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The Transistor as a Switch

The two extreme possible operating points Q_sat and Q_cutoff show how the transistor can act as a switch. 

The operating point Q_sat corresponds to the maximum value of I_C. As I_C increases from zero, the voltage drop across R_C increases, and so from (95) V_CE must decrease until the BE junction becomes forward biased and the transistor saturates. The saturation value of V_CE is denoted V_CE(sat) and is typically 0.2 V. The saturation value of the collector current is

$$I_{C(sat)} = \frac{V_{CC} - V_{CE(sat)} }{R_C} .$$ (96)

In order to achieve saturation, V_BB must be large enough to ensure that I_B is greater than

$$\frac{I_{C(sat)}}{\beta}$$

which means that V_BB must satisfy

$$V_{BB} \geq \frac{I_{C(sat)}R_B}{\beta} + 0.7 \ V .$$ (97)

In saturation mode, the transistor is on, allowing maximum current to flow.

At the other extreme, Q_cutoff the transistor is cutoff. This requires I_B=0, and hence I_C=0. The cutoff value of V_CE is

 

V_CE(cutoff) = V_CC . (98)

Of course, in this mode the transistor is off. With the simple model this requires $V_{BB} \leq 0$.

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