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Ohm's Law taking into account temperature coefficient of resistance

G

Griff

Jan 1, 1970
0
Dear boffins

How does one calculate the current flowing in a resistor if one wishes
to take the temperature coefficient of resistance into account ?

ie we know the Voltage applied to the resistor, we know the TCR of the
resistor, and we know the initial values of resistance and
temperature.

I have looked around to find a straightforward formula for this, but
can't find one anywhere.

Help on this would be much appreciated,

cheers

Griff
 
M

Michael Varney

Jan 1, 1970
0
Griff said:
Dear boffins

How does one calculate the current flowing in a resistor if one wishes
to take the temperature coefficient of resistance into account ?

Non linear effects. Learn about tensors first.

ie we know the Voltage applied to the resistor, we know the TCR of the
resistor, and we know the initial values of resistance and
temperature.

I have looked around to find a straightforward formula for this, but
can't find one anywhere.

Help on this would be much appreciated,

www.google.com "how to use google"
 
J

John Popelish

Jan 1, 1970
0
Griff said:
Dear boffins

How does one calculate the current flowing in a resistor if one wishes
to take the temperature coefficient of resistance into account ?

ie we know the Voltage applied to the resistor, we know the TCR of the
resistor, and we know the initial values of resistance and
temperature.

I have looked around to find a straightforward formula for this, but
can't find one anywhere.

Help on this would be much appreciated,

cheers

Griff

You replace R in the formula, E=I*R, with R(t), which is resistance as
a function of temperature. Any such function you can express is an
approximation of the actual resistance and the kind of function that
is a good enough approximation depends on your accuracy needs, and the
range of temperature. For very narrow ranges, a linear function might
work.

E.G. R(t)=Rbase + (T-Tref)*deltaR/degree
http://scienceworld.wolfram.com/physics/ThermalResistanceCoefficient.html
http://www.utc.edu/Faculty/Tatiana-Allen/Temp.html
page 8 of:
http://www.utc.edu/Faculty/Tatiana-Allen/Temp.html

This approximation can be improved by including terms for higher
powers of temperature (search [polynomial curve fit]).

For for many conductive materials, an exponential approximation is
more accurate, since a %/degree description is closer to reality than
increment/degree.

see page 9 of:
http://www.utc.edu/Faculty/Tatiana-Allen/Temp.html
 
M

Mark Folsom

Jan 1, 1970
0
Griff said:
Dear boffins

How does one calculate the current flowing in a resistor if one wishes
to take the temperature coefficient of resistance into account ?

ie we know the Voltage applied to the resistor, we know the TCR of the
resistor, and we know the initial values of resistance and
temperature.

I have looked around to find a straightforward formula for this, but
can't find one anywhere.

Help on this would be much appreciated,

Are you trying to take account of the resistance heating with a particular
applied voltage? If so, you need to have rate of heat dissipation versus
temperature. When the voltage is first applied, the temperature will change
rapidly, and then it will asymptotically approach an equilibrium
temperature. If you have an approximation for the heat dissipation, then
the best approach is probably to do some kind of numerical approximation by
difference equations.

Mark Folsom

Mark Folsom
 
R

Rich Grise

Jan 1, 1970
0
If you know the TCR, isn't "The resistance at temp. T
equals the resistance at 68F times X" provided? Then
just do that, and use that value as the resistance
in I = E/R.

Cheers!
Rich
 
G

Griff

Jan 1, 1970
0
Thanks to all of you for taking the trouble to reply (even Mr.Varney,
who I suggest might benefit himself from going to Google and typing in
"how to be helpful".)

Rich, I take your point but I now also understand the arguments put
forward by Mark and John, ie that it's not straightforward to
ascertain a temperature "T" to plug into the formula. You know the
ambient temperature when you switch the voltage on, but then: (1) the
temperature of the resistor will go up as the current flows through
it, and (2) the resistance of the resistor will go up or down as a
consequence of the temperature changing, thus varying the current and
bringing rule (1) back into play again, etc.

Clearly it's a judgement call as to how accurately one wants to model
the situation - I'll go away and think about it.

- griffph
 
M

Michael Varney

Jan 1, 1970
0
Griff said:
Thanks to all of you for taking the trouble to reply (even Mr.Varney,
who I suggest might benefit himself from going to Google and typing in
"how to be helpful".)

I taught you how to fish rather than giving you a fish.
I'll go away and think about it.


Good for you.
 
H

hanson

Jan 1, 1970
0
Michael Varney said:
I taught you how to fish rather than giving you a fish.
Wow, you are holier than Jesus........ahahaha...AHAHAHA......
How did you escape from the cross?.....in a bus?.....
hahahaha......ahahanson
 
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