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Question: The current gain of a common emitter transistor circuit shown in figure is \(120\). Draw the dc load...

The current gain of a common emitter transistor circuit shown in figure is 120120. Draw the dc load line and mark the Q point in it. (VBE{V_{BE}} to be ignored).

Explanation

Solution

The way to solve these types of questions are using the standard formulas of DC Load line and Voltage Divider Bias formulas. We have been given the value of β= 120\beta = {\text{ 120}} . Vcc= 25V{{\text{V}}_{cc}} = {\text{ 25V}}.Now we have to find the Q point using the formula of Vcc = Vce + icRc{V_{cc}}{\text{ = }}{V_{ce}}{\text{ + }}{{\text{i}}_c}{{\text{R}}_c} and Vce= VccicRc{V_{ce}} = {\text{ }}{V_{cc}} - {{\text{i}}_c}{{\text{R}}_c} . We have all the required values given. Whenever not given, assume that Vbe=0.7V{V_{be}} = 0.7V .

Complete step by step answer:
We know that by the formula of voltage divider Bias:
Vbb ibRbVbe=0(1){V_{bb}} - {\text{ }}{i_b}{R_b} - {V_{be}} = 0 - - - - (1)
Thus, we can get base current from the equation:
ib=VbbVbeRb{i_b} = \dfrac{{{V_{bb}} - {V_{be}}}}{{{R_b}}}
Thus, ib=250.7106{i_b} = \dfrac{{25 - 0.7}}{{{{10}^6}}}
Now, we know that collector current is related to base current by the formula:
ic= βib{i_c} = {\text{ }}\beta {i_b}
hence, ic= 120(250.7106){i_c} = {\text{ 120(}}\dfrac{{25 - 0.7}}{{{{10}^6}}})
Now by using Vcc = Vce + icRc {V_{cc}}{\text{ = }}{V_{ce}}{\text{ + }}{{\text{i}}_c}{{\text{R}}_c}{\text{ }}
Vce= VccicRc{V_{ce}} = {\text{ }}{V_{cc}} - {{\text{i}}_c}{{\text{R}}_c}
Thus Vce=25(ic×5×103){V_{ce}} = 25 - ({{\text{i}}_c} \times 5 \times {10^3}).

Now we can easily plot the Q point by using the values of Vce{V_{ce}} and ic{i_c} , by plotting a graph along them , considering $$$$${V_{ce}}astheXaxisandas the X axis and{i_c}astheYaxis.Wehavetoconsideras the Y axis. We have to consider{i_c}iszerowhenwehavetocalculatethevalueofis zero when we have to calculate the value of{V_{ce}}andthevalueofand the value of{V_{ce}}aszerowhenwehavetofindthevalueofas zero when we have to find the value of{i_c}$ . Now we will get a line passing through the first two points and now according to the given value of Vbe{V_{be}} , we have to point to the Q point.

Note: These sums come only with this one type of questions, just find the variables needed and graph it, if the value of Vbe{V_{be}} is given, draw a line according to it, or else take it as 0.70.7 (remember) and move ahead. The point of intersection is very easy to find as almost all values are given in the question. Note that you remember how collector current, base current, collector emf, base emf are all interrelated to each other.