Power Dissipation across NTC

Can anyone help me, how i can calculate power dissipation across NTC?

hi,
The dissipation will depend upon the actual resistance of the NTC 'thermistor' at a known temperature of the NTC.
Most NTC's do not a have linear resistance versus temperature response.

The power dissipation is W= V * I or W= V^2/R, where V is the voltage across the NTC and R is its resistance.

What is the NTC type.?
E

NTC is 3 letters that mean many things.
Wikipedia lists 8 organizations, 3 telecommunications companies, 3 religions and 3 other things of which one is a Negative Temperature Coefficient resistor.

Hi,

I would have interpreted NTC to mean Negative Temperature Coefficient (Thermistor), same as Eric did. If that is what it really is then the following information should help...

The resistance of a NTC thermistor can be calculated from:
Rx=R1*e^((B*(T1-Tc))/((K+Tc)*(T1+K)))

or in Latex when that is working:
[LATEX]Rx=R1\,{e}^{\,\frac{B\,\left( T1-Tc\right) }{\left( K+Tc\right) \,\left( T1+K\right) }}[/LATEX]

where
Rx is the resistance in Ohms at a known thermistor body temperature Tc in degrees C,
R1 is the resistance in Ohms measured at the temperature T1 in degrees C,
B is the thermistor 'Beta' coefficient, and
K is the difference between the Celsius temperature scale and the Kelvin temperature scale (taken to be the constant 273)

The values B, T1, and R1 are found on the manufacturer's data sheet.

Once the resistance Rx is calculated, the power is easily calculated knowing the voltage across the thermistor:
P=V^2/Rx

or the current through the thermistor:
P=I^2*Rx

For example, say we have a thermistor with a Beta coefficient equal to 4100 at a temperature of 25 degrees C. That makes B=4100 and T1=25. And the resistance at 25 degrees C is 10k. We want to know the power dissipation with 2.5 volts across it at the temperature of 40.8 degrees C. K is always equal to 273.

First we calculate the resistance from the above:
Rx=R1*e^((B*(T1-Tc))/((K+Tc)*(T1+K)))

and plugging in the manufacturer's values and the value for the constant K we have:
Rx=R1*e^((B*(T1-Tc))/((K+Tc)*(T1+K)))
Rx=10000*e^((4100*(25-Tc))/((273+Tc)*(25+273)))

and we can simplify that so we have a formula for this thermistor, or just plug
in the temperature of 40.8 degrees C we get:
Rx=10000*e^((4100*(25-40.8))/((273+40.8)*(25+273)))

and calculating that out we get:
Rx=5002 Ohms approximately.

Now that we have Rx, we can calculate the power knowing V=2.5 from:
P=V^2/Rx

so we have:
P=2.5^2/5002

which gives us:
P=0.00125 watts approximately.

So we have a little more than a milliwatt being dissipated in this thermistor.

In some cases the voltage applied and resulting power dissipation will actually raise the temperature of the thermistor by some amount so that is why the manufacturer usually suggests some maximum power dissipation for more accurate temperature measurement.

hi Al,
I did assume a thermistor with a NTC characteristic.

This Tool will give the OP an idea for the voltage across his 'thermistor' at a given temperature, its easy to use P= V*I or V^2/R
https://www.electro-tech-online.com/tools/ThermPlotV3.php

E
EDIT: I will upgrade the Tool one day to show Power dissipation.