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# [SOLVED]Resistor ppm calculations clarification

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#### d123

Hi,

I just want to press a pushbutton and see an LED light up, but the innocent-looking 27R resistor is making me do hours of maths...

An LED is 30mA maximum current from 0°C to +25°C, fron there to +82°C it falls in a linear fashion to about 7mA, at +85°C the LED forward current drops immediately to 0mA, according to the datasheet graph of If vs. Tambient. Just trying to avoid unpleasant surprises of frying the LED due to resistance change across temperature.

27R, +-1%, +-100 ppm/°C, voltage across resistor = 800mV, temperature range of operation = 0°C to +85°C.

1) How many calculations do I need to do for this resistor? Is my list complete?

2) Would it really change from 27R to ~44R at +80°C Tamb so the current at 800mV wouldn't be ~27mA but would fall to ~18mA?

3) This takes ages just for one resistor. It's not viable to spend two hours calculating per resistor if a circuit has 20 or 50 resistors - is there a shorter method I can do without using online calculators or some free software that calculates all this for me? I have a very basic understanding of using Excel type spreadsheet programs to add or multiply rows only, that's it for my ability to create formulas in such programs.

A:
1) +-1% @ +25°C = 3 calculations
2) Iout @ +25°C then = 3 calculations
3) PD @ +25°C then = 3 calculations

B:
4) Self-heating @ +25°C for above three +-1% resistance values for +100ppm/°C
5) Change of resistance due to self-heating and Tambient rise above +25°C for the three +-1% resistance values for +100ppm/°C.
6) Change of Iout due to result of calculation 5
7) Change of PD due to result of calculation 5

C:
8) Self-heating @ +25°C for - 100ppm/°C
9) Change of resistance due to self-heating and Tambient rise above +25°C for these three resistance values for - 100ppm°/C
10) Change of I out due to self-heating and Tambient rise above +25°C for those three resistance values and - 100ppm/°C
11) Change of PD due to self-heating and Tambient rise above +25°C for those three resistance values and - 100ppm/°C

D (same steps as B but for Tambient fall below +25°C):
12), 13), 14), 15) Same four calculations/steps, three times each for + 100ppm/°C and Tambient fall below +25°C

E (same steps as C but for Tambient fall below +25°C):
16), 17), 18, 19) Same four calculations/steps, three times each for - 100ppm/°C and Tambient fall below +25°C

II have read rather a lot about ppm resistance change over the past years but never see clarification and examples I fully can be certain of about if +-100ppm/°C means + OR - with temperature rise or it means + with temperature rise and - with temperature fall from nominal room temperature.

Not interested in factoring in Johnson noise or any other causes of resistance shift just yet, only how many calculations I should be doing to get ideal, lowest and highest values for the resistor so as to see how much the 27mA may go up or down across temperature range of operation.

Thanks.

#### FvM

##### Super Moderator
Staff member
No idea what you are calculating. $$\Delta\vartheta$$ of 60 degree refers to a maximal resistance variation of 60*10e-6 = 0.6 %, less than the initial resistor tolerance.

d123

### d123

Points: 2

#### KlausST

##### Super Moderator
Staff member
Hi,

you are talking about SOA of the LED. Do you want to operate it at that high temperatures?
If so: What is the highes temperature you wont to operate the LED? Take this current for your calculations.

No need to care about resistor tolerance and resistor tempco. .. because all other drifts and uncertainties will cause much higher error.
(power supply drift, LED forward voltage drift ...)

Show the complete LED circuit.

ppm means "parts per million"
100ppm/°C = 100/1000000 /°C = 1/10000 /°C = 1/100 %/°C = 0.01%/°C.
Without dedicated information it should be the worst case in both directions: i.e. +/-100ppm/°C

Klaus

### d123

Points: 2

#### d123

Hi,

No idea what you are calculating. $$\Delta\vartheta$$ of 60 degree refers to a maximal resistance variation of 60*10e-6 = 0.6 %, less than the initial resistor tolerance.

Thanks. Something annoying/puzzling me is that when I read about conversion of ppm to percentage, they all say 100ppm/°C = 0.01%.

So, I multiply 60°C temperature rise above ambient by 0.01 and get 0.6. Added to a 2°C self-heating rise, 0.02, I multiply the resistor value by 1.62 - this seems totally wrong to me as 27 to 44 Ohms is outlandish.

Your calculation is what I'd expect. I need to review my approach to/interpretation of the calculation as I'm getting some order of magnitude very wrong.

Adding the ppm change % to 1 or subtracting it from 1 is my mistake, is it?
--- Updated ---

Hi Klaus,

Hi,

you are talking about SOA of the LED. Do you want to operate it at that high temperatures?
If so: What is the highes temperature you wont to operate the LED? Take this current for your calculations.

Not especially, just wanted to be aware of temperature and therefore resistor change in value leading to any possible LED current extreme value in case enclosure heated up to 85°C.

No need to care about resistor tolerance and resistor tempco. .. because all other drifts and uncertainties will cause much higher error.
(power supply drift, LED forward voltage drift ...)

Show the complete LED circuit

No schematic yet, same as torch CCS circuit with OA and NMOS...

ppm means "parts per million"
100ppm/°C = 100/1000000 /°C = 1/10000 /°C = 1/100 %/°C = 0.01%/°C.
Without dedicated information it should be the worst case in both directions: i.e. +/-100ppm/°C

Klaus

Thanks, I seem to not be understanding how to calculate this properly yet despite understanding it's 0.01%*°C rise or fall. Must be adding 1 to the product is wrong method. ...How thick am I at understanding basic formulas and the basic maths, it's embarrassing... Back to the calculator...

Thanks.

Last edited:

#### FvM

##### Super Moderator
Staff member
ppm means part per million. 100 ppm is 0.01 %/K You have apparently calculated the 0.6 % relative change correctly, but added 60 % instead.

d123

### d123

Points: 2

#### d123

Hi,

Great, now I see where I've been getting the ppm calculations wrong - not comprehending it's a percentage, a rise or fall of e.g. 1%, a blindness to a wrong decimal place... Thanks a lot.

I'm going to add a couple of schematics. After seeing the LED datasheet SOA graph, I thought an LM35 would make a good temperature-dependent current-limiter voltage reference for the LED torch circuit (instead of a fixed voltage reference to limit the current), if the LM35 or other temperature sensor output were inverted, then I 'got a bee in my bonnet' about how to emulate the LED SOA curve/graph as precisely as I personally could, so yesterday I did an inverting amplifier that inverted 0V to 0.8V quite well - it went from Vin = 0mV to 806mV and Vout = 826mV to 13mV (102mV in = 724mV out, 202mV in = 624mV out, 402mV in = 425mV out, etc). The simulation for idea 1 shows a worse VOut min. than the breadboard version and DMM measured using the same op amp.

I then realised that 'Idea 1' lost a lot of possible LED current by drawing a line along the graph max. and min points, so for 'Idea 2' drew a straight line to where I guess the slope would begin and end with no plateau at 30mA and no vertical drop at 85ºC.

Idea 2 is obviously overkill for a little torch, probably useful for much more powerful circuits or applications, I just wanted to see if it were possible to condition a signal to emulate a graph to a reasonable degree. It's hard to gauge how it matches the LED SOA graph from the simulation maybe as the scale isn't quite the same. At 820mV, T1 is supposed to turn on and drain any voltage from the reference voltage to emulate the datasheet steep drop to 0mA at 85ºC.

Back to not worrying about resistor error...

#### d123

Hi,

Spent the afternoon re-doing the ppm resistance change calculations for 30.3 ohms, +-1%, +-100ppm/°C, 125°C/W. If I got the maths right this time, it really is irrelevant for 30mA - 494 mOhms more or 474 mOhms less, 432uA more or 432uA less across temperature range of interest.

Did another one for 1k, +-5%, +-100ppm, 125°C/W to be able to ask if I am doing the maths correctly now. Did it on paper to save time/phone battery.

Does that look right now, please - the general method and decimal place thing I got wrong before, not the actual calculated results (I wouldn't expect anyone to go through the whole thing, evidently)?

I want to get this reasonably right as I think it is useful for higher-precision circuits and to ensure 'I guess it' ll be okay, more or less' is a tool I depend on less.

Is it reasonable to use 125°C/W as a rule-of-thumb number when datasheets don't specify the thermal resistance?

Is thermal resistance guessable from derating graphs/max. operating temperature and wattage? Or can it only be obtained by requesting the information from the manufacturer?

Thanks.

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