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Category: Datasheet/Specs

I'm referring to the AD538 data sheet [1] (page 12) and the article in Analog Dialogue [2] (page 4 and 6).

In [2] (page 4) is explained that the temperature coefficient is +3330 ppm/K. It is so because the output voltage of the log section is proportional to the absolute temperature, and if we assume T=300K then the temperature coefficient is 1/(300K), which is the same as 3333 ppm/K.

In [1] (page 12) and [2] (page 6) is written that this temperature coefficient is approximated by a 90.9 Ohm resistor (which is assumed to have no temperature coefficient) in series with a 1000 Ohm PTC with +3500 ppm/K.

I think this is wrong. The formula for the temperature coefficient of two resistors in series is:

alpha = alpha1 * R1 / (R1 + R2)      or        R2 = R1 * (alpha1 / alpha - 1)

With R1 = 1000 Ohm, alpha1 = 3500 and alpha = 3330 the series resistor should be R2 = 51 Ohm. Not 90.9 Ohm as suggested.

Please explain why the series resistor is 90.9 Ohm, or what's wrong with my formula.

Thanks,

Michael

added where 3330 ppm/K comes from
[edited by: mkoch at 6:09 AM (GMT -4) on 8 Aug 2023]

1. Can the absolute value section (as shown in figure 17) also be used for the Ix Vx and Iy Vy inputs in figure 13? I mean that three absolute value sections are used for all three inputs x, y and z. Can in this case the diode at pin 11 be omitted? Or with other words: Please explain what's the purpose of the diode in figure 13. Why is the diode required at the Iy input, while it seems the Ix and Iz inputs don't require a diode?

2. In figure 15 the antilog section is mostly unused. If I understood it correctly, only the output amplifier is used. Is it also possible to use only the antilog section, without using the log section? I think in this case we need a temperature compensation at the input of the antilog section. Do you have a sample circuit for this temperature compensation?

• Hello Mkoch,

If we assume that the temperature coefficient for the 90.9ohm metal-film resistor is around 50ppm/C to 100ppm/C (which is what I am seeing for most 90.9ohm resistors available online)

alpha = [(1000 * 3500) + (90.9 * 100)] / 1000 + 90.9

alpha = 3216 ppm/C which is approximately close to 3300ppm/C which is stated on page 12 of the DS.

Maybe that's why the total resistance will become 1.09kohms with 3300ppm/C temp coefficient.

If we reverse the formula and make it the exact 3300ppm/C temp coefficient, temp coefficient for the 90.9ohm metal film resistor will turn out to be 1100ppm/C(but not sure regarding this since DS did not state the actual temp coefficient for the metal film resistor.

I will check this further and let you know how did the DS come up with a 90.9ohm resistor. But for now, let's just correct the formula you used on calculating the temp coefficient for resistors in series.

Best Regards,
Kenneth

• Your formula is correct if both resistors have a temperature coefficient. I was assuming that the 90.9 Ohm resistor has zero temperature coefficient.

I think when a metal film resistor is specified as "50ppm", that does always mean plus or minus 50. Because we don't know where in the -50 to +50 range the temperature coefficient actually is, we should use the average value 0.

See for example here: www.mouser.de/.../Yageo_LR_MFR_1-1714151.pdf

Best Regards,

Michael

• Hello Michael,

Plus or minus only denotes direction of change in resistance (if its increasing or decreasing with temp). During calculation we have to assume only one direction for both resistors. In this case, I assumed it to be positive since the resultant coefficient is also positive.

I used 100ppm for the 90.9ohm resistor since it is the extreme possible temp coefficient that a metal film resistor could have (since it was not indicated on the DS).

But that's only an assumption since it was not indicated on the DS. It is still recommended to follow what's written on the datasheet for correct operation.

Best Regards,
Kenneth