IEEE Power Electronics Magazine - September 2017 - 43

Look at Figure 3, which describes
Now, what is the quasi-static gain of
a first-order passive circuit involvthis circuit for s = 0? In dc conditions,
By using this technique,
ing an injection source-the stimua capacitor becomes an open circuit,
lus-biasing the left side of the netwhile an inductor becomes a short ciryou will appreciate
work. The input signal Vin propagates
cuit. Apply this concept to the Fighow quickly and
through meshes and nodes to form
ure 3 circuit, and redraw it as shown
the response Vout observed across the
in Figure 5. In your head, you cut the
intuitively one can
resistor R 3 . We are interested in derivconnection before R 4, and you see a
determine a particular
ing the transfer function G linking
resistive divider involving R 1 and R 2 .
The Thévenin voltage across R 2 is
Vout to Vin .
transfer function.
To determine the time constant
R
of this example circuit, we will set
(4)
Vth = Vin R +2R .
1
2
the excitation to zero (a 0-V source is
replaced by a short circuit, while a 0-A current source would
The output resistance R th is R 1 paralleled with R 2 . The
be replaced by an open circuit) and remove the capacitor.
complete transfer function, therefore, involves the resistive
Then, we connect mentally an ohmmeter to determine
divider made of R 4 in series with R th and loaded by R 3 . The
the resistance offered by the capacitor terminals. Figure 4
value of rC is off picture since capacitor C 1 is removed in this
guides you in these steps.
dc analysis. You can, therefore, write
If you run the exercise in Figure 4, you see rC in series
V
R
R
with the parallel combination of R 3 with the series-parallel
(5)
G 0 = Vout = R +2R R + R +3 R < R .
in
2
1
4
3
1
2
arrangement of R 4 and R 1 - R 2 . The time constant of this
circuit is simply the product of R and C 1
We are almost there, and we are missing the zeros. As we
wrote in the section "A Quick Introduction to Fast Analytix 1 = 6rC + ^ R 4 + R 1 | | R 2h | | R 3@ C 1 .
(2)
cal Techniques," a zero manifests itself in a circuit by blockWe can show that the pole of a first-order system is the
ing the propagation of the excitation signal and creating an
inverse of its time constant. Therefore,
output null (see Figure 1). If we consider a transformed circuit-in which C 1 is replaced by 1/sC 1 (as shown in
1
1
.
~ p = x1 =
(3)
Figure 6)-what particular condition would imply a nulled
6rC + ^ R 4 + R 1 | | R 2 h | | R 3@ C 1
response when a stimulus biases the network? Having a
nulled response simply means that the current circulating in
R 3 is zero. This is not a short circuit but rather a virtual
The Response
ground if you prefer the analogy.
R1
R4
If we have no current in R 3, then the series connection
Vout(s)
of rC and 1/sC 1 creates a transformed short circuit:
Vin(s)

rC

+

1
Z 1 ^s zh = rC + s C = 0.
z 1

R3

R2
C1

The
Excitation

The root s z is the zero location that we want
1
sZ = - r C ,
C 1

FIG 3 Determining the time constant of a circuit requires you to
set the excitation to zero and look at the resistance offered by the
energy-storing elements temporarily removed from the circuit.

R1
Vin (s) = 0 V
The Excitation
Is Set to Zero

R4

R?

(7)

and it leads to
1
~z = r C .
C 1

R1

(8)

R4

rC
R2

(6)

rC
R3

R2

R3

R = 250 Ω,
ffor Example

FIG 4 After replacing the 0-V source by a short circuit, you determine the resistance seen from the capacitor terminals.

September 2017

z	IEEE POWER ELECTRONICS MAGAZINE

43



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