Thermal voltage of a BJT vs. diode voltage drop

Question

I'm learning about the hydrid-pi model of the BJT transistor:

In this model, the base and the emitter have a resistance \$r_{\pi}\$ in between. The voltage between the base and the emitter is apparently called "thermal voltage" and at room temperature it is approximately \$26mV\$. This voltage divided by the bias current gives the resistance \$r_{\pi}\$.

But normally in circuit analysis it is assumed that a transistors base-emitter junction has the voltage of a diode, approximately \$0.7V\$ across it. Isn't there a contradiction?

Answer 1

There's no contradiction if you're using the model for what's it's intended for.

The hybrid pi model is only useful for the dynamic gain of the transistor (and then at only one specific emitter current), not about the static biassing.

If you want a model for the static biassing, then a more useful one is to replace the input resistor with a diode, and the \$g_m V_{be}\$ current source with \$\beta I_b\$

Answer 2

The thermal voltage of a BJT, \$V_\mathrm{t}\$, is not the default voltage across a junction. It originates from the Shockley diode model from which large signal BJT models are formed:

$$ I = I_\mathrm{s} \left(\mathrm{e}^\frac{V}{N V_\mathrm{t}} - 1 \right) $$ where Is is the saturation current, Vt the thermal voltage, and N the forward emissivity coefficient or ideality factor.

The 0.7 V you mention also can be derived from this model, which is referred to as the forward voltage, when using parameters suitable for a silicon diode (other semiconductors have different forward voltages). Typically a forward current is assumed, lets say 1 mA, which when solving the above equation for V, produces a voltage of around 0.7 V. This voltage is similar for a range of currents due to the exponential nature of the equation, which is why it is commonly stated. To try this out the parameters for a silicon pn junction are around \$ I_\mathrm{s} = 10^{-14}\ \mathrm{A}\$, \$ N = 1\$, and \$ V_\mathrm{t} = 25\ \mathrm{mV} \$.

The thermal voltage actually arises from the junction temperature, given by $$ V_\mathrm{t} = \frac{kT}{q} $$ where q is the charge on an electron, k is Boltzmann's constant, and T the temperature in Kelvin. For room temperature this comes out around 25 mV (depending on how hot you like your room :) ).

A BJT features two pn junctions that are typically modelled using the Shockley model, plus a bit extra. I recommend reading up on the Ebers-Moll model. The Hybrid Pi model is derived from large signal models like the Ebers-Moll model by finding the DC operating point, and from this finding the derivatives of currents with respect to voltages. The derivative of a nonlinear impedance linearises the model around a point, giving transconductances and resistances depending on which derivative is used.