Problems solving in electrical engineering (Theoretical Foundations of Electrical Engineering), circuit analysis (Foundations of Circuit Analysis) and electronics
Main page|| Examples|| Order the problems solution|| Prices|| Guarantees||Cooperation||Contact information
Theoretical foundations of electrical engineering

Solving problems in electrical engineering – Voltage divider

The scheme of the resistance voltage divider:

resistance-voltage divider(1)

Simple resistance-voltage divider consists of two series-connected resistors R1 and R2, connected to the voltage source U. Since the resistors are connected in series, the current through them will be the same according to the first Kirchhoff’s law. The voltage drop (potential drop at the flow of the charge from one point to another point of the circuit) on each resistor according to the Ohm’s law is proportional to the resistance (the current, as stated earlier, is the same):

U=I·R;

For each resistor:

U1=I·R1

U2=I·R2

Having divided the expression for U1 by the expression for U2 we receive:

U1/U2=R1/R2;

Thus, the ratio of voltages U1 и U2 are exactly equal to the ratio of resistances R1 и R2.

Using the equality:

U=U1+U2

We receive the formula connecting output (U2) and input (U) voltage divider:

U2=U·R2/(R2+R1)

It should be noted that the load impedance of the voltage divider should be much more than own resistance of the divider, so that in calculations this resistance which is parallel to R2 can be neglected. To select specific resistances values in practice is usually enough to follow the following algorithm. First it is necessary to define the value of the current divider working without load. This current should be significantly more than current (usually accept excess from 10 times in size) consumed by the load, but, however, specified current should not create surcharge on the voltage source. Based on the current value, we define the value of the overall resistance according to the Ohm’s law R=R1+R2. It is necessary only to take exact values of resistances from the standard series which ratio values are close to the desired voltages ratio and the amount of values is close to the calculated amount. At calculating of the real divider it is necessary to take into account the temperature coefficient of the resistance, tolerance on nominal resistance values, variation range of the input voltage and possible properties changes of the load divider, as well as the maximum dissipated power of resistors – it should exceed the power emitted on them P=I2·(R1+R2), where I – current source without load (in this case the maximum current flows through resistors).

Problems solving in electrical engineering online
Copyright © 2012-2018 ei-help.com. All rights reserved.
->