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Question

Question: In the given circuit what is the maximum value of \( R_S \) for a zener diode to regulate voltage on...

In the given circuit what is the maximum value of RSR_S for a zener diode to regulate voltage on load resistance?

(A)2kΩ\left( A \right)2k\Omega
(B)5kΩ\left( B \right)5k\Omega
(C)6kΩ\left( C \right)6k\Omega
(D)4kΩ\left( D \right)4k\Omega

Explanation

Solution

There is a mistake in the question that the value of load resistance is given in matt it should be in ohm. If calculated by taking its power then the solution will not match with the optio. Hence we will solve by taking load resistance on kiloohm. To satisfy the given condition the voltage across the load must be equal to the enor voltage. Then using ohm’s law in the further step we can calculate the value of RSR_S .

Complete answer:
As per the problem we need to calculate the maximum value of RSR_S for a zener diode to regulate voltage on load resistance.
Now the zener diode to regulate voltage on load resistance we can write it as,
VRL=VZ(1)V_{R_L} = V_Z \ldots \ldots \left( 1 \right)
Where,
The voltage across the load resistance is represent as VRLV_{R_L}
And the voltage across zener diode is represented as VZVZ
Now in place of VRLV_{R_L} we can write,
VRL=IRL(2)V_{R_L} = I_{R_L} \ldots \ldots \left( 2 \right)
Now using ohm's law on current value we will get,
I=VReqI = \dfrac{V}{{\operatorname{R} {\text{eq}}}}
Here the load resistance and RSR_S are connected in series,
Hence the equivalent resistance be,
Req=RS+RL\operatorname{R} {\text{eq}} = R_S + R_L
Now putting the value in the above current equation we will get,
I=VRS+RLI = \dfrac{V}{{R_S + R_L}}
Putting this value in the above equation (2)\left( 2 \right) we will get,
VRL=VRS+RLRL(3)V_{R_L} = \dfrac{V}{{R_S + R_L}}R_L \ldots \ldots \left( 3 \right)
Putting equation (3)\left( 3 \right) in place of equation (1)\left( 1 \right) we will get,
VRS+RLRL=VZ\dfrac{V}{{R_S + R_L}}R_L = V_Z
From the figure we know,
V=30VV = 30V
RL=1kΩRL = 1k\Omega
VZ=6VV_Z = 6V
Now putting these values we will get,
30VRS+1kΩ×1kΩ=6V\dfrac{{30V}}{{R_S + 1k\Omega }} \times 1k\Omega = 6V
Cancelling the terms we will get,
5kΩRS+1kΩ=1\dfrac{{5k\Omega }}{{R_S + 1k\Omega }} = 1
Rearranging the above equation we will egt,
5kΩ=RS+1kΩ5k\Omega = R_S + 1k\Omega
Hence the value of RS=4kΩR_S = 4k\Omega .
Therefore the correct option is (D)\left( D \right) .

Note:
Here the zener diode is also called voltage regulator only reverse biased because it is specially made to have a reverse voltage breakdown at a specific voltage. Note that if the zener diode is biased in the forward direction it behaves like a normal signal passing the current increasing linearly with voltage, now in this case it will not act as a voltage regulator.