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Question 5.17: Calculate the dc bias voltages and currents for the Darlingt...

Calculate the dc bias voltages and currents for the Darlington configuration of Fig. 5.76 .

5.76
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\beta_{D}=\beta_{1} \beta_{2}=(50)(100)= 5 0 0 0

 

I_{B_{1}}=\frac{V_{C C}-V_{B E_{1}}-V_{B E_{2}}}{R_{B}+\beta_{D} R_{E}}=\frac{18 V -0.7 V -0.7 V }{3.3 M \Omega+(5000)(390 \Omega)}

 

=\frac{18 V -1.4 V }{3.3 M \Omega+1.95 M \Omega}=\frac{16.6 V }{5.25 M \Omega}= 3 . 1 6 \mu A

 

I_{C_{2}} \cong I_{E_{\gamma}}=\beta_{D} I_{B_{1}}=(5000)(3.16 mA )=15.80 mA

 

V_{C_{1}}=V_{C_{2}}=18 V

 

V_{E_{2}}=I_{E_{2}} R_{E}=(15.80 mA )(390 \Omega)= 6 . 1 6 V

 

V_{B_{1}}=V_{E_{2}}+V_{B E_{1}}+V_{B E_{2}}=6.16 V +0.7 V +0.7 V =7.56 V

 

V_{C E_{2}}=V_{C C}-V_{E_{2}}=18 V -6.16 V = 1 1 . 8 4 V

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