- Voltage divider
In

electronics , a**voltage divider**(also known as a**potential divider**) is a simplelinear circuit that produces an outputvoltage ("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

resistor s in series or apotentiometer . It is commonly used to create a reference voltage, and may also be used as a signal attenuator at low frequencies.**General case**A voltage divider referenced to ground is created by connecting two

impedance s in series, as shown in Figure 1. The input voltage is applied across the series impedances Z_{1}and Z_{2}and the output is the voltage across Z_{2}.Z_{1}and Z_{2}may be composed of any combination of elements such asresistor s,inductor s andcapacitor s.Applying

Ohm's Law , the relationship between the input voltage, V_{in}, and the output voltage, V_{out}, can be found::$V\_mathrm\{out\}\; =\; frac\{Z\_2\}\{Z\_1+Z\_2\}\; cdot\; V\_mathrm\{in\}$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::$H\; =\; frac\; \{V\_\{out\{V\_\{in\; =\; frac\{Z\_2\}\{Z\_1+Z\_2\}$In general this transfer function is a complex,

rational function offrequency .**Resistive divider**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::$V\_mathrm\{out\}\; =\; frac\{R\_2\}\{R\_1+R\_2\}\; cdot\; V\_mathrm\{in\}$As in the general case, "R"

_{1}and "R"_{2}may be any combination of series/parallel resistors.**Examples****Resistive divider**As a simple example, if "R"

_{1}= "R"_{2}then:$V\_mathrm\{out\}\; =\; frac\{1\}\{2\}\; cdot\; V\_mathrm\{in\}$As a more specific and/or practical example, if "V"

_{out}=6V and "V"_{in}=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 usingalgebra , "R"_{2}must be twice the value of "R"_{1}.To solve for R1::$R\_1\; =\; frac\{R\_2\; cdot\; V\_mathrm\{in\{V\_mathrm\{out\; -\; R\_2$

To solve for R2::$R\_2\; =\; frac\{R\_1\}\{frac\{V\_mathrm\{in\{V\_mathrm\{out-1\}$

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}**Low-pass RC filter**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 thecapacitance of the capacitor, "j" is theimaginary unit , and "ω" (omega) is theradian frequency of the input voltage.This divider will then have the voltage ratio::$\{V\_mathrm\{out\}\; over\; V\_mathrm\{in\; =\; \{Z\_mathrm\{2\}\; over\; Z\_mathrm\{1\}\; +\; Z\_mathrm\{2\; =$1 over j omega C} over {1 over j omega C} + R} = {1 over 1 + j omega R C}.

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 ight| = frac {1} {sqrt { 1 + ( omega R C )^2 } } .

**Inductive divider:**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 divider:**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.

**Loading effect**The voltage output of a voltage divider is not fixed but varies according to the load. To obtain a reasonably stable output voltage the output current should be a small fraction of the input current. The drawback of this is that most of the input current is wasted as heat in the resistors.

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

The

gain of anamplifier 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. cite book

author=A.S. Sedra and K.C. Smith

title=Microelectronic circuits

year= 2004

pages=§1.5 pp. 23-31

publisher=Oxford University Press

edition=Fifth Edition

location=New York

isbn= 0-19-514251-9

url=http://worldcat.org/isbn/0-19-514251-9] 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 anopen 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 ashort-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 oncurrent 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/orcurrent division , as described next.**Input loading**A general voltage source can be represented by a Thévenin equivalent circuit with Thévenin series impedance "R

_{S}". For a Thévenin driver, the "input" voltage v_{i}is reduced from v_{S}byvoltage division to a value::$v\_i\; =\; v\_S\; frac\; \{R\_\{in\; \{R\_S\; +\; R\_\{in\; ,$ where "R_{in}" is the amplifier input resistance, and the overall gain is reduced below the idealized gain by the same voltage division factor.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}bycurrent division to a value::$i\_i\; =\; i\_S\; frac\; \{R\_\{S\; \{R\_S\; +\; R\_\{in\; ,$ where "R_{in}" is the amplifier input resistance, and the overall gain is reduced below the gain estimated using an ideal driver by the same current division factor.More generally, complex frequency-dependent impedances can be used instead of the driver and amplifier resistances.

**Output loading**For a finite load, "R

_{L}" an output voltage is reduced byvoltage division by the factor "R_{L}/ ( R_{L}+ R_{out})", where "R_{out}" is the amplifier output resistance. Likewise, as the term "short-circuit" implies, the output current delivered to a load "R_{L}" is reduced bycurrent division by the factor "R_{out}/ ( R_{L}+ R_{out})". The overall gain is reduced below the gain estimated using an ideal load by the same current division factor. More generally, complex frequency-dependent impedances can be used instead of the load and amplifier resistances.**Loaded gain - voltage amplifier case**Including both the input and output voltage division factors for the voltage amplifier of Figure 4, the ideal voltage gain "A

_{v}" realized with an ideal driver and an open-circuit load is reduced to the**loaded gain**"A_{loaded}":::$A\_\{loaded\}\; =frac\; \{v\_L\}\; \{v\_S\}\; =\; frac\; \{R\_\{in\; \{R\_S+R\_\{in$ $frac\; \{R\_\{L\; \{R\_\{out\}+R\_\{L\; A\_v\; .$

The resistor ratios in the above expression are called the

**loading factors**.**Unilateral versus bilateral amplifiers**Figure 3 and the associated discussion refers to a unilateral amplifier. In a more general case where the amplifier is represented by a two port, the input resistance of the amplifier depends on its load, and the output resistance on the source impedance. The loading factors in these cases must employ the true amplifier impedances including these bilateral effects.For example, taking the unilateral voltage amplifier of Figure 3, the corresponding bilateral two-port network is shown in Figure 4 based upon g-parameters. [

*The g-parameter two port is the only one of the standard four choices that has a voltage-controlled voltage source on the output side.*] Carrying out the analysis for this circuit, the voltage gain with feedback "A_{fb}" is found to be::$A\_\{fb\}\; =\; frac\; \{v\_L\}\{v\_S\}\; =\; frac\; \{A\_\{loaded\; \{1+\; \{eta\}(R\_S/R\_L)\; A\_\{loaded\; .$

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 [*Often called the "improvement factor" or the "desensitivity factor".*] ( 1 + β (R_{S}/ R_{L}) A_{loaded}), which is typical ofnegative 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.**Applications****Reference voltage**Voltage dividers are often used to produce stable reference voltages. The term "reference voltage" implies that little or no current is drawn from the divider output node by an attached load. Thus, use of the divider as a reference requires a load device with a high input impedance to avoid

**loading**the divider, that is, to avoid disturbing its output voltage. A simple way of avoiding loading (for low power applications) is to simply input the reference voltage into the non-inverting input of anop-amp buffer.**Voltage source**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::$V\_mathrm\{out\}\; =\; frac\; \{R\_2\; |\; R\_mathrm\{L\; \{\; R\_1+R\_2\; |\; R\_mathrm\{L\}\; \}\; cdot\; V\_mathrm\{in\}$ $=\; frac\; \{R\_2\}\{R\_1\; left(\; 1\; +\; frac\; \{R\_2\}\{R\_mathrm\{L\}\; \}\; ight)\; +\; R\_2\; \}\; cdot\; V\_mathrm\{in\}\; ,$

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.**References****Further reading*** Paul Horowitz and Winfield Hill, "The Art of Electronics", Cambridge University Press, 1989.

**External links*** [

*http://www.electriccircuits.net/book,6,chapter,319,lesson,1464,voltage_divider.aspx Lesson In Voltage divider*]

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