Theory of Dynamic Characteristics II


The charge control equations are given for forward active mode of transistor as:

iC = QFF – dQBC/dt

iB = QFBF + dQF/dt + dQBC/dt + dQBE/dt

iE = QFF + QFBF + dQF/dt + dQBE/dt

Where, all currents and charges are intrinsically taken as instantaneous functions of time.

Delay Time, td

The above equations do not apply to calculate the delay time, as the transistor is not in forward active mode throughout this time. The delay time is, though, very small in relation to the other switching times and is therefore ignored.

Fall-Time, tf : Transistor Turn-On

The fall-time, tf, for the output voltage is obtained by resolving the charge control equations for this condition. To resolve the above charge control equations in this concern, the currents require being associated in terms of circuit parameters. As the emitter terminal is grounded, this can’t be done for the emitter current and therefore its charge control equation is of little advantage. Throughout transistor turn-on in the forward active mode, the base of transistor can be taken as sitting at the potential of VBE ON and hence the base current can be supposed constant throughout this period as shown in figure below.

Therefore, iB = (VCC - VBEON)/RB

And, dQBE/dt = CBE (dVBE/dt) = 0

Figure: Constant Voltage on Base throughout Transistor Turn-On

In this case, the charge control equations are simplified to:

iC = QFF - dQBC/dt

iB = (QFBF) + (dQF/dt) + (dQBC/dt) = (VCC – VBE ON)/RB

The equation for iB is expressed in terms of circuit parameters and is therefore clearly the one to be solved. Though, since a solution for ic is needed, substitutions for QF and dQF/dt terms in terms of ic should be found. Let first consider a figure shown below:

Figure: The Effect of Collector-Base Junction Capacitance

From the circuit:

dQBC/dt = CBC (dVBC/dt) = CBC d/dt[VBE ON – Vo(t)]

Therefore, dQBC/dt = - CBC [dVo(t)]/dt


Vo(t) = VCC - iCRC


dVo/dt = -RC (diC/dt)


dQBC/dt = CBC RC (diC/dt)……………… (a)

Secondly, consider:

iC = QFF - dQBC/dt

QF = τF iC + τF (dQBC/dt)

And hence,

QF = τF iC + τF CBC RC (diC/dt)…………. (b)

dQF/dt = τF (diC/dt) + τF CBC RC (d2iC/dt2)

All through turn-on, the collector current starts to rise towards βF IB however is limited on reaching saturation to ICMAX << βFIB. The increase in collector current can be seen to contain an almost constant rate throughout this time as shown in figure below.

Then, if (diC/dt) ≈ constant => (d2iC/dt2) → 0 and avoiding the second order derivative gives:

dQF/dt ≈ τF (diC/dt)……………. (c)

Figure: Rise in Collector Current throughout Transistor Turn-On

Substituting the equations (a), (b) and (c) to replace the charge terms in the equation for base current provides:


As τFBF = τFFτF = 1/βF then, this equation can be rearranged as:

(1/βF) iC + [(1/βF) CBC RC + τF + CBC RC] (diC/dt) = (VCC - VBEON)/RB

On multiplying by βF gives:

iC + (CBC RC + βF τF + βF CBC RC) (diC/dt) = βF [(VCC - VBE ON)/RB]

If βF >> 1 then (βF +1) CBC RC ≈ βF CBC RC and hence:

iC + βFF + CBC RC) (diC/dt) = βF [(VCC - VBE ON)/RB]

This is a first order linear differential equation in ic which can be solved by taking the Laplace Transform. Note that the Laplace Transforms of constant, K and a derivative of a function are as follows:

L {K} = K/s

L {dy/dt} = s L {y} – y (t = 0) = sY(s) – y (t = 0)

And hence,

IC(s) + βFF + CBC RC) [s IC (s) - iC (t = 0)] = βF [(VCC - VBE ON)/sRB]

iC(t = 0) is the initial value of collector current throughout turn-on that should be taken as zero. Then,


The solution can be easily obtained by taking Inverse Laplace Transform where the Inverse Laplace transforms of the term:

This equation exhibits that in linear forward active region, throughout transistor turn-on, the collector current increases exponentially from zero towards the value of   βF [(VCC - VBE ON)/RB)]

Time constant of the expression can be seen to depend on transistor properties βF, τF and CBC and also the value of the external resistor RC.

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