Consider the circuit shown in the figure below-Assume that all the transistors are perfectly matched with finite gain an early voltage VA- - and - 25- If the current flowing through the resistance R1 is equal to 10 mA- then the value of load current IL- is approximately equal tomA-

Applying KCL at base of transistor T_3. I=I_C1+I_B3 I_B3=I_E3β+1 I_E3=I_B1+I_B2 All the transistors are matched and β_1=β_2=β_3=β ∴ I_C1=I_C2=β I_B2 (∵ ...

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Consider the ...

Question

Consider the circuit shown in the figure below.

$Assume that all the transistors are perfectly matched with finite gain an early voltage V_{A} = \infty and β = 25 . If the current flowing through the resistance R_{1} is equal to 10 mA. then the value of load current I_{L}, is approximately equal to______mA.$

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Solution

$Applying KCL at base of transistor T_{3} .$

$I = I_{C 1} + I_{B 3}$

$I_{B 3} = \frac{I_{E 3}}{β + 1}$

$I_{E 3} = I_{B 1} + I_{B 2}$

All the transistors are matched and

$β_{1} = β_{2} = β_{3} = β$

$∴ I_{C 1} = I_{C 2} = β I_{B 2} (∵ V_{B E 1} = V_{B E 2}$

$and I_{B 1} = I_{B 2}$

Thus, we have,

$I_{E 3} = 2 I_{B 2}$

$So, I_{B 3} = \frac{2 I_{B 2}}{β + 1}$

$Thus, I = I_{C 1} + \frac{2 I_{B 2}}{β + 1} = I_{C 2} + \frac{2 I_{C 2}}{β (β + 1)}$

$- I_{L} = I_{C 2} = \frac{1}{1 + \frac{2}{β (β + 1)}}$

$- I_{L} = \frac{10 mA}{1 + \frac{2}{25 \times 26}} = 9.97 m A t h u s I_{L} = - 9.97 mA$

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Consider the circuit shown in the figure below. Assume that all the transistors are perfectly matched with finite gain an early voltage VA=∞ and β=25. If the current flowing through the resistance R1 is equal to 10 mA. then the value of load current IL, is approximately equal to______mA.