Consider the circuit shown in the figure below-Both the transistors Q1 and Q2 are matched with - 50- The value of VBE - 0-7 V- the value of current I0 is equal to mA

The circuit can be redrawn as For current mirror circuit, I_reff =V_CC V_BER = 10 0.737× 10^3 = 0.251 mA Now, I_G= I_reff ( 1+2β )= 0.2511+250=0.241 mA ...

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Consider the circuit shown in the figure below:

Both the transistors $Q_{1}$ and $Q_{2}$ are matched with $β = 50$ . The value of $V_{B E} = 0.7 V$ , the value of current $I_{0}$ is equal to ____ mA
0.241

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Solution

The correct option is A 0.241
The circuit can be redrawn as

For current mirror circuit,

$I_{r e f f} = \frac{V_{C C} - V_{B E}}{R} = \frac{10 - 0.7}{37 \times 10^{3}} = 0.251 m A$

Now,

$I_{G} = \frac{I_{r e f f}}{(1 + \frac{2}{β})} = \frac{0.251}{1 + \frac{2}{50}} = 0.241 m A$

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Consider the circuit shown in the figure below: Both the transistors Q1 and Q2 are matched with β=50. The value of VBE=0.7 V, the value of current I0 is equal to ____ mA0.241

The correct option is A 0.241The circuit can be redrawn as For current mirror circuit, Ireff=VCC−VBER=10−0.737×103=0.251 mA Now, IG=Ireff(1+2β)=0.2511+250=0.241 mA

Consider the circuit shown in the figure below:

Both the transistors $Q_{1}$ and $Q_{2}$ are matched with $β = 50$ . The value of $V_{B E} = 0.7 V$ , the value of current $I_{0}$ is equal to ____ mA
0.241

The correct option is A 0.241
The circuit can be redrawn as

For current mirror circuit,

$I_{r e f f} = \frac{V_{C C} - V_{B E}}{R} = \frac{10 - 0.7}{37 \times 10^{3}} = 0.251 m A$

Now,

$I_{G} = \frac{I_{r e f f}}{(1 + \frac{2}{β})} = \frac{0.251}{1 + \frac{2}{50}} = 0.241 m A$