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The transistor in Figure 4-19 has the following maximum ratings: Pomax) = 800 mW, VCE(max) = 15 V, and lcomax) = 100 mA. Determine the maximum value to which Vec can be adjusted without exceeding a rating. Which rating would be exceeded first? EXAMPLE 4-6 FIGURE 4-19 RLOkn Re Poc 100 Vec 22 k Solution First, find Ig so that you can determine le VBB – VRE 5 V – 0.7 V 22 kn Ic = Boch = (100)(195 µA) = 19.5 mA Iç is much less than Icmax) and ideally will not change with Vcc. It is determined only Is = = 195 µA by /g and Bpc The voltage drop across Re is VR = IRc = (19.5 MAX1.0kfl) = 19.5 V %3D Now you can determine the value of Vec when VCE = VCE(max) = 15 V. 12 VRc = Vcc - VeE So, Vecomax) = VCEmax) + (Vr. = 15 V + 19.5 V = 34.5 V Vcc can be increased to 34.5 V, under the existing conditions, before Vee(max) is exceeded. However, at this point it is not known whether or not Poymax) has been exceeded. Pp = Vemaxy/c = (15 V)(19.5 mA) = 293 mW Since Ppymax) is 800 mW, it is not exceeded when Vcc = 34.5 V. So, VCEimax) = 15 V is the limiting rating in this case. If the base current is removed causing the transistor to tum off, VCE(max) will be exceeded first because the entire supply voltage, Vcc. will be dropped across the transistor. Related Problem The transistor in Figure 4-19 has the following maximum ratings: Pp(max) = 500 mW, VCEmax) = 25 V, and Icimax) = 200 mA. Determine the maximum value to which Vcc can be adjusted without exceeding a rating. Which rating would be exceeded first?

The transistor in Figure 4-19 has the following maximum ratings: Pomax) = 800 mW, VCE(max) = 15 V, and lcomax) = 100 mA. Determine the maximum value to which Vec can be adjusted without exceeding a rating. Which rating would be exceeded first? EXAMPLE 4-6 FIGURE 4-19 RLOkn Re Poc 100 Vec 22 k Solution First, find Ig so that you can determine le VBB – VRE 5 V – 0.7 V 22 kn Ic = Boch = (100)(195 µA) = 19.5 mA Iç is much less than Icmax) and ideally will not change with Vcc. It is determined only Is = = 195 µA by /g and Bpc The voltage drop across Re is VR = IRc = (19.5 MAX1.0kfl) = 19.5 V %3D Now you can determine the value of Vec when VCE = VCE(max) = 15 V. 12 VRc = Vcc - VeE So, Vecomax) = VCEmax) + (Vr. = 15 V + 19.5 V = 34.5 V Vcc can be increased to 34.5 V, under the existing conditions, before Vee(max) is exceeded. However, at this point it is not known whether or not Poymax) has been exceeded. Pp = Vemaxy/c = (15 V)(19.5 mA) = 293 mW Since Ppymax) is 800 mW, it is not exceeded when Vcc = 34.5 V. So, VCEimax) = 15 V is the limiting rating in this case. If the base current is removed causing the transistor to tum off, VCE(max) will be exceeded first because the entire supply voltage, Vcc. will be dropped across the transistor. Related Problem The transistor in Figure 4-19 has the following maximum ratings: Pp(max) = 500 mW, VCEmax) = 25 V, and Icimax) = 200 mA. Determine the maximum value to which Vcc can be adjusted without exceeding a rating. Which rating would be exceeded first?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Solve the related problems in this examples.
EXAMPLE 4-3 Sketch an ideal family of collector curves for the circuit in Figure 4-11 for Ig = 5 µA to 25 µA in 5 µA increments. Assume Bpc = 100 and that VCE does not exceed breakdown. FIGURE 4-11 Re Ic Poc = 100 Vcc Solution Using the relationship lc = Bocls, values of Ic are calculated and tabulated in Table 4-1. The resulting curves are plotted in Figure 4-12. TABLE 4-1 Ic 5 µA 10 μΑ 15 μΑ 0.5 mA 1.0 mA 1.5 mA 20 μΑ 2.0 mA 25 μΑ 2.5 mA Ic (mA) 2.5- = 25 µA 2.0- h = 20 µA 15- 4 = 15 µA 1.0- 4= 10 µA 0.5- = 5 µA Ver o 0,7 V A FIGURE 4-12 Related Problem Where would the curve for Ig 0 appear on the graph in Figure 4-12, neglecting col- lector leakage current?
Transcribed Image Text:EXAMPLE 4-3 Sketch an ideal family of collector curves for the circuit in Figure 4-11 for Ig = 5 µA to 25 µA in 5 µA increments. Assume Bpc = 100 and that VCE does not exceed breakdown. FIGURE 4-11 Re Ic Poc = 100 Vcc Solution Using the relationship lc = Bocls, values of Ic are calculated and tabulated in Table 4-1. The resulting curves are plotted in Figure 4-12. TABLE 4-1 Ic 5 µA 10 μΑ 15 μΑ 0.5 mA 1.0 mA 1.5 mA 20 μΑ 2.0 mA 25 μΑ 2.5 mA Ic (mA) 2.5- = 25 µA 2.0- h = 20 µA 15- 4 = 15 µA 1.0- 4= 10 µA 0.5- = 5 µA Ver o 0,7 V A FIGURE 4-12 Related Problem Where would the curve for Ig 0 appear on the graph in Figure 4-12, neglecting col- lector leakage current?
A certain transistor is to be operated with VCE = 6 V. If its maximum power rating is 250 mW, what is the most collector current that it can handle? EXAMPLE 4-5 Poxmax) 250 mW Ic = VCE Solution = 41.7 mA 6 V This is the maximum current for this particular value of VCE. The transistor can handle more collector current if VCE is reduced, as long as Ppma) and Icimax) are not exceeded. Related Problem If Pp(max) = 1 W, how much voltage is allowed from collector to emitter if the transis- tor is operating with Ie = 100 mA? 11 EXAMPLE 4-6 The transistor in Figure 4-19 has the following maximum ratings: Pp(max) = 800 mW, VCE(max) = 15 V, and Icimat) = 100 mA. Determine the maximum value to which Vcc can be adjusted without exceeding a rating. Which rating would be exceeded first? FIGURE 4-19 Re Boc= 100 22 kn Solution First, find /n so that you can determine le. 5V - 0.7 V 22 k Ic = Bpcls = (100)(195 µA) = 19.5 mA VBB - VBE Ig = = 195 µA Rp Iç is much less than Icmax) and ideally will not change with Vcc. It is determined only by Ig and Bpc- The voltage drop across Rc is VR. = ICRC = (19.5 mA)(1.0 kfN) = 19.5 V Now you can determine the value of Vcc when VCE = VCE(max) = 15 V. 12 VRC = Vcc - VCE So, Vocmax) = VeTimay) + (VR. = 15 V + 19,5 V = 34,5 V Vcc can be increased to 34.5 V, under the existing conditions, before VCE(max) is exceeded. However, at this point it is not known whether or not Pp(max) has been exceeded. Pp = VCE(max)/c = (15 V)(19.5 mA) = 293 mW Since Ppimax) is 800 mW, it is not exceeded when Vcc = 34.5 V. So, VCE(max) = 15 V is the limiting rating in this case. If the base current is removed causing the transistor to turn off, VCEmax) will be exceeded first because the entire supply voltage, Vc. will be dropped across the transistor. Related Problem The transistor in Figure 4-19 has the following maximum ratings: Ppmax) = 500 mW, VCEmax) = 25 V, and Icmax) = 200 mA. Determine the maximum value to which Vcc can be adjusted without exceeding a rating. Which rating would be exceeded first?
Transcribed Image Text:A certain transistor is to be operated with VCE = 6 V. If its maximum power rating is 250 mW, what is the most collector current that it can handle? EXAMPLE 4-5 Poxmax) 250 mW Ic = VCE Solution = 41.7 mA 6 V This is the maximum current for this particular value of VCE. The transistor can handle more collector current if VCE is reduced, as long as Ppma) and Icimax) are not exceeded. Related Problem If Pp(max) = 1 W, how much voltage is allowed from collector to emitter if the transis- tor is operating with Ie = 100 mA? 11 EXAMPLE 4-6 The transistor in Figure 4-19 has the following maximum ratings: Pp(max) = 800 mW, VCE(max) = 15 V, and Icimat) = 100 mA. Determine the maximum value to which Vcc can be adjusted without exceeding a rating. Which rating would be exceeded first? FIGURE 4-19 Re Boc= 100 22 kn Solution First, find /n so that you can determine le. 5V - 0.7 V 22 k Ic = Bpcls = (100)(195 µA) = 19.5 mA VBB - VBE Ig = = 195 µA Rp Iç is much less than Icmax) and ideally will not change with Vcc. It is determined only by Ig and Bpc- The voltage drop across Rc is VR. = ICRC = (19.5 mA)(1.0 kfN) = 19.5 V Now you can determine the value of Vcc when VCE = VCE(max) = 15 V. 12 VRC = Vcc - VCE So, Vocmax) = VeTimay) + (VR. = 15 V + 19,5 V = 34,5 V Vcc can be increased to 34.5 V, under the existing conditions, before VCE(max) is exceeded. However, at this point it is not known whether or not Pp(max) has been exceeded. Pp = VCE(max)/c = (15 V)(19.5 mA) = 293 mW Since Ppimax) is 800 mW, it is not exceeded when Vcc = 34.5 V. So, VCE(max) = 15 V is the limiting rating in this case. If the base current is removed causing the transistor to turn off, VCEmax) will be exceeded first because the entire supply voltage, Vc. will be dropped across the transistor. Related Problem The transistor in Figure 4-19 has the following maximum ratings: Ppmax) = 500 mW, VCEmax) = 25 V, and Icmax) = 200 mA. Determine the maximum value to which Vcc can be adjusted without exceeding a rating. Which rating would be exceeded first?
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