Units of BJT parameters

Question

Let's say the model is the Ebers-Moll model, but this obviously extends to other models:

$$ I_{\mathrm{b}} = \frac{I_{\mathrm{s}}}{\beta_{\mathrm{f}}}\left(\mathrm{e}^{\frac{V_{\mathrm{be}}}{NV_{\mathrm{t}}}} - 1\right) + \frac{I_{\mathrm{s}}}{\beta_{\mathrm{r}}}\left(\mathrm{e}^{\frac{V_\mathrm{bc}}{NV_{\mathrm{t}}}} - 1\right)\label{eq:bjt-base} $$ $$ I_{\mathrm{c}} = I_{\mathrm{s}}\left(\mathrm{e}^{\frac{V_{\mathrm{be}}}{NV_{\mathrm{t}}}} - 1\right) - I_{\mathrm{s}}\frac{\beta_{\mathrm{r}} + 1}{\beta_{\mathrm{r}}}\left(\mathrm{e}^{\frac{V_\mathrm{bc}}{NV_{\mathrm{t}}}} - 1\right)\label{eq:bjt-collector} $$

Most parameters of the model have obvious units e.g. \$I_\mathrm{s}\$ is the saturation current. There are several which catch me out though, the units of:

• \$N\$ - Ideality factor/emissivity coefficient;
• \$\beta\$ - current gain.

As the current gain is a gain, is it given in the same way a voltage gain is given, i.e. instead of (V/V), (A/A)? I've never seen this notation.

Is it possible that the emissivity coefficient is unitless?

Answer 1

Is it possible that the emissivity coefficient is unitless?

Yes, it doesn't have any unit. In general, functions like exp, log, ln, trigonometric functions like sin, cos, tg, etc. all don't take dimensional operands. So in your formula of Is or Ib since Vbe and Vt both appear in the operand of exp function, and since they have units of voltage, both units cancel out and N has to be unitless.

As the current gain is a gain, is it given in the same way a voltage gain is given, i.e. instead of (V/V), (A/A)?

Yes, it's A/A since \$\beta=\frac{I_c}{I_b}\$, so it's unitless. Again it can be seen from the equation. \$I_s\$ has unit of current, so \$\beta\$ must have no unit, otherwise it wouldn't cancel out with the denominator as the parenthesis has no unit already. In other words, you have something like this for both equations:

$$ [I]=\frac{[I]}{[N]} \times [exp-1]$$

Since function exp-1 has no unit, N has to be unitless to allow both sides to meet in unit.

Answer 2

Yes, current gain is A/A.

Factors and coefficients are often (maybe always?) unitless.

In this case, yes, the Ideality factor is unitless.