Exam 2: Bipolar Junction Transistors BJTS

$(a)The BJTs in the circuits of Fig. 6.2.1 have \beta=100 . (a) For the circuit in Fig. 6.2.1(a), find V_{E}, I_{B} , and V_{C} . (b) For the circuit in Fig. 6.2.1(b), find I_{B}, I_{C} , and V_{C} .$ (a)The BJTs in the circuits of Fig. 6.2.1 have $\beta=100$ . (a) For the circuit in Fig. 6.2.1(a), find $V_{E}, I_{B}$ , and $V_{C}$ . (b) For the circuit in Fig. 6.2.1(b), find $I_{B}, I_{C}$ , and $V_{C}$ .

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Question 1

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(a)
$(a) Figure 6.2.2(a) The analysis is illustrated in Fig. 6.2.2(a). We have assumed that Q is operating in the active mode, and the results obtained justify this assumption: V_{C} is lower than V_{B} , keeping the CBJ reverse biased. The results are V_{E}=+0.7 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA}, \quad V_{C}=-2.03 \mathrm{~V} (b) Q cannot be in Active Mode; must be in Saturation Figure 6.2.2(b) The analysis in Fig. 6.2.2(b) is based on the assumption that the transistor is operating in the active mode. The results obtained show that this assumption is not justified: V_{C} at -425 \mathrm{~V} implies that the CBJ has a forward bias of 425.7 \mathrm{~V} ! Thus, the transistor must be in saturation. The corresponding analysis is shown in Fig. 6.2.2(c) and the results are I_{B}=0.43 \mathrm{~mA}, \quad I_{C}=0.47 \mathrm{~mA}, \quad V_{C}=+0.3 \mathrm{~V}$

Figure 6.2.2(a)
The analysis is illustrated in Fig. 6.2.2(a). We have assumed that $Q$ is operating in the active mode, and the results obtained justify this assumption: $V_{C}$ is lower than $V_{B}$ , keeping the CBJ reverse biased. The results are
$V_{E}=+0.7 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA}, \quad V_{C}=-2.03 \mathrm{~V}$
(b)
$(a) Figure 6.2.2(a) The analysis is illustrated in Fig. 6.2.2(a). We have assumed that Q is operating in the active mode, and the results obtained justify this assumption: V_{C} is lower than V_{B} , keeping the CBJ reverse biased. The results are V_{E}=+0.7 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA}, \quad V_{C}=-2.03 \mathrm{~V} (b) Q cannot be in Active Mode; must be in Saturation Figure 6.2.2(b) The analysis in Fig. 6.2.2(b) is based on the assumption that the transistor is operating in the active mode. The results obtained show that this assumption is not justified: V_{C} at -425 \mathrm{~V} implies that the CBJ has a forward bias of 425.7 \mathrm{~V} ! Thus, the transistor must be in saturation. The corresponding analysis is shown in Fig. 6.2.2(c) and the results are I_{B}=0.43 \mathrm{~mA}, \quad I_{C}=0.47 \mathrm{~mA}, \quad V_{C}=+0.3 \mathrm{~V}$

$Q$ cannot be in Active Mode; must be in Saturation
Figure 6.2.2(b)
$(a) Figure 6.2.2(a) The analysis is illustrated in Fig. 6.2.2(a). We have assumed that Q is operating in the active mode, and the results obtained justify this assumption: V_{C} is lower than V_{B} , keeping the CBJ reverse biased. The results are V_{E}=+0.7 \mathrm{~V}, \quad I_{B}=0.01 \mathrm{~mA}, \quad V_{C}=-2.03 \mathrm{~V} (b) Q cannot be in Active Mode; must be in Saturation Figure 6.2.2(b) The analysis in Fig. 6.2.2(b) is based on the assumption that the transistor is operating in the active mode. The results obtained show that this assumption is not justified: V_{C} at -425 \mathrm{~V} implies that the CBJ has a forward bias of 425.7 \mathrm{~V} ! Thus, the transistor must be in saturation. The corresponding analysis is shown in Fig. 6.2.2(c) and the results are I_{B}=0.43 \mathrm{~mA}, \quad I_{C}=0.47 \mathrm{~mA}, \quad V_{C}=+0.3 \mathrm{~V}$ The analysis in Fig. 6.2.2(b) is based on the assumption that the transistor is operating in the active mode. The results obtained show that this assumption is not justified: $V_{C}$ at $-425 \mathrm{~V}$ implies that the CBJ has a forward bias of $425.7 \mathrm{~V}$ ! Thus, the transistor must be in saturation. The corresponding analysis is shown in Fig. 6.2.2(c) and the results are
$I_{B}=0.43 \mathrm{~mA}, \quad I_{C}=0.47 \mathrm{~mA}, \quad V_{C}=+0.3 \mathrm{~V}$

$The transistor in the circuit of Fig. 6.3.1 has \beta= 100. Find the values of the four resistors so that V_{B}=+5 \mathrm{~V}, I_{1}=0.1 \mathrm{~mA}, I_{E}=1 \mathrm{~mA} , and V_{C}= +6 \mathrm{~V} .$ The transistor in the circuit of Fig. 6.3.1 has $\beta=$ 100. Find the values of the four resistors so that $V_{B}=+5 \mathrm{~V}, I_{1}=0.1 \mathrm{~mA}, I_{E}=1 \mathrm{~mA}$ , and $V_{C}=$ $+6 \mathrm{~V}$ .

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Question 2

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$The complete solution is performed directly on the circuit diagram in Fig. 6.3.2. The solution is based on the fact that the transistor is operating in the active mode because V_{C}>V_{B} . The results are \begin{aligned} R_{B 1} & =100 \mathrm{k} \Omega, & & R_{B 2}=55.6 \mathrm{k} \Omega, \\ R_{E} & =4.3 \mathrm{k} \Omega, & & R_{C}=9.1 \mathrm{k} \Omega \end{aligned}$

The complete solution is performed directly on the circuit diagram in Fig. 6.3.2. The solution is based on the fact that the transistor is operating in the active mode because $V_{C}>V_{B}$ . The results are
$\begin{aligned}R_{B 1} & =100 \mathrm{k} \Omega, & & R_{B 2}=55.6 \mathrm{k} \Omega, \\R_{E} & =4.3 \mathrm{k} \Omega, & & R_{C}=9.1 \mathrm{k} \Omega\end{aligned}$

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

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Question 3

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

$Figure 6.5.1 Design the circuit in Fig. 6.5.1 to obtain I_{1}= 0.1 \mathrm{~mA}, V_{B 1}=+4 \mathrm{~V}, I_{E 1}=1 \mathrm{~mA}, V_{C 1}=+6 \mathrm{~V} , I_{E 2}=2 \mathrm{~mA} , and V_{C 2}=+4 \mathrm{~V} . Let \beta_{1}=\beta_{2}=100 and \left|V_{B E}\right|=0.7 \mathrm{~V} . Specify the required values of all resistors.$ Figure 6.5.1 Design the circuit in Fig. 6.5.1 to obtain $I_{1}=$ $0.1 \mathrm{~mA}, V_{B 1}=+4 \mathrm{~V}, I_{E 1}=1 \mathrm{~mA}, V_{C 1}=+6 \mathrm{~V}$ , $I_{E 2}=2 \mathrm{~mA}$ , and $V_{C 2}=+4 \mathrm{~V}$ . Let $\beta_{1}=\beta_{2}=100$ and $\left|V_{B E}\right|=0.7 \mathrm{~V}$ . Specify the required values of all resistors.

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Question 4

$The BJT in the circuit in Fig. 6.6.1 has I_{S}= 10^{-15} \mathrm{~A} at room temperature. (a) Find V_{O} at room temperature where V_{T} is 25 \mathrm{mV} . (b) If the temperature increases by 50^{\circ} \mathrm{C} , what will V_{O} become? (c) If, alternatively, the temperature decreases by 50^{\circ} \mathrm{C} , what will V_{O} become?$ The BJT in the circuit in Fig. 6.6.1 has $I_{S}=$ $10^{-15} \mathrm{~A}$ at room temperature. (a) Find $V_{O}$ at room temperature where $V_{T}$ is $25 \mathrm{mV}$ . (b) If the temperature increases by $50^{\circ} \mathrm{C}$ , what will $V_{O}$ become? (c) If, alternatively, the temperature decreases by $50^{\circ} \mathrm{C}$ , what will $V_{O}$ become?

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Question 5

$Figure 6.4.1 The BJTs in the circuit of Fig. 6.4.1 have \beta=100 and \left|V_{C E \text { sat }}\right|=0.3 \mathrm{~V} . Find v_{O} for each of the following cases: (a) v_{I}=0 \mathrm{~V} (b) v_{I}=+5 \mathrm{~V} (c) v_{I}=-5 \mathrm{~V} (d) v_{I}=+10 \mathrm{~V} (e) v_{I}=-10 \mathrm{~V}$ Figure 6.4.1 The BJTs in the circuit of Fig. 6.4.1 have $\beta=100$ and $\left|V_{C E \text { sat }}\right|=0.3 \mathrm{~V}$ . Find $v_{O}$ for each of the following cases: (a) $v_{I}=0 \mathrm{~V}$ (b) $v_{I}=+5 \mathrm{~V}$ (c) $v_{I}=-5 \mathrm{~V}$ (d) $v_{I}=+10 \mathrm{~V}$ (e) $v_{I}=-10 \mathrm{~V}$

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Exam 2: Bipolar Junction Transistors BJTS

(a)The BJTs in the circuits of Fig. 6.2.1 have β=100\beta=100β=100 . (a) For the circuit in Fig. 6.2.1(a), find VE,IBV_{E}, I_{B}VE​,IB​ , and VCV_{C}VC​ . (b) For the circuit in Fig. 6.2.1(b), find IB,ICI_{B}, I_{C}IB​,IC​ , and VCV_{C}VC​ .

The transistor in the circuit of Fig. 6.3.1 has β=\beta=β= 100. Find the values of the four resistors so that VB=+5 V,I1=0.1 mA,IE=1 mAV_{B}=+5 \mathrm{~V}, I_{1}=0.1 \mathrm{~mA}, I_{E}=1 \mathrm{~mA}VB​=+5 V,I1​=0.1 mA,IE​=1 mA , and VC=V_{C}=VC​= +6 V+6 \mathrm{~V}+6 V .

Figure 6.1.1 The BJT in the circuit in Fig. 6.1.1 has β=100\beta=100β=100 . Find VE,VCV_{E}, V_{C}VE​,VC​ , and IBI_{B}IB​ for the three cases: (a) VB=0 VV_{B}=0 \mathrm{~V}VB​=0 V , (b) VB=3 VV_{B}=3 \mathrm{~V}VB​=3 V , and (c) VB=5 VV_{B}=5 \mathrm{~V}VB​=5 V .

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