# Voltage Divider Supply

Whatsapp
A voltage divider is a simple circuit that can reduce voltage. It distributes the input voltage among the components of the circuit. The best example of a voltage divider is two resistors connected in series, with the input voltage applied across the resistor pair and the output voltage taken from a point between them.
Voltage divider in a voltage regulator for adjustment of output voltage.
When the term loaded is used to describe voltage divider supply, it refers to the application of an element, network, or system to a supply that draws current from the supply.
In other words, the loading down of a system is the process of introducing elements that will draw current from the system. The heavier the current, the greater is the loading effect. Recall from internal resistance of voltage source that the application of a load can affect the terminal voltage of a supply due to the internal resistance.
Fig. 1: Voltage divider supply.

Through a voltage divider network such as that in [Fig .1], a number of different terminal voltages can be made available from a single supply. Instead of having a single supply of $120 V$, we now have terminal voltages of $100 V$ and $60 V$ available - a wonderful result for such a simple network. However, there can be disadvantages. One is that the applied resistive loads can have values too close to those making up the voltage divider network.
In general,
for a voltage divider supply to be effective, the applied resistive loads should be significantly larger than the resistors appearing in the voltage divider network.
To demonstrate the validity of the above statement, let us now examine the effect of applying resistors with values very close to those of the voltage divider network.

In [Fig. 2], resistors of $20 Ω$ have been connected to each of the terminal voltages. Note that this value is equal to one of the resistors in the voltage divider network and very close to the other two.
Fig. 2: Voltage divider supply with loads equal to the average value
of the resistive elements that make up the supply.
Voltage $V_a$ is unaffected by the load $R_{L_1}$ since the load is in parallel with the supply voltage E. The result is $V_a = 120 V$, which is the same as the no-load level. To determine $V_b$, we must first note that $R_3$ and $R_{L_3}$ are in parallel and
$$R'_3 = R_3 || R_{L_3} = 30 Ω || 20 Ω= 12 Ω.$$
The parallel combination gives
$$R'_2 = (R_2 + R'_3) || R_{L_2} = (20 Ω + 12 Ω) || 20 Ω$$
$$= 32 Ω || 20 Ω = 12.31 Ω$$
Applying the voltage divider rule gives
$$V_b = {(12.31 Ω)(120 V) \over 12.31 Ω + 10 Ω} = 66.21 V$$
versus $100 V$ under no-load conditions.
Voltage $Vc$ is
$$Vc = {(12 Ω)(66.21 V) \over 12 Ω + 20 Ω} = 24.83 V$$
versus $60 V$ under no-load conditions.
The effect of load resistors close in value to the resistor employed in the voltage divider network is, therefore, to decrease significantly some of the terminal voltages. If the load resistors are changed to the $1 kΩ$ level, the terminal voltages will all be relatively close to the no-load values. The analysis is similar to the above, with the following results:
$$V_a = 120 V$$
$$V_b = 98.88 V$$
$$V_c = 58.63 V$$
If we compare current drains established by the applied loads, we find for the network in [Fig. 2] that
$$I_{L_2} = {V_{L_2} \over R_{L_2}} ={66.21 V \over 20 Ω }= 3.31 A$$
and for the 1 kΩ level,
$$I_{L_2} = {98.88 V \over 1 kΩ} = 98.88 mA < 0.1 A$$
As demonstrated above, the greater the current drain, the greater is the change in terminal voltage with the application of the load.

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250