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Find VE and IE for the circuit given below
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written 3.5 years ago by |
Answer:
Given : $R_B=270\,k\Omega,\,\,R_E=1.5\,k\Omega,\,\,V_{EE}=5V,\,\,\beta=100$
To Find: $V_E=?,\,\,I_E=?$
Solution:
Applying Kirchhoff's voltage law to the input circuit of the above figure, will result in
$-I_BR_B-V_{BE}-I_ER_E+V_{EE}=0$
but $I_E=(\beta+1)I_B$
and $V_{EE}-V_{BE}-(\beta+1)I_BR_E-I_BR_B=0$
with $I_B=\dfrac{V_{EE}-V_{BE}}{R_B+(\beta+1)R_E}\,\,\,\,\,\,.....(1)$
Substituting values to the equation (1) yields,
$I_B=\dfrac{5\,V-0.7\,V}{270\,k\Omega+(101)(1.5\,k\Omega)}$
$=\dfrac{4.3\,V}{270\,k\Omega+151.5\,k\Omega}=\dfrac{4.3\,V}{421.5\,k\Omega}$
$=10.2\,\mu A$
Hence,
$I_E=(\beta+1)I_B=(101)(10.2\,\mu A)$
$=1.03\,mA$
$\therefore \,\,V_E=I_ER_E=(1.03\,mA)(1.5\,k\Omega)=1.545\,V$
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