0195136047.pdf

424 DIGITAL CIRCUITS

VCC = 5 V

VT = 0.7

1

5

4

3

2

1

0.2 to 0.3 V

2345

IB sat = 50

60 μA

50 μA

40 μA

30 μA

20 μA

10 μA

0.5

Saturation

Cutoff

1

Vi

Vi

VO = vCE

RC

RB

RB

vBE

vBE, V

iC

iB

iB, μA

vCE, V

iC, mA

(a)

(b)

Slope = −

− −

+

−

+

+

vCE

−

+

+

1

1

2

2

VCE cutoff VCC

VCC

VCC

RC

VCE sat = Vsat

IC cutoff = ICEO iB =^0

IC sat

Figure 9.1.2BJT inverter switch.(a)Circuit withnpnswitching transistor.

(b)Typical operation using load lines.

Thus, the transistor behaves like an ideal switch, as shown in Figure 9.1.3. It can be shown that

saturation will occur when

Vi−VT

RB

>

VCC−Vsat

βRC

or Vi>(VCC−Vsat)

RB

βRC

+VT (9.1.6)

The power dissipated in the transistorp=vCEiC+vBEiB∼=vCEiCis very small or approximately

zero in either cutoff or saturation. It should be noted, however, that power is expended in switching

from one state to the other, going through the linear or active region.