# BJT: why does Ic increase when temperature increases? [closed]

When the temperature of an BJT increases, the thermal voltage $$V_T = \frac{k_BT}{q}$$ where $$\k_B\$$ is the Boltzmann constant, $$\T\$$ is the absolute temperature of the junction, $$\q\$$ is the elementary charge, should increase. Therefore collector current which is equal to $$I_C=I_s\exp(V_{in}/V_T)$$ should decrease but it does not. Why is this?

## closed as unclear what you're asking by Bimpelrekkie, Warren Hill, Finbarr, Voltage Spike, Dwayne ReidFeb 23 at 19:05

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• What is $V_{in}$ ? Maybe you mean $V_{BE}$ ? Also a schematic is needed showing you how you're using the transistor. Think about $V_{BE}$ as well, does it stay constant as temperature changes? Also note that transistors work on physics and not formulas. The formulas describe the behavior, they do not define it. – Bimpelrekkie Feb 20 at 19:46
• Is is also temperature-dependent. – Dave Tweed Feb 20 at 19:48
• Indeed, $I_s$ is very temperature dependent. – Bimpelrekkie Feb 20 at 19:49
• Is doubles every 8 degC (typical silicon bjt) – sstobbe Feb 20 at 20:00
• This may be better asked on a Physics site. It is not about electronics design but the physical behaviour of a component. – Warren Hill Feb 21 at 13:27

I've discussed this issue at least four times here. Maybe more. It's due to the temperature variation in the saturation current for the device, which has an opposite sign and greater magnitude than the Shockley equation provides.

## Shockley Equation

As I'm sure you can find repeated, in many places, the simple Shockley equation that operates reasonably well for a BJT in active mode:

$$I_\text{C}=I_{\text{SAT}\left(T\right)}\left(e^{\left[\frac{V_\text{BE}}{\eta \:V_T}\right]}-1\right)\\ \text{where the thermal voltage is }V_T=\frac{k\: T}{q}\label{eq1}\tag{Shockley equation}$$

(The emission coefficient, $$\\eta\$$, is usually taken to be 1, by default.)

The saturation current $$\I_{\text{SAT}\left(T\right)}\$$ is effectively a $$\y\$$-axis intercept point found by projecting the log-chart plot of collector (or base) current vs $$\V_\text{BE}\$$ voltage. See this chart for a little clarity about that: (Note that Region II is the usual "operation" area when in active mode.)

It's also pretty obvious that the additional -1 term in the equation removes this offset and brings the results into the general expectation of zero collector current when the base-emitter junction voltage is $$\V_\text{BE}=0\:\text{V}\$$.

Solving the above for $$\V_\text{BE}\$$:

$$V_\text{BE}=\eta\:V_T\:\operatorname{ln}\left(\frac{I_\text{C}} {I_{\text{SAT}\left(T\right)}}+1\right)\approx \eta\:\frac{k\: T}{q}\:\operatorname{ln}\left(\frac{I_\text{C}} {I_{\text{SAT}\left(T\right)}}\right)$$

From the $$\\ref{eq1}\$$, we can get:

$$\frac{\text{d}\: I_\text{C}}{\text{d}\: T}\approx -\frac{V_\text{BE}\:I_\text{C}}{\eta\:V_T\:T}$$

Because of the sign here, you'd expect that the collector current to decline with increasing absolute temperature. (This equation only quantifies the expectation you'd get from a cursory examination of the $$\\ref{eq1}\$$.)

So it's a good question you ask.

## Saturation Current

Perhaps you noticed my use of $$\I_{\text{SAT}\left(T\right)}\$$, earlier? As it turns out, the temperature dependence of $$\I_{\text{SAT}\left(T\right)}\$$ is huge. The approximate equation looks like this:

$$I_{\text{SAT}\left(T\right)}=I_{\text{SAT}\left(T_\text{nom}\right)}\cdot\left[\left(\frac{T}{T_\text{nom}}\right)^{3}e^{\frac{E_g}{k}\cdot\left(\frac{1}{T_\text{nom}}-\frac{1}{T}\right)}\right]$$

$$\E_g\$$ is the effective energy gap (in eV) and $$\k\$$ is Boltzmann's constant (in appropriate units.) $$\T_\text{nom}\$$ is the temperature at which the equation was calibrated, of course, and $$\I_{\text{SAT}\left(T_\text{nom}\right)}\$$ is the extrapolated saturation current at that calibration temperature.

This formula heavily depends upon fundamental thermodynamics theory and the Boltzmann factor (do not confuse this with the Bolzmann constant $$\k\$$), which you can easily look up and is above represented by the factor: $$\e^{\frac{E_g}{k}\cdot\left(\frac{1}{T_\text{nom}}-\frac{1}{T}\right)}\$$. (It's based on the simple ratio of the numbers of states at different temperatures; really no more complex than fair dice used in elementary probability theory. Perhaps the best introduction to the Boltzmann factor is C. Kittel, "Thermal Physics", John Wiley & Sons, 1969, chapters 1-6 in particular.)

(Note that the power of 3 used in the equation above is actually a problem, because of the temperature dependence of diffusivity, $$\\frac{k T}{q} \mu_T\$$. And even that, itself, ignores the bandgap narrowing caused by heavy doping. In practice, the power of 3 is itself turned into a model parameter.)

Once you have included the equation exposing the temperature-dependence of the BJT saturation current into the $$\\ref{eq1}\$$ above, you find that the computation of $$\\frac{\text{d}\: I_\text{C}}{\text{d}\: T}\$$ now has a positive sign instead of a negative one. And this is what explains the effect you noted in your question.

I'll leave the derivation of newly arriving equation for $$\\frac{\text{d}\: I_\text{C}}{\text{d}\: T}\$$ to you, though.

For example,

• hFE can increase 50% from +25'C to 125'C if it is operating below 10% of its Abs Max current. (See Fig 1)

• Note Fig 3 should read Vce (not Vbe) saturation vs Temp which is also NTC (neg. temp coeff.) so if Ic is saturated with a load near Rce= Vce/Ice, the Ic will also increase. Although technically correct since this curve is due to Vbe where Vbe = Vce+Vbc in saturation mode.

• GBW also increases with temperature and increases AC current if it is BW or slew rate limited on applied signal.