Consider the diode circuit shown in the figure below-The diode has a cut in voltage V-0-7 V- otherwise ideal- If an input voltage of 10 V is applied to the circuit then the value of VoV1 is equal to mV-V

Drawing the equivalent model we get, ∴ I=10 ( 5) 0.74.7+2.2× 10^ 3A=2.072 mA Now, V_1= IR= 2.072× 4.7= 9.74 V and V_0= 5+V_2 = 5+IR_2 = 5+2.072× 2.2= 0.44 V ...

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Question

Consider the diode circuit shown in the figure below:

The diode has a cut in voltage $V_{γ} = 0.7 V,$ otherwise ideal. If an input voltage of $10 V$ is applied to the circuit then the value of $\frac{V_{o}}{V_{1}}$ is equal to ____ $m V / V$
45.17

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Solution

The correct option is A 45.17
Drawing the equivalent model we get,

$∴ I = \frac{10 - (- 5) - 0.7}{4.7 + 2.2} \times 10^{- 3} A = 2.072 m A$
Now, $V_{1} = - I R = - 2.072 \times 4.7 = - 9.74 V$ and $V_{0} = - 5 + V_{2}$
$= - 5 + I R_{2}$
$= - 5 + 2.072 \times 2.2 = - 0.44 V$
$∴ \frac{V_{0}}{V_{1}} = \frac{- 0.44}{- 9.74} = 45.17 m V / V$

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Consider the diode circuit shown in the figure below: The diode has a cut in voltage Vγ=0.7 V, otherwise ideal. If an input voltage of 10 V is applied to the circuit then the value of VoV1 is equal to ____ mV/V45.17

The correct option is A 45.17Drawing the equivalent model we get, ∴ I=10−(−5)−0.74.7+2.2×10−3A=2.072 mA Now, V1=−IR=−2.072×4.7=−9.74 V and V0=−5+V2 =−5+IR2 =−5+2.072×2.2=−0.44 V ∴ V0V1=−0.44−9.74=45.17 mV/V

Consider the diode circuit shown in the figure below:

The diode has a cut in voltage $V_{γ} = 0.7 V,$ otherwise ideal. If an input voltage of $10 V$ is applied to the circuit then the value of $\frac{V_{o}}{V_{1}}$ is equal to ____ $m V / V$
45.17