Consider the circuit shown in the figure below-Assuming all the transistors to be matched and working in active region with -1-2-100-3 99 and VBEon-0-7- If the value of I0-100 mA then the value of resistance R1 is equal to -

Now, I_reff=I_Cl+I_B3 and since I_B1=I_B2 and I_0=k_1=k_2 (∵ transistors are matched) nbsp; ∴ I_E3=2I_B2 Also, I_E3=(1 β_3)I_B3 Thus, I_reff=IC1+I_E3(1+β_3)= ...

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Byju's Answer

Standard XII

Physics

Transistor: Working

Consider the ...

Question

Consider the circuit shown in the figure below:

Assuming all the transistors to be matched and working in active region with $β = β_{1} = β_{2} = 100, β_{3} 99$ and $V_{B E (o n)} = 0.7$ . If the value of $I_{0} = 100 m A$ then the value of resistance $R_{1}$ is equal to ___ $Ω$ .

86

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Solution

The correct option is A 86

Now, $I_{r e f f} = I_{C l} + I_{B 3}$
and since $I_{B 1} = I_{B 2}$ and $I_{0} = k_{1} = k_{2}$ ( $∵$ transistors are matched)
$∴ I_{E 3} = 2 I_{B 2}$
Also, $I_{E 3} = (1 - β_{3}) I_{B 3}$
Thus, $I_{r e f f} = I C 1 + \frac{I_{E 3}}{(1 + β_{3})} = I_{C 1} + \frac{2 I_{B 2}}{(1 + β_{3})}$
We can write is as, $I_{r e f f} = I_{C 2} + \frac{2 I_{C 2}}{β (1 + β_{3})} = I_{0} [1 + \frac{2}{β (1 + β_{3})}]$

$∴ I_{r e f f} = 100 [1 + \frac{2}{100 \times 100}] \approx 0.1 A$
Now,
$V^{'} = V_{B E 5} + V_{B e 2} = 0.7 + 0.7 = 1.4 V$

$∴ R_{1} = \frac{10 - 1.4}{0.1} = 86 Ω$

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Consider the circuit shown in the figure below: Assuming all the transistors to be matched and working in active region with β=β1=β2=100,β399 and VBE(on)=0.7. If the value of I0=100 mA then the value of resistance R1 is equal to ___Ω. 86

Consider the circuit shown in the figure below:

Assuming all the transistors to be matched and working in active region with $β = β_{1} = β_{2} = 100, β_{3} 99$ and $V_{B E (o n)} = 0.7$ . If the value of $I_{0} = 100 m A$ then the value of resistance $R_{1}$ is equal to ___ $Ω$ .

86