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 Home Electronics lessons General electronics Voltage divider

 Voltage divider

# Voltage divider

In electronics, a voltage divider is a simple device designed to create a voltage (Vout) which is proportional to another voltage (Vin). It is commonly used to create a reference voltage, and may also used as a signal attenuator at low frequencies. Voltage dividers are also known by the terms resistor divider and potential divider.

## Resistor divider

A voltage divider referenced to ground is created by connecting two resistors as shown in the following diagram:

The output voltage Vout is related to Vin as follows:

$V_\mathrm{out} = \frac{R_2}{R_1+R_2} \cdot V_\mathrm{in}$

It may be useful to note that R1 and R2 may each be comprised of many resistors in series.

As a simple example, if R1 = R2 then

$V_\mathrm{out} = \frac{1}{2} \cdot V_\mathrm{in}$

As a more specific and/or practical example, if Vout=6V and Vin=9V (both commonly used voltages), then:

$\frac{V_\mathrm{out}}{V_\mathrm{in}} = \frac{R_2}{R_1+R_2} = \frac{6}{9} = \frac{2}{3}$

and by solving using algebra, R2 must be twice the value of R1.

Any ratio between 0 and 1 is possible. That is, using resistors alone it is not possible to either reverse the voltage or increase Vout above Vin

## Voltage divider as a voltage source

While voltage dividers may be used to produce very precise reference voltages, they make very poor voltage sources. This is because if a load is connected between the output voltage and ground the effective resistance between Vout and ground decreases. A change in the resistance of R2 changes the load voltage, an undesirable situation for a voltage source.

In terms of the above equation, if current flows into a load resistance (through Vout), that load resistance must be considered in parallel with R2 to determine the voltage at Vout. In this case, the voltage at Vout is calculated as follows:

$V_\mathrm{out} = \frac{R_2 \| R_\mathrm{L}}{R_1+R_2 \| R_\mathrm{L}} \cdot V_\mathrm{in} = \frac{R_2}{R_1+R_2+\frac{R_1R_2}{R_\mathrm{L}}} \cdot V_\mathrm{in}$

where RL is a load resistor in parallel with R2.

Note that for high impedence loads it is possible to use a voltage divider as a voltage source, as long as R1 and R2 have very small values compared to the load. This technique is rarely used, as the power disipated in such a divider would be considerable.

## Use of voltage dividers

Voltage dividers are often used to produce stable reference voltages. These reference voltages may be used at a device with a high input impedence, such as an op-amp without fear of loading the divider. Alternatively, the reference voltage may be used to set the voltage being produced by a voltage source. A simple way of doing this (for low power applications) is to simply input the reference voltage into the non-inverting input of an op-amp buffer.

## Impedance divider

A voltage divider is usually thought of as two resistors, but for electronics signals at a given frequency capacitors, inductors, or any combined impedance can be used. For general impedances Z1 and Z2, the voltage becomes

$V_\mathrm{out} = \frac{Z_2}{Z_1+Z_2} \cdot V_\mathrm{in}$

For instance, a divider can be made with a resistor and capacitor:

The resistor's impedance is simply its resistance:

ZR = R

The capacitor's impedance is a large resistance at low frequencies and a low resistance at high frequencies. The exact formula is:

$Z_\mathrm{C} = {1 \over j \omega C}$

where j is the imaginary unit, and ω is frequency in radians per second. This divider will then have the voltage ratio:

${V_\mathrm{out} \over V_\mathrm{in}} = {Z_\mathrm{C} \over Z_\mathrm{C} + Z_\mathrm{R}} = {{1 \over j \omega C} \over {1 \over j \omega C} + R} = {1 \over 1 + R j \omega C}$

The ratio then depends on frequency, in this case decreasing as frequency increases. This circuit is, in fact, a basic (first-order) lowpass filter, or, in the world of audio, a treble-cut filter. The ratio contains an imaginary number, and actually contains both the amplitude and phase shift information of the filter. To extract just the amplitude ratio, calculate the magnitude of the ratio, or just use the reactance of the capacitor instead of the impedance.

## References

• Paul Horowitz and Winfield Hill, The Art of Electronics, Cambridge University Press, 1989.