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## Q. 33.5

The transistor of Fig. 33-11a has $β_{dc}$ = 300. Calculate $I_B, I_C, V_{CE}$, and $P_D$.

## Verified Solution

Figure 33-11b shows the same circuit with grounds. The base current equals

$I_B=\frac{V_{BB}-V_{BE} }{R_B}=\frac{10 V-0.7V}{1 M\Omega }=9.3 \mu A$

The collector current is

$I_C = \beta _{dc}I_B = (300)(9.3 \mu A) = 2.79 mA$

and the collector-emitter voltage is

$V_{CE} = V_{CC} – I_CR_C = 10 V – (2.79 mA)(2 k\Omega) = 4.42 V$

The collector power dissipation is

$P_D = V_{CE}I_C = (4.42 V)(2.79 mA) = 12.3 mW$

Incidentally, when both the base and the collector supply voltages are equal, as in Fig. 33-11b, you usually see the circuit drawn in the simpler form of Fig. 33-11c.