Question

(13 points) For the transistor in the circuit shown below, VBE(on) = 0.7 V, Vce(sat) = 0.2 V, and B = 100. Determine the value of the collector current. +5 V R=1.2 ko R =33 k2 -5 V

Answer 1

  Given

Solution

• 
  VBE(on) = 0.7V
  
• $\large V_{CE}(sat)=0.2V$
• $\large \beta=100$

The circuit is:

First of all we have to find the region of operation the BJT.

45V Re=1.2 km IC Rg=33 kr VE IB + VBE -5V

To find the region of operation we have to first find the value of  $\large V_{CE}$ using KVL in the loop of base and emitter .

• If $\large V_{CE} \leq V_{CE}(sat.)$

Then the  BJT will  be in saturation region .

And if the BJT is working in saturation region then we will apply KVL in the outer loop using  $\large V_{CE} = V_{CE}(sat.)=0.2V$

to find the collector current $\large I_c$ .

#To find the region of operation:-

Applying KVL in the loop of base and emitter:

$\large 0-I_BR_B-V_{BE}-(-5)=0$

$\large =\gg I_B\times33\times10^3=5-0.7$

$\large =\gg I_B=\frac{5-0.7}{33\times10^3}\;\;A$

$\large =\gg I_B=\frac{4.3\times10^{-3}}{33}\;\;A$

=> IB = 130.3 x 10-6 A

$\large =\gg {\color{Blue} I_B=130.3}\;\;\mu A$

Since we know that $\large I_C=\beta I_B$

$\large =\gg I_C=100\times 130.3\times10^{-6}\;\;A$

$\large =\gg {\color{Blue} I_C=13.03\times10^{-3}}\;\;A$

Now applying  KVL in the outer  loop :

$\large 5-I_CR_C-V_{CE}-(-5)=0$

$\large =\gg 5-1.2\times10^3\times13.03\times10^{-3}-V_{CE}+5=0$

$\large =\gg V_{CE}=10-15.636\;\;\;V$

$\large =\gg {\color{Blue} V_{CE}=-5.636}\;\;V$

Since $\large =\gg V_{CE}=-5.636\;\;\;V<V_{CE}(sat.)=0.2$

$\large {\color{Blue} V_{CE}<V_{CE}(sat.)}$

So, the BJT is working in the saturation region.

Now apply KVL in the outer loop  to find  $\large I_C$

since BJT is working in saturation region so to find I_C the value of  $\large V_{CE}=V_{CE}(sat.)=0.2\;\;V$

So, the KVL equation in the outer loop :

$\large 5-I_CR_C-V_{CE}(sat.)-(-5)=0$

$\large =\gg 5-I_C\times1.2\times10^3-0.2+5=0$

$\large =\gg I_C=\frac{10-0.2}{1.2\times10^3}\;\;\;A$

$\large =\gg I_C=\frac{9.8}{1.2}\times10^{-3}\;\;\;A$

$\large =\gg {\color{Blue} I_C=8.166\times10^{-3}}\;\;\;A{\color{Red} .........Answer}$

$\large {\color{Red} OR}$

$\large {\color{Blue} I_C=8.166}\;\;\;mA{\color{Red} .........Answer}$