Voltage Divider : What is it?
Voltage divider is a common, very commonly used Electronic Circuit. Voltage division has various applications, it works on very simple principal but is highly efficient. High Efficiency makes it more desirable to use.
Working Principal of Voltage Divider:
Voltage divider works on a Simple Principal, if two resistors are connected in series across a voltage source there is a division of the total voltage. This division factor is completely dependent on value of resistances. If both the resistors are of same value, Voltage at Junction of both resistors is half of voltage source.
Above figure illustrates basic voltage divider circuit. The output of above circuit(Vo) is:
Vo= (R2/(R1+R2)) * V
If R1 & R2 are equal resistances, Vo= V/2.
Applications of a Voltage Divider:
Voltage Divider is very commonly used in many electronic circuits. It finds application wherein a specific voltage is only to be used from Supply Voltage. Eg 5V is required by the supply is of 10V. It is used in Transistor Biasing, A-D Converters for Reference Voltage Levels, HV Instrumentation like HV Voltmeters and many more.
Note: Voltage Divider can not be used as a DC Supply Source. As it can not provide high current which a DC Source provides. It has poor Load Regulation, that is if you sink more current through Voltage Divider Voltage Levels drop significantly.
So Voltage Dividers can only be used where current requirement is not more than few mA. Eg: Transistor Biasing,Instrumentation Systems, As reference voltage levels.
Designing a Voltage Divider:
In order to design a voltage divider we need to know only two basic things,Supply voltage and Required Voltage.
Lets Solve an Example wherein an Instrumentation system needs to measure 0-1000V. 1000V/1kV is a very high potential and no meter can sustain such large potential difference. So we need to attenuate/ convert 0-1000V to 0-10V in a well defined ratio. Once conversion is done we can use any meter with full scale deflection of 10V
Step 1: Vo is 10 V & Vin is 1000 V
Select any Resistance R1, such that R1>> R2
Let R1= 1MΩ
∴ 10 = (R2/1MΩ+R2) * 1000
∴1/100 = R2/ 1MΩ+R2
∴ R2= 1MΩ/99
Now two resistors R1= 1MΩ & R2= 10kΩ can be used to implement above problem.
But, Power Dissipation needs to be calculated for selecting a Resistor, improper selection may lead to over heating and damaging the circuit.
Rt= R1+R2= 1.1MΩ
I = 1000/1.1MΩ = 1mA
P= V * I = 1000 * 1 mA = 1 W
As power dissipation is 1W resistors with power rating more than 1 W should be used.
If power rating is more than 1W, resistors get bulky and are not found in local markets, in such a problem you can use 3 resistors and take output Vo from junction of second and third resistor.
In above figure 3 resistors are used instead of 2, if we use R1 & R2 of two 1 MΩ each in above example
Consider, R1& R2 as single resistor in series, so R3= 2MΩ/99 =20kΩ
Rt= 2.2MΩ, I= 1000/2.2MΩ = 0.5 mA
So now, P = 1/2 W
Resistors of 1/2 W are commonly available and can be used in your application. N number of resistors can be used in this way to handle high currents.