Dynamic Impedance of a π-Model
We analyze the input impedance Zin of a π-model consisting of two capacitors C1 , C2 connected to ground, and a resistor R in between. The operating
frequency is given by ω = 2π
dt .
Impedance Components
The impedance of the capacitors is:
ZC1 =
1
,
jωC1
ZC2 =
1
jωC2
The total input impedance Zin is then:
−1
1
1
Zin = ZC1 ∥ (R + ZC2 ) =
+
ZC1
R + ZC2
Define:
A=
1
= jωC1 ,
ZC1
B=
Then:
Zin =
1
1
jωC2
=
=
1
R + ZC2
1
+
jωRC2
R + jωC
2
1
1
=
jωC2
A+B
jωC1 + 1+jωRC
2
Multiply numerator and denominator by 1 + jωRC2 :
1 + jωRC2
jωC1 (1 + jωRC2 ) + jωC2
1 + jωRC2
=
jω(C1 + C2 + jωRC1 C2 )
Zin =
Let us write the denominator:
D = jω(C1 + C2 + jωRC1 C2 ) = jω(C1 + C2 ) − ω 2 RC1 C2
Then:
Zin =
1 + jωRC2
−ω 2 RC1 C2 + jω(C1 + C2 )
Let real and imaginary parts of the denominator be:
x = −ω 2 RC1 C2 ,
y = ω(C1 + C2 )
Then the magnitude is:
p
1 + (ωRC2 )2
1 + jωRC2
p
|Zin | =
=
x + jy
x2 + y 2
p
1 + (ωRC2 )2
=p
ω 4 R2 C12 C22 + ω 2 (C1 + C2 )2
This is the final expression for the magnitude of the input impedance.
1
Special Cases
Case 1: C1 = 0
Then ZC1 → ∞, so no current flows into that branch. The impedance becomes:
Zin = R + ZC2 = R +
1
jωC2
s
2
R2 +
|Zin | =
1
ωC2
Case 2: C2 = 0
Then ZC2 → ∞, the current cannot pass through R, and input sees only ZC1 :
1
jωC1
Zin = ZC1 =
|Zin | =
1
ωC1
Case 3: R = 0
Then the two capacitors are in parallel:
Zin = ZC1 ∥ ZC2 =
1
1
1
∥
=
jωC1 jωC2
jω(C1 + C2 )
|Zin | =
1
ω(C1 + C2 )
Case 4: C1 = ∞
Then ZC1 = 0 (short to ground), so:
Zin = 0
Case 5: C2 = ∞
Then ZC2 = 0 (short to ground), so:
Zin = ZC1 ∥ R =
1
1
+
ZC1
R
|Zin | = q
−1
=
1
(ωC1 )2 + R12
2
1
jωC1 +
R
−1
Case 6: R = ∞
Then R is open, and ZC2 is disconnected. The impedance is:
Zin = ZC1 =
1
,
jωC1
3
|Zin | =
1
ωC1