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For the diode formula: $$I_D = I_S(e^{V_D/V_T} - 1)$$ (Where \$I_D\$ is the current of the diode)

Why is \$V_T\$ approximately always 26 mV in a lot of problems?

I can't find the part of the book that says this. It says \$V_T = {kT \over q}\$ but how is that always 26 mV?

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    \$\begingroup\$ +1 Good question. And good answer. - You could have plugged in values to see what happens BUT fauling that, asking is better than ognoring the unknown. \$\endgroup\$
    – Russell McMahon
    Commented Oct 12, 2015 at 11:55

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K is Boltzmann's constant, which is (ahem) constant.

q is the charge of an electron. Similarly constant.

T is temperature. It's not constant, but humans exist in a fairly narrow range of temperatures around 300K. So 25 or 26mV is a reasonable value for room temperature Vt.

By the way, the 't' in Vt stands for 'thermal' so that should remind you that it is a function of (absolute) temperature.

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  • \$\begingroup\$ thank you! so the voltage in a diode has to do with temperature! \$\endgroup\$ Commented Oct 12, 2015 at 7:05
  • \$\begingroup\$ @KennyTruong that's why they are used as temperature sensors in quite a few devices. \$\endgroup\$
    – Arsenal
    Commented Oct 12, 2015 at 8:55
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    \$\begingroup\$ Yes, however Is is also a function of temperature and there are a couple of other factors in the equation for a real diode (resistance and ideality), but ratios of Vf at different currents for a diode-connected transistor or matched transistors can make an excellent temperature sensor. \$\endgroup\$ Commented Oct 12, 2015 at 9:19
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    \$\begingroup\$ In case someone is too lazy to plug in the values themselves, taking 20 deg as the room temperature, \$ V_T = \frac{kT}{q} = \frac{1.380648521 ×10^{−23} J . K^{-1} \times 293.15 K}{1.60217662 × 10^{-19} coulombs} = 0.0252617\$ joules per coulomb or \$25.26 \ mV\$ \$\endgroup\$
    – Aditya P
    Commented Oct 22, 2018 at 17:01

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