Desing a circuit of BJT in a voltage divider network with a Vcc=28V with a Transistor TIP41, Icmax=4.5A and have to work in a Q-point of 2/3 of Icmax, Ve=1/4 Vcc. Use commercial value resistors.
Well done Marlyn got your other extra half point and the last one, now you can let your classmate do some of the exersices if they want and you can do something more challenging.
2 comments:
Si usamos unas resistencias de:
RC = 5 Ohms.,
R1 = 100 Ohms. y
R2 = 150 Ohms.
Y si
VCC = 28V.
Icmax = 4.5 A.
2/3Icmax = IC = 3 A.
VE = 1/4VCC = 7V.
Entonces
IB = IC/B = (3 A/30) = 0.1 A.
IE = IB(B+1) = (0.1)(31) = 3.1 A.
RE = VE/IE = (7V/3.1A) = 2.258 Ohms.
VCE = VCC-[(B*IB)*RC]-[(B+1)*IB*RE]
VCE = 28V-[30*0.1*5]-[31*0.1*2.258] = 6.0002V.
RTH = (R1*R2)/(R1+R2)
RTH = (100*150)/(100+150) = 60 Ohms.
VTH = (VCC*R2)/(R1+R2)
VTH = (28*150)/(100+150) = 16.8V.
VB = VTH = 16.8V.
VBE = 0.7V.
VC = VCC-(IC*RC) = 28-(3*5) = 13V.
y es todo. :)
Well done Marlyn got your other extra half point and the last one, now you can let your classmate do some of the exersices if they want and you can do something more challenging.
Congrats
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