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voltage divider: see potentiometer.

The Columbia Electronic Encyclopedia Copyright © 2004.

Licensed from Columbia University Press

Licensed from Columbia University Press

In electronics, a voltage divider (also known as a potential divider) is a simple linear circuit that produces an output voltage (V_{out}) that is a fraction of its input voltage (V_{in}). Voltage division refers to the partitioning of a voltage among the components of the divider.

The formula governing a voltage divider is similar to that for a current divider, but the ratio describing voltage division places the selected impedance in the numerator, unlike current division where it is the unselected components that enter the numerator.

A simple example of a voltage divider consists of two resistors in series or a potentiometer. It is commonly used to create a reference voltage, and may also be used as a signal attenuator at low frequencies.

Applying Ohm's Law, the relationship between the input voltage, V_{in}, and the output voltage, V_{out}, can be found:

- $$

Proof:

- $V\_mathrm\{in\}\; =\; Icdot(Z\_1+Z\_2)$

- $V\_mathrm\{out\}\; =\; Icdot\; Z\_2$

- $I\; =\; V\_mathrm\{in\}cdotfrac\; \{1\}\{Z\_1+Z\_2\}$

- $V\_mathrm\{out\}\; =\; V\_mathrm\{in\}\; cdotfrac\; \{Z\_2\}\{Z\_1+Z\_2\}$

The transfer function (also known as the divider's voltage ratio) of this circuit is simply:

- $$

In general this transfer function is a complex, rational function of frequency.

A resistive divider is a special case where both impedances, Z_{1} and Z_{2}, are purely resistive (Figure 2).

Substituting Z_{1} = R_{1} and Z_{2} = R_{2} into the previous expression gives:

- $$

As in the general case, R_{1} and R_{2} may be any combination of series/parallel resistors.

- $$

As a more specific and/or practical example, if V_{out}=6V and V_{in}=9V (both commonly used voltages), then:

- $$

To solve for R1:

- $$

To solve for R2:

- $$

Any ratio between 0 and 1 is possible. That is, using resistors alone it is not possible to either reverse the voltage or increase V_{out} above V_{in}

Consider a divider consisting of a resistor and capacitor as shown in Figure 3.

Comparing with the general case, we see Z_{1} = R and Z_{2} is the impedance of the capacitor, given by

- $Z\_2\; =\; jX\_\{mathrm\{C\}\}\; =frac\; \{1\}\; \{j\; omega\; C\}$ $,$

where X_{C} is the reactance of the capacitor, C is the capacitance of the capacitor, j is the imaginary unit, and ω (omega) is the radian frequency of the input voltage.

This divider will then have the voltage ratio:

- $$

The product of τ (tau) = RC is called the time constant of the circuit.

The ratio then depends on frequency, in this case decreasing as frequency increases. This circuit is, in fact, a basic (first-order) lowpass 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, that is:

- $left|\; frac\; \{V\_mathrm\{out\}\}\; \{V\_mathrm\{in\}\}\; right|\; =\; frac\; \{1\}\; \{sqrt\; \{\; 1\; +\; (omega\; R\; C\; )^2\; \}\; \}\; .$

Inductive dividers split DC input according to resistive divider rules above.

Inductive dividers split AC input according to inductance:

Vout = Vin * [L2 / (L1 + L2 ) ]

The above equation is for ideal conditions. In the real world the amount of mutual inductance will alter the results.

Capacitive dividers do not pass DC input.

For a AC input a simple capacitive equation is:

Vout = Vin * [C1 / (C1 + C2 ) ] = Voltage across capacitor 2

Capacitive dividers are limited in current by the capacitance of the elements used.

This effect is opposite to resistive division and inductive division.

The following example describes the effect when a voltage divider is used to drive an amplifier:

The gain of an amplifier generally depends on its source and load terminations, so-called loading effects that reduce the gain. The analysis of the amplifier itself is conveniently treated separately using idealized drivers and loads, and then supplemented by the use of voltage and current division to include the loading effects of real sources and loads. The choice of idealized driver and idealized load depends upon whether current or voltage is the input/output variable for the amplifier at hand, as described next. For more detail on types of amplifier based upon input/output variables, see classification based on input and output variables.

In terms of sources, amplifiers with voltage input (voltage and transconductance amplifiers) typically are characterized using ideal zero-impedance voltage sources. In terms of terminations, amplifiers with voltage output (voltage and transresistance amplifiers) typically are characterized in terms of an open circuit output condition.

Similarly, amplifiers with current input (current and transresistance amplifiers) are characterized using ideal infinite impedance current sources, while amplifiers with current output (current and transconductance amplifiers) are characterized by a short-circuit output condition,

As stated above, when any of these amplifiers is driven by a non-ideal source, and/or terminated by a finite, non-zero load, the effective gain is lowered due to the loading effect at the input and/or the output. Figure 3 illustrates loading by voltage division at both input and output for a simple voltage amplifier. (A current amplifier example is found in the article on current division.) For any of the four types of amplifier (current, voltage, transconductance or transresistance), these loading effects can be understood as a result of voltage division and/or current division, as described next.

- $v\_i\; =\; v\_S\; frac\; \{R\_\{in\}\}\; \{R\_S\; +\; R\_\{in\}\}\; ,$

In the same manner, the ideal input current for an ideal driver i_{i} is realized only for an infinite-resistance current driver. For a Norton driver with current i_{S} and source impedance R_{S}, the input current i_{i} is reduced from i_{S} by current division to a value

- $i\_i\; =\; i\_S\; frac\; \{R\_\{S\}\}\; \{R\_S\; +\; R\_\{in\}\}\; ,$

More generally, complex frequency-dependent impedances can be used instead of the driver and amplifier resistances.

- $A\_\{loaded\}\; =frac\; \{v\_L\}\; \{v\_S\}\; =\; frac\; \{R\_\{in\}\}\; \{R\_S+R\_\{in\}\}$ $frac\; \{R\_\{L\}\}\; \{R\_\{out\}+R\_\{L\}\}\; A\_v\; .$

The resistor ratios in the above expression are called the loading factors.

- $A\_\{fb\}\; =\; frac\; \{v\_L\}\{v\_S\}\; =\; frac\; \{A\_\{loaded\}\}\; \{1+\; \{beta\}(R\_S/R\_L)\; A\_\{loaded\}\}\; .$

That is, the ideal current gain A_{i} is reduced not only by the loading factors, but due to the bilateral nature of the two-port by an additional factor (1 + β (R_{S} / R_{L} ) A_{loaded} ), which is typical of negative feedback amplifier circuits. The factor β (R_{S} / R_{L} ) is the voltage feedback provided by the current feedback source of current gain β A/A. For instance, for an ideal voltage source with R_{S} = 0 Ω, the current feedback has no influence, and for R_{L} = ∞ Ω, there is zero load current, again disabling the feedback.

While voltage dividers may be used to produce precise reference voltages (that is, when no current is drawn from the reference node), they make poor voltage sources (that is, when current is drawn from the reference node). The reason for poor source behavior is that the current drawn by the load passes through resistor R_{1}, but not through R_{2}, causing the voltage drop across R_{1} to change with the load current, and thereby changing the output voltage.

In terms of the above equations, if current flows into a load resistance R_{L} (attached at the output node where the voltage is V_{out}), that load resistance must be considered in parallel with R_{2} to determine the voltage at V_{out}. In this case, the voltage at V_{out} is calculated as follows:

- $V\_mathrm\{out\}\; =\; frac\; \{R\_2\; |\; R\_mathrm\{L\}\}\; \{\; R\_1+R\_2\; |\; R\_mathrm\{L\}\; \}\; cdot\; V\_mathrm\{in\}$ $=\; frac\; \{R\_2\}\{R\_1\; left(1\; +\; frac\; \{R\_2\}\{R\_mathrm\{L\}\; \}\; right)\; +\; R\_2\; \}$

where R_{L} is a load resistor in parallel with R_{2}. From this result it is clear that V_{out} is decreased by R_{L} unless R_{2} // R_{L} ≈ R_{2} , that is, unless R_{L} >> R_{2}.

In other words, for high impedance loads it is possible to use a voltage divider as a voltage source, as long as R_{2} has very small value compared to the load. This technique leads to considerable power dissipation in the divider.

A voltage divider is commonly used to set the DC bias of a common emitter amplifier, where the current drawn from the divider is the relatively low base current of the transistor.

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

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Last updated on Monday September 29, 2008 at 19:53:47 PDT (GMT -0700)

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