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## Q. 3.3

The manufacturer’s specification for a 2N3904 transistor shows that $\beta _{DC}$ has a range from 100 to 300 (given in Fig. 3–25). Assume a 2N3904 is used in the base- biased circuit shown in Figure 3–15. Compute the minimum and maximum collector current based on this specification. (Note that this is effectively the same circuit that was solved with load line analysis in Figure 3–13 except it now is shown with a single power supply.)    ## Verified Solution

The base-emitter junction is forward-biased, causing a 0.7 V drop. The voltage across $R_{B}$ is

$V_{R_{B} } = V_{CC} – V_{BE} = 12 V – 0.7 V = 11.3 V$

The base current can be found by applying Ohm’s law to the base resistor.

$I_{B} = \frac{V_{R_{B} } }{R_{B} } = \frac{11.3 V}{1.0 M\Omega } = 11.3 \mu A$

With linear operation, the collector current is $\beta _{DC}$ times larger than the base current. Therefore, the minimum collector current is

$I_{C(min)} = \beta _{DC} I_{B} = (100)(11.30) \mu A)$ =1.13 mA

The maximum collector current is

$I_{C(max)} = \beta _{DC} I_{B} = (300)(11.30 \mu A)$ =3.39 mA

Notice that a 300% change in $\beta _{DC}$ caused a 300% change in collector current.