How to Calculate a Voltage Drop Circuit

Written by pauline gill
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How to Calculate a Voltage Drop Circuit
A digital multimeter makes measuring electronic circuits easy. (digital multimeter 3 image by dinostock from

Voltage drop series circuits are used in direct and alternating current (DC and AC) devices to allow lower voltages than those supplied to the circuit to be used for specific purposes within the device. For example, a device with a 24-volt DC battery may have an 18-volt motor, 3.4-volt lights and a 5-volt control logic chip which all need steady supply voltages. The principle that allows calculating the parameters of these circuits is Kirchhoff's Voltage Law which states that the summation of all voltages within a series circuit must equal zero.

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  1. 1

    Define the application. In the example above, a power implement with an 18-volt motor, 3.4-volt LED (light emitting diode) indicator lights and a 5-volt controller chip has a 24-volt rechargeable battery. Current to each of the devices is regulated by transistors that are referenced to a precision voltage drop circuit consisting of three resistors in series with the battery. With a knowledge of the voltage needs, you can calculate this voltage drop reference circuit.

    How to Calculate a Voltage Drop Circuit
    Putting selected resistors into a series circuit provides several levels of voltage drop. (resistors image by Albert Lozano from
  2. 2

    Calculate an overall voltage drop resistance and current flow through the series circuit. As this is a reference voltage circuit, only very small currents are required to regulate the transistors, so set the total circuit resistance at 10,000 ohms (10k-ohm). Solve for current by dividing supply voltage (Vs) by total resistance (Rt). Substituting, 24 volts/10,000 ohms = 2.4 milliamps (0.0024 amps).

    How to Calculate a Voltage Drop Circuit
    Circuit boards use groups of resistors to modify voltage. (circuit board image by dwags from
  3. 3

    Calculate the proportion of total circuit resistance needed for the 3.4-volt reference for the LED lights, since this is the lowest voltage, requiring the lowest resistance. The equation will be R lights = Rt(10k-ohms) X 3.4-volt lights/24-volt supply = 1416 ohms (1.416 k-ohms).

  4. 4

    Calculate the proportion of the total circuit resistance needed for the 5-volt logic chip. The equation will be R chip = Rt(10k-ohms) X 5-volt chip/24-volt supply = 2,083 ohms (2.083 k-ohms). Subtracting the first 1.416k-ohms resistor for the lights, all that is needed is a 2.083 minus 1.416 = 0.667k ohm or a 667-ohm resistor to provide the 2.083k-ohm series resistance.

  5. 5

    Calculate the proportion of the total circuit resistance that will be needed to provide the 18-volt reference voltage for the motor. This will be the highest resistance required and is solved by the equation R motor = Rt(10k-ohms) X 18-volt motor/24-volt supply = 10 k-ohm X 0.75 = 7.5 k-ohms. Subtracting the 2.083 k-ohm resistance you already have, this will be a 7.5 k minus 2.083 k = 5.417 k-ohm resistor.

  6. 6

    Calculate the summation of all three resistors which constitute a proportion of the total voltage drop of 24 volts over 10 k-ohms for the circuit. The equation: 1.416 k-ohms for the lights + 0.667 k-ohms for the chip + 5.417 k-ohms for the motor = 7.5 k-ohms total.

  7. 7

    Determine the value of the final resistor that will bring total resistance up to 10 k-ohms by subtracting 7.5 k ohms from 10 k-ohms = 2.5 k-ohms (2,500 ohms).

  8. 8

    Determine the intermediate voltages between each of the resistors in series (1.416 k + 0.667 k + 5.417 k + 2.5 k = 10 k compared to the negative of the 24-volt supply (-24-volts). This results in voltage readings of +3.4-volts, +5-volts, +18-volts and +24-volts respectively, and the circuit resolves to zero per Kirchhoff's Voltage Law.

Tips and warnings

  • Using high impedance resistor voltage drop circuits conserves power.
  • Installing resistors with too low an overall circuit ohm value can result in too much current flowing, and smoking and burning of electronic components.

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