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Temperature Compensation

Rules of thumb - diode

Diodes forward drop goes down as temperature goes up
For a given current, the V/I curve shifts down by approximately 2 mV/°C.

$I = I_s\left(e^\frac{qV}{nkT} - 1\right)$

where:

I = the net current flowing through the diode.

I_s = "dark saturation current", or is the reverse bias saturation current (or scale current) the diode leakage current density in the absence of light. I_s approximately doubles for every 10 °C increase in temperature - this temperature effect dominates the one in the exponent and results in the voltage drop with temperature. I_s is related to the area of the junction and just how it was made.

V = applied voltage across the terminals of the diode.

q = absolute value of electron charge.

k = Boltzmann's constant.

T = absolute temperature (K).

n = ideality factor, a number between 1 and 2 which typically increases as the current decreases.

It is possible to pull the thermal voltage VT out of the equation - VT is approximately 25.85 mV at 300 K, a temperature close to "room temperature" commonly used in device simulation software. This voltage you will see in Zero_Gain_Amps.

$V_\text{T} = \frac{k T}{q} \,$

But β increases with increasing temperature

Simplified trans-conductance of a Transistor

$G_m = \frac{1}{r_e} \cong \frac{qi_c}{kT}$

$r_e \cong \frac{kT}{qi_c}$

Transistor current ratios

Doubling the collector current will cause an increase of about 18mv in V_be or:

$\Delta V_{be} = V_T\ \ln \frac{I_{c1}}{I_{c2}}$

Where , $V_T \cong 26mV$ at room temperature

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