Inverter Fall Time:The equation obtained explaining the collector current as a function of time is as follows:
But, as VO = VCC - iC RC, then during the drop of the output voltage of inverter, we have:
This gives the impression that output voltage is heading from VCC towards a big, negative voltage, however, VCC – σ (VCC - VBEON) as can be seen in figure below:
Figure: Output Voltage of Transistor Inverter for Transistor Turn-On
Though, the transistor reaches saturation when iC = ICMAX. At this point, the output voltage reaches its least value VCESAT.If VCC >> VBE ON then VCC - VBE ON ≈ VCC and as (βF RC)/RB = σu then the expression for Vo(t) can be as follows:
To get the fall-time for inverter, the times at which the output voltage reaches 90% and 10% correspondingly of the supply voltage should be determined.Then, at t = t90, Vo(t) = 0.9VCC
Usually, βF =50, RC =1kΩ, CBC = 0.5pF and τF =0.2ns that gives:CBC RC = 0.5 x 10-12 x 103 = 0.5 ns whereas τF = 0.2nsThen,βF (τF + CBC RC) = 50 (0.5 + 0.2) x 10-9 = 35 nsNote that both external load and internal properties of the transistor are uniformly significant in determining the switching speed of inverter.When no overdrive were utilized, σu = 1then ln [(1 – 0.1)/(1 – 0.9)] =ln 9 = 2.2However with an overdrive factor, σu = 5then, then ln [(5– 0.1)/(5 – 0.9)] =ln 1.195 = 0.18And hence, tf = 35 x 10-9 x 0.18 = 6.3nsNote that the base overdrive significantly decreases the switching time throughout turn-on of the transistor.Storage Time, tSDuring storage time, the transistor operates in saturation mode. For this whole period, the collector current remains at its maximum value, ICMAX and the output voltage of inverter remains low, in spite of the fact that input voltage has switched from high to low. This time is spent eliminating the surplus overdrive charge from the base region as illustrated in figure shown below.
Figure: Storage Time Due to Removal of Overdrive Charge
In order to model this procedure accurately, a more complete charge control model is required that takes account of charge flow in both forward and reverse directions via the transistor. This can be shown that the forward charge concentration consists of a solution of the form:
Where, τs is termed to as the saturation time constant, usually 20-25ns. This can be seen thus that, while the base overdrive speeds up the transistor turn-on, it slows down the transistor turn-off.Rise-Time, tRDuring this time, the transistor again operates in forward active mode, however with the transistor turning off and collector current falling from ICMAX to zero. The similar charge control equations apply as in case of transistor turn-on. The difference is that, this time charge is being eliminated from the base and as a result, the base current is negative as shown in figure below.
Figure: Elimination of Stored Charge during Transistor Turn-Off
When the base-emitter voltage is taken as constant at VBE ON whereas in the forward active mode then iB = - (VBE ON/RB) and the charge control equations become:iC = QF/TF – dQBC/dtiB = QF/TBF + dQF/dt + dQBC/dt = - (VBE ON/RB)The similar substitutions apply as in case of the fall-time analysis and hence the equation for base current, that has to be solved to get an expression for collector current, is as follows:
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