).
For example mosfets will have higher Rdson, but diodes will have lower forward drop. So it really depends on the circuit in question.
Probably the difference in power loss from 25C to 100C is small.

----------------------------------------------------------
I don't have time to look into this but you are not having 3k watts in D2. We are doing something wrong.
 

crutschow

Well-Known Member
Most Helpful Member
#18
If I force all parts to be 100 C, then I just want to see power dissipation (w losses).
Spice cannot tell you that.
It can only tell you the dissipation due to the circuit operation.

What do you mean (w losses)?
What losses?

To determine what you want, you would need to know the thermal resistance of the device, junction to ambient.
 

crutschow

Well-Known Member
Most Helpful Member
#20
I mean the power dissipation/losses.
Okay, I'm confused about what you are trying to do.
You cannot learn anything about power dissipation by changing the temperature of the device in simulation.
Power dissipation is a function of the power dissipated in the device by the circuit voltages and currents.
That's what causes the junction temperature to change.
And the only way you can find the actual junction temperature is to know the devices power dissipation and it's thermal resistance to ambient.
Is that clear?

Okay, lets work backwards as you seem intent on doing.
Let's say the junction temperature of a device is 100°C at a room ambient of 25°C, with a junction to ambient thermal resistance of 10°C/W.
That means the device is dissipating (100-25)/10 = 7.5W.
Note you can only make that calculation if you know the device's thermal resistance. and that's not part of the Spice models.
 

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