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Please forget the "why" for now, but I am attempting to determine the temperature of a 1N4148WS-E3 part by doing two measurements at different current setpoints. I've done some tests and some calculations and I don't understand where the error is coming from.

The theory

Derived from the Shockley diode equation, the following formula should give me the absolute temperature of the diode:

$$T=\frac{\Delta V_D}{\ln\left (\frac{1}{N}\right )}\cdot\frac{q}{n\cdot k}$$

Source: This video by Arda Yilmaz

Where:

  • ΔVD: the difference between two measured voltages over the diode
  • N: the ratio between the two measured currents corresponding to those voltages
  • q: Charge of an electron, 1.602176634 × 10-19 coulomb
  • k: Boltzmann constant, 1.380649 × 10-23
  • n: ideality factor of the diode, this is 2.60607 according to Vishay

The measurement

I've measured voltage and current over/through this diode using two setpoints. The measurement data I got is:

  • U1: 0.5819 V, I1: 582.8 µA
  • U2: 0.4886 V, I2: 78.3 µA

To get this data, I used two multimeters, a 1 kΩ resistor in series with the diode and a lab voltage supply to set the setpoint. The current measurement is in series with the resistor, while the voltage measurement is directly over the diode.

schematic

simulate this circuit – Schematic created using CircuitLab

Feeding this into the formula, ΔVD is 0.0933 V and N is (78.3/582.8 =) 0.134351

This results is T = 207 K which is obviously incorrect. I'm expecting something close to room temperature.

Does anybody have any idea where this error is coming from?

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  • \$\begingroup\$ Try taking a few more points and see if they track exponentially; and take a few at another temperature (as well calibrated as you can; can be tricky without a thermal chamber), see if you can solve for \$n\$. Depending on where you got the parameters from, the manufacturer may be prioritizing different parts of the curve, or have other parameters that modify the basic equation. \$\endgroup\$
    – Tim Williams
    Mar 23 at 14:25
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    \$\begingroup\$ I think all you have done is prove that the ideality factor is under 2. See this web page for your information: 2n3904blog.com/1n4148-diode-forward-biased-i-v-curve See also this discussion of IF: electronics.stackexchange.com/questions/559667/… \$\endgroup\$
    – Andy aka
    Mar 23 at 14:35
  • \$\begingroup\$ I don't think you can use the same value of n for all 1N4148s ever produced. The fact that this value is quoted to 7 sig. figs. is a big red flag to me about that source, anyway. Even devices from the same batch will vary. A typical SPICE model uses 1.9, this answer calculates it to be 2.3 from Vishay's own data. Your best bet is to make some measurements and find n for your particular device. \$\endgroup\$
    – Simon Fitch
    Mar 23 at 14:43
  • \$\begingroup\$ Look at Linear Technology's application note 137 on how to do temperature measurement. A transistor is used for the sensor and \$ V_{be} \$ is measured at two different currents. Also beware of self heating of the sensor. \$\endgroup\$
    – qrk
    Mar 23 at 17:17
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    \$\begingroup\$ Did you cover the diode to prevent light from making it conduct like a photo-diode? \$\endgroup\$
    – Audioguru
    Mar 23 at 17:18

1 Answer 1

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Ideality factor is a strong function of temperature (and varies a lot from unit-to-unit on top of that). I suggest you use a diode-connected transistor, which has an ideality factor very close to 1 (and is known to work well in this application).

enter image description here

A 10:1 ratio in measuring currents works well (yielding a change of about 200uV/K), but it's also possible to use three currents and eliminate the effect of the resistive component in the "diode" behavior.

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