A simple NPN bipolar transistor circuit that consists of an NPN bipolar

A simple npn bipolar transistor circuit that consists

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A simple NPN bipolar transistor circuit that consists of an NPN bipolar transistor, R B and R C is given in Fig. 2.25 . If V CC = 15 V, V IN = 5 V and b = 100, calculate I B , I C . What mode will this transistor operate in? The fi rst task is to write Kirchoff s Voltage Law (KVL) around the BE and CE junctions as shown in Eqs. 2.6 and 2.7 . Therefore, V IN ¼ R B I B þ V BE ð 2 : 6 Þ V CC ¼ R C I C þ V CE ð 2 : 7 Þ Test for cut-off mode: Set I B = I C = 0 A. This produces V BE = 5 V and V CE = 15 V, and it proves that the transistor does not operate in cut-off region. Test for active mode, set V BE = 0.7 V and solve I B from Eq. 2.6 . Thus, I B ¼ ð V IN ± V BE Þ R B ¼ ð 5 ± 0 : 7 Þ 10 K X ¼ 0 : 43 mA Then, I C = b I B = 100 (0.43 mA) = 43 mA. Substitute I C = 43 mA into Eq. 2.7 and solve for V CE . Thus, V CE = V CC - R C I C = 15 - 0.1 K Ω (43 mA) = 15 - 4.3 = 10.7 V This value is greater than 0.2 V and proves that the transistor operates in the active region as shown in Fig. 2.26 . In this fi gure, Eq. 2.7 is superimposed on top of the transistor I C -V CE characteristics in order to show where exactly this transistor operates at (quiescent point, Q). V CC = 15V R C = 100Ω V CE V IN = 5V V BE I B I C R B = 10KΩ Fig. 2.25 A simple NPN bipolar circuit 46 2 Diode and Bipolar Transistor Circuits
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However, the location of the Q point is not optimal in this fi gure, and as it needs to be closer to the midpoint between 0 and V CC such as V CEQ = V CC /2 = 7.5 V to be more suitable for analog applications. From Fig. 2.26 , when I B = 0.43 mA and V CE = 7.5 V, I C becomes 43 mA. Thus, R C ¼ ð V CC ± V CEQ Þ I C ¼ ð 15 ± 7 : 5 Þ 0 : 043 ¼ 174 X instead of 100 X : The new load line is plotted in Fig. 2.26 with dashed lines. Example 2.2 Another bipolar transistor circuit is given in Fig. 2.27 . In this schematic, I C is selected to be 20 mA. The circuit still uses V IN = 5 V and V CC = 15 V for its operation. What are the values of R B and R C such that this transistor operates in the middle of the active region at V CEQ = 7.5 V? I C V CE 0 I B = 0.43mA 15V Q 10.7V I C = 43mA I CMAX = 15V/100Ω = 150mA V CE = V CC R C I C (load line) 7.5V R C = 100Ω R C = 174Ω Fig. 2.26 The quiescent point of the transistor V CC = 15V R C V CE V IN = 5V V BE I B I C = 20mA R B Fig. 2.27 A bipolar circuit whose I C = 20 mA, but its R C and R B are unknown 2.7 Bipolar Transistor Circuits 47
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For this design question, one must refer to the manufacturer s datasheet and fetch the transistor I C -V CE characteristics as shown in Fig. 2.28 . The load line from Eq. 2.7 is superimposed on top of the I C -V CE characteristics, and I B becomes 2 mA at V CEQ = 7.5 V when I C = 20 mA. Since the transistor operates in active region, b = I C /I B becomes 10. Rewriting Eq. 2.7 yields: R C ¼ ð V CC ± V CEQ Þ I C ¼ ð 15 ± 7 : 5 Þ 0 : 020 ¼ 375 X Rewriting Eq. 2.6 yields R B ¼ ð V IN ± V BE Þ I B ¼ ð 5 ± 0 : 7 Þ 0 : 002 ¼ 2 : 15 K X However, there may be an instance where the load line may not necessarily intersect any of the I C -V CE curves at V CEQ = 7.5 V provided by the manufacturer, and the situation becomes similar to the case in Fig. 2.29 . In this instance, R C is still calculated using Eq. 2.7 since V CEQ and I CQ have not changed.
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