• introductory math
• algorithmic math

II. Algorithmic Math

The BJT switching process depends on the initial condition in the Base Charge and the applied input voltage signal.We shall use these definitions:

• Q_bo = amount of excess base charge at the time of switching
• Q_eos = Base charge at the "edge-of-saturation", i.e., at the boundary between the active-mode and the saturation-mode.
• V₁ = input low; V₂ = input high
• I_sat = collector current at saturation.
• t_s = saturation time constant.
• b = forward current gain.

1) Q_bo = 0 (zero excess base charge)
i) If the Emitter junction is fully depleted cut-off:
t_d = R_B (C_je+C_jc) ln[(V₂ -V₁ )/(V₂ - 0.7)]
ii) If the Emitter junction is less than max. reverse bias:
only part of t_d.
V_I, i_B, i_C
2) 0 =< Q_bo < Q_eos : Base charge is in the forward active-mode
Q_b(t) = I_B2 [1 - exp(-t / t )] + Q_bo until Q_b reaches Q_eos , where
t = (R_B||r_p) (C_p + C_m) with the Miller effect ignored, or
t = (R_B||r_p) (C_p + C_m - A_vC_m ) with the Miller effect included.

Here, because the device is in the active-mode operation and the operating point changes continuously, the collector current, diffusion charge, and the voltage gain change also (r_p, C_p , A_v). So an estimated value need be used for these parameters.

V_I, i_B, i_C

3) Q_eos =< Q_bo < t_sI_B2 : The transistor is in saturation mode
Q_b(t) = Q_bo + ( t_sI_B2 - Q_bo ) [ 1 - exp(-t / t_s) ]. Note that the time constant is
different from that of the active-mode.

V_I, i_B, i_C

1) Q_eos < Q_bo = < t_sI_B2 : transistor is in the saturation mode.
Q_b(t) = ( Q_bo - t_sI_B1 ) exp(-t / t_s) + t_sI_B1 , until base charge reaches Q_eos . The saturation-delay time is t_sd = t_sln[(Q_bo - t_sI_B1)/(Q_eos - t_sI_B1)]

V_I, i_B, i_C

2) 0 < Q_bo =< Q_eos : transistor is in forward-active mode.
Q_b(t) = Q_bo exp(-t /t ). Note that Q_eos = t_s ( I_sat / b ) in which formula the (active-mode) time constant is taken the same as the saturation-mode time constant. This inconsistency needs to be removed.

V_I, i_B, i_C

3) Q_bo = 0 : transistor is in cut-off mode.
i) If Emitter junction is in the middle of being charged.
Emitter junction is charged by the increasing reverse bias.
ii) Emitter junction is fully depleted.
No further change or charging.
V_I, i_B, i_C