Figure 6.1.1
The BJT in the circuit in Fig. 6.1.1 has $\beta=100$. Find $V_{E}, V_{C}$, and $I_{B}$ for the three cases:
(a) $V_{B}=0 \mathrm{~V}$, (b) $V_{B}=3 \mathrm{~V}$, and (c) $V_{B}=5 \mathrm{~V}$.

Question

Quizplus · Accepted Answer

The Answer is (a)
$(a) Figure 6.1.2(a) The solution is illustrated in Fig. 6.1.2(a). The transistor is cut off with V_{E}=0 \mathrm{~V}, \quad V_{C}=+10 \mathrm{~V}, \quad I_{B}=0 \mathrm{~mA} (b) The solution is illustrated in Fig. 6.1.2(b). The transistor is operating in the active region with the following results: V_{E}=+2.3 \mathrm{~V}, \quad V_{C}=+4.3 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA} (c) The analysis performed in Fig. 6.1.2(c) shows that the transistor cannot be operating in the active region. Thus, it must be saturated. The corresponding analysis is shown in Fig. 6.1.2(d) with the following results: V_{E}=+4.3 \mathrm{~V}, \quad V_{C}=+4.6 \mathrm{~V}, \quad I_{B}=1.07 \mathrm{~mA}$

Figure 6.1.2(a)
The solution is illustrated in Fig. 6.1.2(a). The transistor is cut off with

V_{E}=0 \mathrm{~V}, \quad V_{C}=+10 \mathrm{~V}, \quad I_{B}=0 \mathrm{~mA}

(b)
$(a) Figure 6.1.2(a) The solution is illustrated in Fig. 6.1.2(a). The transistor is cut off with V_{E}=0 \mathrm{~V}, \quad V_{C}=+10 \mathrm{~V}, \quad I_{B}=0 \mathrm{~mA} (b) The solution is illustrated in Fig. 6.1.2(b). The transistor is operating in the active region with the following results: V_{E}=+2.3 \mathrm{~V}, \quad V_{C}=+4.3 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA} (c) The analysis performed in Fig. 6.1.2(c) shows that the transistor cannot be operating in the active region. Thus, it must be saturated. The corresponding analysis is shown in Fig. 6.1.2(d) with the following results: V_{E}=+4.3 \mathrm{~V}, \quad V_{C}=+4.6 \mathrm{~V}, \quad I_{B}=1.07 \mathrm{~mA}$

The solution is illustrated in Fig. 6.1.2(b). The transistor is operating in the active region with the following results:

V_{E}=+2.3 \mathrm{~V}, \quad V_{C}=+4.3 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA}

(c)
$(a) Figure 6.1.2(a) The solution is illustrated in Fig. 6.1.2(a). The transistor is cut off with V_{E}=0 \mathrm{~V}, \quad V_{C}=+10 \mathrm{~V}, \quad I_{B}=0 \mathrm{~mA} (b) The solution is illustrated in Fig. 6.1.2(b). The transistor is operating in the active region with the following results: V_{E}=+2.3 \mathrm{~V}, \quad V_{C}=+4.3 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA} (c) The analysis performed in Fig. 6.1.2(c) shows that the transistor cannot be operating in the active region. Thus, it must be saturated. The corresponding analysis is shown in Fig. 6.1.2(d) with the following results: V_{E}=+4.3 \mathrm{~V}, \quad V_{C}=+4.6 \mathrm{~V}, \quad I_{B}=1.07 \mathrm{~mA}$
$(a) Figure 6.1.2(a) The solution is illustrated in Fig. 6.1.2(a). The transistor is cut off with V_{E}=0 \mathrm{~V}, \quad V_{C}=+10 \mathrm{~V}, \quad I_{B}=0 \mathrm{~mA} (b) The solution is illustrated in Fig. 6.1.2(b). The transistor is operating in the active region with the following results: V_{E}=+2.3 \mathrm{~V}, \quad V_{C}=+4.3 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA} (c) The analysis performed in Fig. 6.1.2(c) shows that the transistor cannot be operating in the active region. Thus, it must be saturated. The corresponding analysis is shown in Fig. 6.1.2(d) with the following results: V_{E}=+4.3 \mathrm{~V}, \quad V_{C}=+4.6 \mathrm{~V}, \quad I_{B}=1.07 \mathrm{~mA}$

The analysis performed in Fig. 6.1.2(c) shows that the transistor cannot be operating in the active region. Thus, it must be saturated. The corresponding analysis is shown in Fig. 6.1.2(d) with the following results:

V_{E}=+4.3 \mathrm{~V}, \quad V_{C}=+4.6 \mathrm{~V}, \quad I_{B}=1.07 \mathrm{~mA}

Figure 611 the BJT in the Circuit in Fig β=100\beta=100β=100 Find

Figure 611
the BJT in the Circuit in Fig $\beta=100$ Find