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In [[electronics]], the '''Miller effect''' accounts for the increase in the equivalent input [[capacitance]] of an inverting voltage [[amplifier]] due to amplification of the capacitance between the input and output terminals. The additional input capacitance due to the Miller effect is given by
[[Image:Current division example.svg|thumbnail|250px|Figure 1: Schematic of an electrical circuit illustrating current division. Notation ''R<sub>T<sub>.</sub></sub>''  refers to the ''total'' resistance of the circuit to the right of resistor ''R<sub>X</sub>''.]]
:<math>C_{M}=C (1-A_v)\ ,</math>
where ''A<sub>v</sub>''  is the gain of the amplifier, which is a negative number because it is ''inverting'', and ''C'' is the feedback capacitance.


Although the term ''Miller effect'' normally refers to capacitance, the amplifier input impedance is modified by the Miller effect by any impedance connected between the input and another node exhibiting gain. These properties of the Miller effect are generalized in '''Miller's theorem'''.
In [[electronics]], a '''current divider ''' is a simple [[linear]] [[Electrical network|circuit]] that produces an output [[Electric current|current]] (''I''<sub>X</sub>) that is a fraction of its input current (''I''<sub>T</sub>). The splitting of current between the branches of the divider is called '''current division'''. The currents in the various branches of such a circuit divide in such a way as to minimize the total energy expended.


The formula describing a current divider is similar in form to that for the [[voltage divider]]. However, the ratio describing current division places the impedance of the unconsidered branches in the [[numerator]], unlike voltage division where the considered impedance is in the numerator. To be specific, if two or more [[Electrical impedance|impedance]]s are in parallel, the current that enters the combination will be split between them in inverse proportion to their impedances (according to [[Ohm's law]]). It also follows that if the impedances have the same value the current is split equally.


==Resistive divider==
A general formula for the current ''I<sub>X</sub>'' in a resistor ''R<sub>X</sub>'' that is in parallel with a combination of other resistors of total resistance ''R<sub>T</sub>'' is (see Figure 1): 
:<math>I_X = \frac{R_T}{R_X+R_T}I_T \ </math>
where ''I<sub>T</sub>'' is the total current entering the combined network of ''R<sub>X</sub>'' in parallel with ''R<sub>T</sub>''. Notice that when ''R<sub>T</sub>'' is composed of a [[Series_and_parallel_circuits#Parallel_circuits|parallel combination]] of resistors, say ''R<sub>1</sub>'', ''R<sub>2</sub>'', ... ''etc.'', then the reciprocal of each resistor must be added to find the total resistance ''R<sub>T</sub>'':
:<math> \frac {1}{R_T} = \frac {1} {R_1} + \frac {1} {R_2} + \frac {1}{R_3} + ... \ . </math>


== History ==
==General case==
The Miller effect was named after [[John Milton Miller]].<ref name=Miller/> When Miller published his work in 1920, he was working on [[vacuum tube]] triodes, however the same theory applies to more modern devices such as bipolar and MOS [[transistors]].
Although the resistive divider is most common, the current divider may be made of frequency dependent [[Electrical impedance|impedance]]s. In the general case the current I<sub>X</sub> is given by:
:<math>I_X = \frac{Z_T} {Z_X+Z_T}I_T \ ,</math>


== Derivation ==
==Using Admittance==
{{Image|Miller effect.PNG|right|350px|Miller's theorem}}
Instead of using [[Electrical impedance|impedance]]s, the current divider rule can be applied just like the [[voltage divider]] rule if [[admittance]] (the inverse of impedance) is used.
[[Image:Miller cir.png|right|frame|Figure 1: Circuit for deriving the Miller effect, with ideal inverting voltage amplifier]]
:<math>I_X = \frac{Y_X} {Y_{Total}}I_T</math>
Consider an ideal inverting voltage [[amplifier]] of gain <math>-A_v</math> with an [[Electrical impedance|impedance]] <math>Z</math> connected between its input and output nodes. The output voltage is therefore <math>V_o =- A_v V_i</math>.  Assuming that the amplifier input draws no current, all of the input current flows through <math>Z</math>, and is therefore given by
Take care to note that Y<sub>Total</sub> is a straightforward addition, not the sum of the inverses inverted (as you would do for a standard parallel resistive network). For Figure 1, the current I<sub>X</sub> would be
:<math>I_X = \frac{Y_X} {Y_{Total}}I_T = \frac{\frac{1}{R_X}} {\frac{1}{R_X} + \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}I_T</math>


:<math>I_i = \frac{V_i - V_o}{Z} = \frac{V_i (1 + A_v)}{Z}</math>.
===Example: RC combination===
[[Image:Low pass RC filter.PNG|thumbnail|220px|Figure 2: A low pass RC current divider]]
Figure 2 shows a simple current divider made up of a [[capacitor]] and a resistor. Using the formula above, the current in the resistor is given by:


The input impedance of the circuit is
::<math> I_R = \frac {\frac{1}{j \omega C}} {R + \frac{1}{j \omega C} }I_T </math>
:::<math> = \frac {1} {1+j \omega CR} I_T  \ , </math>
where ''Z<sub>C</sub> = 1/(jωC) '' is the impedance of the capacitor.


:<math>Z_{in} = \frac{V_i}{I_i} = \frac{Z}{1+A_v}.</math>
The product ''τ = CR'' is known as the [[time constant]] of the circuit, and the frequency for which ωCR = 1 is called the [[corner frequency]] of the circuit. Because the capacitor has zero impedance at high frequencies and infinite impedance at low frequencies, the current in the resistor remains at its DC value ''I<sub>T</sub>'' for frequencies up to the corner frequency, whereupon it drops toward zero for higher frequencies as the capacitor effectively [[short-circuit]]s the resistor. In other words, the current divider is a [[low pass filter]] for current in the resistor.


If Z represents a capacitor with impedance <math>Z = \frac{1}{s C}</math>, the resulting input impedance is  
==Loading effect==
[[Image:Current division.PNG|thumbnail|300px|Figure 3: A current amplifier (gray box) driven by a Norton source (''i<sub>S</sub>'', ''R<sub>S</sub>'') and with a resistor load ''R<sub>L</sub>''. Current divider in blue box at input (''R<sub>S</sub>'',''R<sub>in</sub>'') reduces the current gain, as does the current divider in green box at the output (''R<sub>out</sub>'',''R<sub>L</sub>'')]]
The gain of an amplifier generally depends on its source and load terminations. Current amplifiers and transconductance amplifiers are characterized by a short-circuit output condition, and current amplifiers and transresistance amplifiers are characterized using ideal infinite impedance current sources. When an amplifier is terminated by a finite, non-zero termination, and/or driven by a non-ideal source, the effective gain is reduced due to the '''loading effect''' at the output and/or the input, which can be understood in terms of current division.


:<math>Z_{in} = \frac{1}{s C_{M}} \quad \mathrm{where} \quad C_{M}=C (1+A_v).</math>
Figure 3 shows a current amplifier example. The amplifier (gray box) has input resistance ''R<sub>in</sub>'' and output resistance ''R<sub>out</sub>'' and an ideal current gain ''A<sub>i</sub>''. With an ideal current driver (infinite Norton resistance) all the source current ''i<sub>S</sub>'' becomes input current to the amplifier. However, for a [[Norton's theorem|Norton driver]] a current divider is formed at the input that reduces the input current to


Thus the effective or '''Miller capacitance''' ''C<sub>M</sub>'' is the physical ''C'' multiplied by the factor <math>(1+A_v)</math><ref name=Spencer/>.
::<math>i_{i} = \frac {R_S} {R_S+R_{in}} i_S \ , </math>


== Effects ==
which clearly is less than ''i<sub>S</sub>''. Likewise, for a short circuit at the output, the amplifier delivers an output current ''i<sub>o</sub>'' = ''A<sub>i</sub> i<sub>i</sub>'' to the short-circuit. However, when the load is a non-zero resistor ''R<sub>L</sub>'', the current delivered to the load is reduced by current division to the value:


As most amplifiers are inverting (i.e. <math>A_v < 0</math>), the effective capacitance at their inputs is increased due to the Miller effect. This can lower the bandwidth of the amplifier, reducing its range of operation to lower frequencies. The tiny junction and stray capacitances between the base and collector terminals of a [[Darlington transistor]], for example, may be drastically increased by the Miller effects due to its high gain, lowering the high frequency response of the device.
::<math>i_L = \frac {R_{out}} {R_{out}+R_{L}} A_i  i_{i} \ . </math>


It is also important to note that the Miller capacitance is the capacitance seen looking into the input.  If looking for all of the [[RC time constant]]s (poles) it is important to include as well the capacitance seen by the output.  The capacitance on the output is often neglected since it sees <math>{C}({1-1/A_v})</math> and amplifier outputs are typically low impedance.  However if the amplifier has a high impedance output, such as if a gain stage is also the output stage, then this RC can have a [[open-circuit time constant method|significant impact]] on the performance of the amplifier.  This is when [[pole splitting]] techniques are used. 
Combining these results, the ideal current gain ''A<sub>i</sub>'' realized with an ideal driver and a short-circuit load is reduced to the '''loaded gain''' ''A<sub>loaded</sub>'':


The Miller effect may also be exploited to synthesize larger capacitors from smaller ones. One such example is in the stabilization of [[negative feedback amplifier|feedback amplifiers]], where the required capacitance may be too large to practically include in the circuit. This may be particularly important in the design of [[integrated circuit]], where capacitors can consume significant area, increasing costs.
::<math>A_{loaded} =\frac {i_L} {i_S} =  \frac {R_S} {R_S+R_{in}}</math> <math> \frac {R_{out}} {R_{out}+R_{L}} A_i  \ . </math>


===Mitigation===
The resistor ratios in the above expression are called the '''loading factors'''. For more discussion of loading in other amplifier types, see [[Voltage division#Loading effect|loading effect]].
The Miller effect may be undesired in many cases, and approaches may be sought to lower its impact. Several such techniques are used in the design of amplifiers.


A current buffer stage may be added at the output to lower the gain <math>A_v</math> between the input and output terminals of the amplifier (though not necessarily the overall gain). For example, a [[common base]] may be used as a current buffer at the output of a [[common emitter]] stage, forming a [[cascode]]. This will typically reduce the Miller effect and increase the bandwidth of the amplifier.
===Unilateral versus bilateral amplifiers===
[[Image:H-parameter current amplifier.PNG|thumbnail|300px|Figure 4: Current amplifier as a bilateral two-port network; feedback through dependent voltage source of gain β V/V]]
Figure 3 and the associated discussion refers to a [[Electronic amplifier#Unilateral or bilateral|unilateral]] amplifier. In a more general case where the amplifier is represented by a [[two-port network|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 current amplifier of Figure 3, the corresponding bilateral two-port network is shown in Figure 4 based upon [[Two-port network#Hybrid parameters (h-parameters)| h-parameters]].<ref name=H_port/> Carrying out the analysis for this circuit, the current gain with feedback ''A<sub>fb</sub>'' is found to be


Alternatively, a voltage buffer may be used before the amplifier input, reducing the effective source impedance seen by the input terminals. This lowers the <math>RC</math> time constant of the circuit and typically increases the bandwidth.
::<math> A_{fb} = \frac {i_L}{i_S} = \frac {A_{loaded}} {1+ {\beta}(R_L/R_S) A_{loaded}} \ . </math>


== Impact on frequency response ==
That is, the ideal current gain ''A<sub>i</sub>'' is reduced not only by the loading factors, but due to the bilateral nature of the two-port by an additional factor<ref>Often called the ''improvement factor'' or the ''desensitivity factor''.</ref> ( 1 + β (R<sub>L</sub> / R<sub>S</sub> ) A<sub>loaded</sub> ), which is typical of [[negative feedback amplifier]] circuits. The factor β (R<sub>L</sub> / R<sub>S</sub> ) is the current feedback provided by the voltage feedback source of voltage gain β V/V. For instance, for an ideal current source with ''R<sub>S</sub>'' = ∞ Ω, the voltage feedback has no influence, and for ''R<sub>L</sub>'' = 0 Ω, there is zero load voltage, again disabling the feedback.
[[Image:Miller before transform.PNG|thumbnail|250px|Figure 2: Operational amplifier with feedback capacitor ''C<sub>C</sub>''.]]
[[Image:Miller after transform.PNG|thumbnail|250px|Figure 3: Circuit of Figure 2 transformed using Miller's theorem, introducing the '''Miller capacitance''' on the input side of the circuit.]]
Figure 2 shows an example of Figure 1 where the impedance coupling the input to the output is the coupling capacitor ''C<sub>C</sub>''. A [[Thévenin's theorem|Thévenin voltage]] source ''V<sub>A</sub>'' drives the circuit with Thévenin resistance ''R<sub>A</sub>''. At the output a parallel ''RC''-circuit serves as load. (The load is irrelevant to this discussion: it just provides a path for the current to leave the circuit.) In Figure 2, the coupling capacitor delivers a current jω''C<sub>C</sub>( v<sub>i</sub> - v<sub>o</sub> )'' to the output circuit.
 
Figure 3 shows a circuit electrically identical to Figure 2 using Miller's theorem. The coupling capacitor is replaced on the input side of the circuit by the Miller capacitance ''C<sub>M</sub>'', which draws the same current from the driver as the coupling capacitor in Figure 2. Therefore, the driver sees exactly the same loading in both circuits. On the output side, a dependent current source in Figure 3 delivers the same current to the output as does the coupling capacitor in Figure 2. That is, the ''R-C''-load sees the same current in Figure 3 that it does in Figure 2.
 
In order that the Miller capacitance draw the same current in Figure 3 as the coupling capacitor in Figure 2, the Miller transformation is used to relate ''C<sub>M</sub>'' to ''C<sub>C</sub>''. In this example, this transformation is equivalent to setting the currents equal, that is
::<math>\  j\omega C_C ( v_i - v_O ) = j \omega C_M v_i, </math>
or, rearranging this equation
:: <math> C_M = C_C \left( 1 + \frac {v_o} {v_i} \right )  = C_C (1 + A_v). </math>
This result is the same as ''C<sub>M</sub>'' of the ''Derivation Section''.
 
The present example with ''A<sub>v</sub>'' frequency independent shows the implications of the Miller effect, and therefore of  ''C<sub>C</sub>'', upon the frequency response of this circuit, and is typical of the impact of the Miller effect (see, for example, [[common source]]). If ''C<sub>C</sub>'' = 0 F, the output voltage of the circuit is simply ''A<sub>v</sub> v<sub>A</sub>'', independent of frequency. However, when ''C<sub>C</sub>'' is not zero, Figure 3 shows the large Miller capacitance appears at the input of the circuit. The voltage output of the circuit now becomes
 
::<math> v_o =- A_v v_i = A_v \frac {v_A} {1+j \omega C_M R_A}, </math>
 
and rolls off with frequency once frequency is high enough that ω ''C<sub>M</sub>R<sub>A</sub>'' ≥ 1. It is a [[low-pass filter]]. In analog amplifiers this curtailment of frequency response is a major implication of the Miller effect. In this example, the frequency ω''<sub>3dB</sub>'' such that ω''<sub>3dB</sub>'' ''C<sub>M</sub>R<sub>A</sub>'' = 1 marks the end of the low-frequency response region and sets the [[Bandwidth (signal processing)|bandwidth]] or [[cutoff frequency]] of the amplifier.
 
It is important to notice that the effect of ''C''<sub>M</sub> upon the amplifier bandwidth is greatly reduced for low impedance drivers (''C''<sub>M</sub> ''R''<sub>A</sub> is small if ''R''<sub>A</sub> is small). Consequently, one way to minimize the Miller effect upon bandwidth is to use a low-impedance driver, for example, by interposing a [[voltage follower]] stage between the driver and the amplifier, which reduces the apparent driver impedance seen by the amplifier.
 
The output voltage of this simple circuit is always ''A<sub>v</sub> v<sub>i</sub>''. However, real amplifiers have output resistance. If the amplifier output resistance is included in the analysis, the output voltage exhibits a more complex frequency response and the impact of the frequency-dependent current source on the output side must be taken into account.<ref name = PoleSplitting/> Ordinarily these effects show up only at frequencies much higher than the [[roll-off]] due to the Miller capacitance, so the analysis presented here is adequate to determine the useful frequency range of an amplifier dominated by the Miller effect.
 
===Miller approximation===
This example also assumes ''A<sub>v</sub>'' is frequency independent, but more generally there is  frequency dependence of the amplifier contained implicitly in ''A<sub>v</sub>''. Such frequency dependence of ''A<sub>v</sub>'' also makes the Miller capacitance frequency dependent, so interpretation of ''C<sub>M</sub>'' as a capacitance becomes more difficult. However, ordinarily any frequency dependence of ''A<sub>v</sub>'' arises only at frequencies much higher than the roll-off with frequency caused by the Miller effect, so for frequencies up to the Miller-effect roll-off of the gain, ''A<sub>v</sub>'' is accurately approximated by its low-frequency value. Determination of ''C<sub>M</sub>'' using ''A<sub>v</sub>'' at low frequencies is the so-called '''Miller approximation'''.<ref name=Spencer/> With the Miller approximation, ''C<sub>M</sub>'' becomes frequency independent, and its interpretation as a capacitance at low frequencies is secure.


==References and notes==
==References and notes==
{{reflist|refs=
{{reflist|refs=
<ref name=Miller>
{{cite journal |author=John M. Miller |title=Dependence of the input impedance of a three-electrode vacuum tube upon the load in the plate circuit |journal=Scientific Papers of the Bureau of Standards |volume=15 |issue= 351 |pages=pp. 367-385 |year=1920 |url=http://books.google.com/books?id=7u8SAAAAYAAJ&pg=PA367&lpg=PA367}}
</ref>
<ref name=Spencer>
{{cite book
|author=R.R. Spencer and M.S. Ghausi
|title=Introduction to electronic circuit design.
|year= 2003
|page=533
|publisher=Prentice Hall/Pearson Education, Inc.
|location=Upper Saddle River NJ
|isbn=0-201-36183-3
|url=http://worldcat.org/isbn/0-201-36183-3}}
</ref>
<ref name = PoleSplitting>See article on [[pole splitting]].</ref>


<ref name=H_port>>The [[Two-port network#Hybrid parameters (h-parameters)|h-parameter two port]] is the only two-port among the four standard choices that has a current-controlled current source on the output side.</ref>


}}
}}

Latest revision as of 03:07, 22 November 2023


The account of this former contributor was not re-activated after the server upgrade of March 2022.


Figure 1: Schematic of an electrical circuit illustrating current division. Notation RT. refers to the total resistance of the circuit to the right of resistor RX.

In electronics, a current divider is a simple linear circuit that produces an output current (IX) that is a fraction of its input current (IT). The splitting of current between the branches of the divider is called current division. The currents in the various branches of such a circuit divide in such a way as to minimize the total energy expended.

The formula describing a current divider is similar in form to that for the voltage divider. However, the ratio describing current division places the impedance of the unconsidered branches in the numerator, unlike voltage division where the considered impedance is in the numerator. To be specific, if two or more impedances are in parallel, the current that enters the combination will be split between them in inverse proportion to their impedances (according to Ohm's law). It also follows that if the impedances have the same value the current is split equally.

Resistive divider

A general formula for the current IX in a resistor RX that is in parallel with a combination of other resistors of total resistance RT is (see Figure 1):

where IT is the total current entering the combined network of RX in parallel with RT. Notice that when RT is composed of a parallel combination of resistors, say R1, R2, ... etc., then the reciprocal of each resistor must be added to find the total resistance RT:

General case

Although the resistive divider is most common, the current divider may be made of frequency dependent impedances. In the general case the current IX is given by:

Using Admittance

Instead of using impedances, the current divider rule can be applied just like the voltage divider rule if admittance (the inverse of impedance) is used.

Take care to note that YTotal is a straightforward addition, not the sum of the inverses inverted (as you would do for a standard parallel resistive network). For Figure 1, the current IX would be

Example: RC combination

Figure 2: A low pass RC current divider

Figure 2 shows a simple current divider made up of a capacitor and a resistor. Using the formula above, the current in the resistor is given by:

where ZC = 1/(jωC) is the impedance of the capacitor.

The product τ = CR is known as the time constant of the circuit, and the frequency for which ωCR = 1 is called the corner frequency of the circuit. Because the capacitor has zero impedance at high frequencies and infinite impedance at low frequencies, the current in the resistor remains at its DC value IT for frequencies up to the corner frequency, whereupon it drops toward zero for higher frequencies as the capacitor effectively short-circuits the resistor. In other words, the current divider is a low pass filter for current in the resistor.

Loading effect

Figure 3: A current amplifier (gray box) driven by a Norton source (iS, RS) and with a resistor load RL. Current divider in blue box at input (RS,Rin) reduces the current gain, as does the current divider in green box at the output (Rout,RL)

The gain of an amplifier generally depends on its source and load terminations. Current amplifiers and transconductance amplifiers are characterized by a short-circuit output condition, and current amplifiers and transresistance amplifiers are characterized using ideal infinite impedance current sources. When an amplifier is terminated by a finite, non-zero termination, and/or driven by a non-ideal source, the effective gain is reduced due to the loading effect at the output and/or the input, which can be understood in terms of current division.

Figure 3 shows a current amplifier example. The amplifier (gray box) has input resistance Rin and output resistance Rout and an ideal current gain Ai. With an ideal current driver (infinite Norton resistance) all the source current iS becomes input current to the amplifier. However, for a Norton driver a current divider is formed at the input that reduces the input current to

which clearly is less than iS. Likewise, for a short circuit at the output, the amplifier delivers an output current io = Ai ii to the short-circuit. However, when the load is a non-zero resistor RL, the current delivered to the load is reduced by current division to the value:

Combining these results, the ideal current gain Ai realized with an ideal driver and a short-circuit load is reduced to the loaded gain Aloaded:

The resistor ratios in the above expression are called the loading factors. For more discussion of loading in other amplifier types, see loading effect.

Unilateral versus bilateral amplifiers

Figure 4: Current amplifier as a bilateral two-port network; feedback through dependent voltage source of gain β V/V

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 current amplifier of Figure 3, the corresponding bilateral two-port network is shown in Figure 4 based upon h-parameters.[1] Carrying out the analysis for this circuit, the current gain with feedback Afb is found to be

That is, the ideal current gain Ai is reduced not only by the loading factors, but due to the bilateral nature of the two-port by an additional factor[2] ( 1 + β (RL / RS ) Aloaded ), which is typical of negative feedback amplifier circuits. The factor β (RL / RS ) is the current feedback provided by the voltage feedback source of voltage gain β V/V. For instance, for an ideal current source with RS = ∞ Ω, the voltage feedback has no influence, and for RL = 0 Ω, there is zero load voltage, again disabling the feedback.

References and notes

  1. ↑ >The h-parameter two port is the only two-port among the four standard choices that has a current-controlled current source on the output side.
  2. ↑ Often called the improvement factor or the desensitivity factor.
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