Converting Temperature Coefficients to ppm/(deg C) Made Simple

[KillerZ]

I am not sure if this is the right place to ask this. I am wondering how to convert to ppm/(deg C). I have a temperature coefficient I found and I am trying to convert the number to ppm/(deg C). The number I have is 0.000000059 A/(deg C) so do I just multiply it by 10^6 which gives me 0.059 ppm/(deg C). Is this correct?

 

[berkeman]

I am not sure if this is the right place to ask this. I am wondering how to convert to ppm/(deg C). I have a temperature coefficient I found and I am trying to convert the number to ppm/(deg C). The number I have is 0.000000059 A/(deg C) so do I just multiply it by 10^6 which gives me 0.059 ppm/(deg C). Is this correct?

That looks right. What's the "A" mean?

 

[KillerZ]

The A is amps.

 

[berkeman]

The A is amps.

That's different then. You need to reference this change to the nominal value, in order to reference it as ppm.

Like, if the nominal current is 1A, and that is the tempco, then your previous calculation is right. But if the nominal current is 2A, then the tempco in ppm is less.

What is the nominal current that you want to reference, in order to express the tempco as ppm instead of an absolute number? What is the application?

 

[KillerZ]

I have a simple current mirror made of BJTs and the output current needs to be 100uA and I am trying to balance a resistors positive temperature coefficient (1060 ppm) with the negative temperature coefficient of the BJTs to temperature compensate it. I am sure that the fractional temperature coefficient I found of 0.000000059 A/(deg C) is correct. So I guess the nominal current would be the 100uA.

Iref = 100uA
R = 43K

$$\frac{dIref}{dT} = \frac{1}{R}\frac{dR}{dT}(Iref) + \frac{1}{R}\frac{dVbe}{dT}$$
$$\frac{dIref}{dT} = (1060 ppm/deg C)(100uA) + \frac{1}{43K}(-2mV/deg C)$$
$$\frac{dIref}{dT} = 0.000000059 A/(deg C)$$

 

[berkeman]

I have a simple current mirror made of BJTs and the output current needs to be 100uA and I am trying to balance a resistors positive temperature coefficient (1060 ppm) with the negative temperature coefficient of the BJTs to temperature compensate it. I am sure that the fractional temperature coefficient I found of 0.000000059 A/(deg C) is correct. So I guess the nominal current would be the 100uA.

"fractional temperature coefficient I found of 0.000000059 A/(deg C)"

Where did you find this? It seems very low to me, but maybe I'm not calibrated yet.

Also, your current mirror is missing something important (well, two matching things). I would think it would have a much worse tempco without the missing bits...

 

[KillerZ]

I updated the previous post but it must have happened after you replied:

Iref = 100uA
R = 43K

$$\frac{dIref}{dT} = \frac{1}{R}\frac{dR}{dT}(Iref) + \frac{1}{R}\frac{dVbe}{dT}$$
$$\frac{dIref}{dT} = (1060 ppm/deg C)(100uA) + \frac{1}{43K}(-2mV/deg C)$$
$$\frac{dIref}{dT} = 0.000000059 A/(deg C)$$

I think I might have figured it out. This value 0.000000059 A/(deg C) needs to be divided by 100uA then multiplied by 10^6 which would give me a temperature coefficient of 594.88 ppm/deg C. 0.000000059 A/(deg C) is only dIref/dT. Because temp co of a resistor is (dR/dT)(1/R)(10^6) ppm/deg C then (dIref/dT)(1/Iout)(10^6) ppm/deg C.