c° Copyright 2008. W. Marshall Leach, Jr., Professor, Georgia Institute of Tec hnology, School of Electricaland Computer Engineering.Useful Circuit TheoremsImpedance of a Two-Terminal RC Netw orkConsider any two-terminal RC network. If the impedance at dc is not infinite, the impedance can be writtenZ = Rdc1+τzs1+τpswhere Rdcisthedcresistanceofthecircuit,τois its open-circuit time constant, and τzis its short-circuittime constant.Figure 1: Example two-terminal circuits.For example applications of the theorem, consider the circuits shown in Fig. 1. In order from (a) to (d),the impedances are given byZ = R11+RCsZ =(R1+ R2)1+R1kR2Cs1+R2CsZ =(R1+ R2)1+(R1kR3+ R2) Cs1+(R2+ R3) CsZ =[R1+ R2k (R3+ R4)]1+(R1kR2+ R3) kR4Cs1+(R2+ R3) kR4CsAlthough the theorem is strictly valid for circuits containing only one capacitor, it can be applied tocircuits containing more than one capacitor if the adjacent poles and zeroes in the transfer function are wellremoved, preferable by a decade or more. Consider the circuit shown in Fig. 2. Let us assume that C2andC3are open circuits in the frequency range in which C1is active, C1is a short circuit and C3is an opencircuit in the frequency range in which C2is active, and C1and C2are short circuits in the frequency rangein which C3becomes active.1Figure 2: Example circuit containing more than one capaictor.With the above information, the impedance in the range where C2and C3are open circuits is given byZ1=(R1+ R2+ R3+ R4)1+R2k (R1+ R3+ R4) C1s1+R2C1s=(R1+ R2+ R3+ R4)1+s/ω21+s/ω1whereω1=1R2C1ω2=11+R2k (R1+ R3+ R4) C1At low frequencies, Z1starts at the value R1+ R2+ R3+ R4and shelv es at high frequencies at the valueR1+ R3+ R4. The impedance in the range where C1is a short circuit and C3is an open circuit is given byZ2=(R1+ R3+ R4)1+R3k (R1+ R4) C2s1+R3C2s=(R1+ R3+ R4)1+s/ω41+s/ω3whereω3=1R3C2ω4=1R3k (R1+ R4) C2A t low frequencies, Z2starts at the value R1+ R3+ R4and shelves at high frequencies at the value R1+ R4.The impedance in the range where C1and C2are short circuits is given byZ3=(R1+ R4)1+R4kR1C3s1+R4C3s=(R1+ R4)1+s/ω61+s/ω5whereω5=1R4C3ω6=1R4kR1C3At lo w frequencies, Z3starts at the value R1+ R4and shelv es at high frequencies at the value R1.The three impedances can be “pieced” together to obtain the overall impedance to obtainZ =(R1+ R2+ R3+ R4)(1 + s/ω2)(1+s/ω4)(1+s/ω6)(1 + s/ω1)(1+s/ω3)(1+s/ω5)This expression is strictly ifω1¿ ω2¿ ω3¿ ω4¿ ω5¿ ω6The straight-line approximation to the Bode magnitude plot for the impedance is shown in Fig. 3.The impedance theorem can be used to write by inspection the transfer function of an inverting op-ampcircuit where the input and feedback impedances contain no more than one capacitor each. Consider thecircuit shown in Figure 4. The voltage gain can be written by inspection to obtainV2V1= −R31+R3C2s÷·(R1+ R2)1+R1kR2C1s1+R2C2s¸= −R3R1+ R21+R2C2s(1 + R1kR2C1s)(1+R3C2s)2Figure 3: Bode magnitude plot for the impedance Z.Figure 4: Inverting op-amp example.3Impedance of an RC Voltage-Divider Net workCase 1 — Capacitor in Shunt ArmConsider the voltage-divider network shown in Fig. 5(a). Let the impedance Z2contain one capacitor andsatisfies the condition for the two-terminal impedance theorem. By voltage division, the gain of the networkis given byV2V1=Z2R1+ Z2=Z2Z3where Z3= R1+ Z2. The impedance Z2can be writtenZ2= R21+τ2ss1+τ2oswhere τ2ois the open-circuit time constant for Z2and τ2sis its short-circuit time constant. The impedanceZ3can be writtenZ3=(R1+ R2)1+τ3ss1+τ3oswhere τ3ois the open-circuit time constant for Z3and τ3sis its short-circuit time constant.Figure 5: Voltage divider networks containing only one capacitor.But the open-circuit time constants for Z2and Z3are the same. Let this be denoted by τo= τ2o= τ3o.thus the two impedances can be writtenZ2= R21+τ2ss1+τosandZ3=(R1+ R2)1+τ3ss1+τosIt follows that the gain of the voltage divider is given byV2V1=R2R1+ R21+τ2ss1+τ3ssNotice that the term 1+τos is canceled. Note also that the gain constant is the circuit gain at dc, the poletime constant τ3sis the time constant calculated with Vi=0, and the zero time constant τ2sis the timeconstan t with Vo=0.Case 2 — Capacitor in Series ArmNow consider the case shown in Fig. 5(b) where Z1con tains one capacitor and satisfies the condition for thetw o-terminal impedance theorem. By voltage division, the gain of the network is given byV2V1=R2Z1+ R2=R2Z3where Z3= Z1+ R2. The impedance Z3can be writtenZ3=(R1+ R2)1+τ3ss1+τ3os4where τ3ois the open-circuit time constant for Z3and τ3sis its short-circuit time constant.It follows that the gain of the voltage divider is given byV2V1=R2R1+ R21+τ3os1+τ3ssBut the open-circuit time constant for Z3is equal to the open-circuit time constant for Z1, i.e. τ3o= τ1o.Th us the gain of the circuit can be writtenV2V1=R2R1+ R21+τ1os1+τ3ssNote that the gain constant is the circuit gain at dc, the pole time constan t τ3sis the time constant calculatedwith Vi=0, and the zero time constant τ1sis the time constant with the Vinode floating, i.e. open circuited.Combined TheoremWe seek to combine the two theorems into one which gives the correct answer for both cases. In the secondcase, the time constant τ1ois the same as the time constant calculated with Vo=0. Thus it follows thatthe two theorems can be combined to obtain the general solutionV2V1= kdc1+τ2s1+τ1swhere kdcisthedcgain,τ1is the time constant with V1=0,andτ2is the time constant with V2=0.As an example, consider the circuit shown in Fig. 6(a). By inspection, the voltage gain can be writtenV2V1=R2+ R3R1+ R2+ R31+(R2kR3+ R4) Cs1+[(R1+ R2) kR3+ R4] CsThe Bode magnitude plot is that of a low-pass shelving function.Figure 6: Example circuits for the voltage-divider theorem.The v oltage gain of the circuit in Fig. 6(b) is given byV2V1=R4R1+ R2+ R41+(R2+ R3) Cs1+[(R1+ R4) kR2+ R3] CsThe Bode magnitude plot is that of a high-pass shelving function.The voltage-divider theorem can be used to write the voltage gain expression for a non-inverting op-ampcircuit. Consider the circuit shown in Fig. 7. The vo ltage divider network in Fig. 6(a) is shown connectedbetween the output of the op amp and its inverting input. Because the op amp has negative feedback, there5is a virtual short between its + and − inputs. Thus the voltage output of the voltage divider is V2= Viandits voltage input is V1= Vo. It follows that the voltage gain of the circuit is given