Voltage Divider Networks

Resistances in series produce ratios of voltages, and these ratios can be tailored to meet certain needs by means of voltage divider networks.

When a voltage divider network is designed and assembled, the resistance values should be as small as possible without causing too much current drain on the battery or power supply. (In practice, the optimum values depend on the nature of the circuit being designed. This is a matter for engineers, and specific details are beyond the scope of this course.) The reason for choosing the smallest possible resistances is that, when the divider is used with a circuit, you do not want that circuit to upset the operation of the divider. The voltage divider “fixes” the intermediate voltages best when the resistance values are as small as the current-delivering capability of the power supply will allow.

General arrangement for a voltage-divider circuit. Illustration for Quiz Questions 19 and 20.

Above figure illustrates the principle of voltage division. The individual resistances are R1, R2, R3, . . . , Rn. The total resistance is R = R1 + R2 + R3 + . . . + Rn. The supply voltage is E, and the current in the circuit is therefore I = E/R. At the various points P1, P2, P3, . . . , Pn, the potential differences relative to the negative battery terminal are E1, E2, E3, . . . , En, respectively. The last voltage, En, is the same as the battery voltage, E. All the other voltages are less than E, and ascend in succession, so that E1 < E2 < E3 < . . . < En. (The mathematical symbol < means “is less than.”) The voltages at the various points increase according to the sum total of the resistances up to each point, in proportion to the total resistance, multiplied by the supply voltage. Thus, the voltage E1 is equal to ER1/R. The voltage E2 is equal to E(R1 + R2)/R. The voltage E3 is equal to E(R1 + R2 + R3)/R. This process goes on for each of the voltages at points all the way up to En = E(R1 + R2 + R3 + . . . + Rn)/R = ER/R = E.

Problem 1

Suppose you are building an electronic circuit, and the battery supplies 9.0 V. The minus terminal is at common (chassis) ground. You need to provide a circuit point where the dc voltage is +2.5 V. Give an example of a pair of resistors that can be connected in a voltage divider configuration, such that +2.5 V appears at some point.


Examine the schematic above diagram. There are infinitely many different combinations of resistances that will work here! Pick some total value, say R = R1 + R2 = 1000 Ω. Keep in mind that the ratio R1:R will always be the same as the ratio E1:E. In this case, E1 = 2.5 V, so E1:E = 2.5/9.0 = 0.28. This means that you want the ratio R1:R to be equal to 0.28. You have chosen to make R equal to 1000 Ω. This means R1 must be 280 Ω in order to get the ratio R1:R = 0.28. The value of R2 is the difference between R and R1. That is 1000 − 280 = 720 Ω. In a practical circuit, you would want to choose the smallest possible value for R. This might be less than 1000 Ω, or it might be more, depending on the nature of the circuit and the currentdelivering capability of the battery. It’s not the actual values of R1 and R2 that determine the voltage you get at the intermediate point, but their ratio.

Problem 2

What is the current I, in milliamperes, drawn by the entire network of series resistances in the situation described in Problem 1 and its solution?
Use Ohm’s Law to get I = E/R = 9.0/1000 = 0.0090 A = 9.0 mA.

Problem 3

Suppose that it is all right for the voltage divider network to draw up to 100 mA of current in the situation shown by problem 1 figure and posed by Problem 1. You want to design the network to draw this amount of current, because that will offer the best voltage regulation for the circuit to be operated from the network. What values of resistances R1 and R2 should you use?

Calculate the total resistance first, using Ohm’s Law. Remember to convert 100 mA to amperes! That means you use the figure I = 0.100 A in your calculations. Then R = E/I = 9.0/0.100 = 90 Ω. The ratio of resistances that you need is R1:R2 = 2.5/9.0 = 0.28. You should use R1 = 0.28 × 90 = 25 Ω. The value of R2 is the difference between R and R1. That is, R2 = R − R1 = 90 − 25 = 65 Ω.