Miller’s Theorem | Miller’s theorem for capacitive Reactance

In this lecture, we are going to learn about Miller’s theorem. How Miller’s theorem can be applied to capacitive reactance? we will discuss all important topics related to Miller’s theorem in this section.

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Miller’s Theorem

  • Consider an arbitrary circuit configuration with N distinct nodes 1,2,3…N  as shown in figure (a).
  • Let the node voltages be V1, V2, V3 …. VN, Where VN = 0 as N is being treated as the reference node or ground node.
  • Let the nodes N1 and N2 be interconnected with the impedance Z. Let the ratio of voltages V2/V1 be denoted by K, where K is a complex number.
Miller's Theorem
  • As shown in the above figure (a) and (b) the results of both circuits will be the same. 
  • Now we will prove how both figures give the same result.

Also Read: What is CMRR? | Common Mode Rejection Ratio

Proof for Miller’s Theorem:

Step 1:

  • Disconnect terminal-1 from Z and connect an impedance Z_1=\frac{Z}{1-K} between terminal-1 and ground as shown in figure (a). 
  • Now let us obtain the expression for I_1 for both the arrangements in figures (a) and (b).
  • From figure (a),

I_1=\frac{V_1-V_2}{Z} ………. Eq.1

  • But as K=\frac{V_2}{V_1} we can substitute V_2=KV_1 in Eq. 1

\therefore I_1 = \frac{V_1-KV_1}{Z}

\therefore I_1 = \frac{V_1}{Z/(1-K)} ……. Eq.2

  • Now consider Figure (b) to write,

I_1 = \frac{V_1}{Z_1} = \frac{V_1}{Z/(1-K)} ……. Eq.3

  • Compare Eq.(2) and Eq.(3) . They give the same value of I_1

Step 2:

  • Now disconnect terminal-2 from the impedance Z and connect an impedance Z_2 = \frac{ZK}{(K-1)} between terminal-2 and ground as shown in figure (b).
  • Now let us obtain an expression for I_2 from the two configurations shown in figures (a) and (b).
  • Consider Figure (a),

I_2=-I_1=\frac{V_2-V_1}{Z} ………. Eq.4

  • But K=\frac{V_2}{V_1}, hence substitute V_1=\frac{V_2}{K} in Eq.4 to get,


\therefore I_2=\frac{V_2(1-1/K)}{Z}

\therefore I_2=\frac{V_2(K-1)}{ZK}

\therefore I_2=\frac{V_2}{ZK/(K-1)}………. Eq.5

  • Now refer to figure (b)  then we can write the value of I_2,

I_2= \frac{V_2}{Z_2}=\frac{V_2}{ZK/(K-1)}………. Eq.6

  • Compare Eq. (5) and Eq. (6). They give the same value of I_2.
  • Thus we obtain similar nodal equations from the configurations of figures (a) and (b) therefore those two networks are equivalent.
  • It should be noted that this theorem is helpful to make calculations if and only if it is possible to find the value of K by some independent means.
  • The transformation of the network shown in figures (a) and (b) is referred to as Miller’s theorem.

Also Read: Power Amplifier | Types of Power Amplifier

Miller’s Theorem for Resistive Impedance:


Miller’s Theorem for Capacitive Reactance

  • From the above analysis, you already know about Miller’s theorem. Now we will extend it for capacitive reactance.
  • The below Figure illustrates the extension of Miller’s theorem for capacitive reactance.
  • As shown in the figure below, C_1 and C_2 are the Miller capacitance and their values are given by,

\boxed{C_1 = C_F(A_V+1)}

\boxed{C_2 = \frac{C_F(A_V+1)}{A_V}}

  • For CE amplifier A_V>>1. Hence A_V+1=A_V. Therefore C_2 can be approximated to C_F.

\therefore C_2\approx C_F


Derivation for Miller’s Theorem for Capacitive Reactance:

  • Referring to the above figure and extending Miller’s theorem to capacitive reactance we get,

X_{C2}=\frac{A_V X_{CF}}{A_V - 1 } and X_{C1}=\frac{X_{CF}}{ 1 - A_V }

But, X_{C2}=\frac{1}{2\pi f C_2} and X_{CF}=\frac{1}{2\pi f C_F}

\therefore \frac{1}{2\pi f C_2} = \frac{A_V}{(A_V - 1) 2\pi f C_F}

\therefore  C_2 = (\frac{A_V - 1}{A_V }) C_F

  • A CE amplifier’s gain A_V is negative. So substitute - A_V in place of A_V to get,

\therefore  C_2 = (\frac{- A_V - 1}{- A_V }) \times C_F = (\frac{A_V + 1}{ A_V }) \times C_F

\therefore \boxed{ C_2 =  ( \frac{A_V + 1}{ A_V } ) \times C_F}     …….Proved.

Similarly, X_{C1}=\frac{1}{2\pi f C_1},

\therefore \frac{1}{2\pi f C_1} = \frac{1}{( 1 - A_V) 2\pi f C_F}

\therefore  C_1 = (1 - A_V)C_F

  • Substitute -A_V n place of A_V we get,

\therefore \boxed{ C_1 = (1 + A_V)C_F }    …….Proved.

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Hello friends, my name is Trupal Bhavsar, I am the Writer and Founder of this blog. I am Electronics Engineer(2014 pass out), Currently working as Junior Telecom Officer(B.S.N.L.) also I do Project Development, PCB designing and Teaching of Electronics Subjects.

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