My R L is giving me the wrong current across the transistor. Help would be appre

Question

My RL is giving me the wrong current across the transistor. Help would be appreciated.

Design a PNP current source for Ic = 1 mA, similar to the NPN current source shown, using a PNP (2N3906) transistor, assume beta = 100. Use the same procedure as above. Remember that while the NPN will sink the current through the load, the PNP should source the current through the load (RL). Determine the RLmax and choose an RL as you did above. Find maximum value for RLmax with supply Vcc=10Vdc. VEC = 0.2V and VE = VB + VEB for this calculation of RLmax maximum. Choose a value for RL that is less than 0.8RLmax and larger than 0.3RLmax. Calculate the voltage Vice with the load resistor at Vcc= 10Vdc

Explanation / Answer

1. Ic =Ie*beta/(beta+1)   that is Ie =101/100*Ic=1.01 mA

Ie = Ue/Re =(Ub-Ube)/Re   that is Re= (Ub-Ube)/Ie =(1.8-0.65)/1.01 m =1.139 K

Ib =Ic/beta =1 m/100 =10 microA

Ir1 =10*Ib =100 mcroA =0.1 mA

R1 =(Vcc-1.8)/0.1 m=(10-1.8)/0.1 m =82 K

Ve =Vb-Vbe =1.8-0.65 =1.15 V

RLmax = (Vcc-Vce -Ve)/Ic =(10-0.2-1.15)/1 m=8.65 K

take RL =0.6*RLmax =5.19 K   standard value RL =5 K

Vce = Vcc-Ic*RL-Ve =10-1m*5 K -1.15 =3.85 V

For the second question:

A rule of electronic design is that a circuit with a NPN transistor is equivalent to a circuit with a (pair) PNP transistor if you reverse the power sorces and reverse the polarity of diodes in circuit. This, since for this PNP circuit beta =100 (the same), diodes are reversed and Vdc =-10 V (reversed), then all the Resistors values are the same as above.

Rb=82 K, Re =1.139 K and RL =5.19 K and Vce =3.85 V

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