Calculating Essential
Charge-Pump Parameters
By Vladimir Vitchev, Customer Service Engineer,
Maxim Integrated Products, Sunnyvale, Calif.
Analysis of a basic charge-pump circuit yields
a model that accurately predicts performance
under a range of operating conditions, while
illustrating a general approach to modeling
other power-supply circuits.
C
apacitive charge-pump circuits are used in many
applications. And though these circuits appear
deceptively simple, engineers working on them
need a thorough understanding of how they
function. By analyzing the model of a basic
charge-pump circuit, it’s possible to derive expressions for
efficiency and output voltage as functions of the pump’s duty
cycle, switching frequency, output and flying capacitances,
switch and other series resistances, and load.
The model derived in this article enables designers to
understand and predict the behavior of charge pumps under
a wide variety of conditions. Mathematical derivations are
shown, both to provide a full understanding of the model and
to provide a general approach for analyzing other complex
power-supply circuits.
Developing the charge-pump model requires some
lengthy derivations of key equations. Space limitations do
not allow us to reproduce all of the derivations here. However, an extended version of this article, available at www.
powerelectronics.com, shows the complete derivations in
their entirety. The online version of this article also includes
a spreadsheet that automates the final expressions derived
1
VG
+
–
R1
C1
1
2
+
C2
+
R2
+
V
–
Fig. 1. The basic charge-pump circuit can
be used for regulated stepdown charge
pumps, inverting regulated charge pumps
and inverting unregulated charge pumps.
Power Electronics Technology July 2006
VG
+
–
here, allowing a quick calculation of essential charge-pump
parameters including output voltage and efficiency.
Charge pumps use a charge-storage element (i.e. a capacitor) to transfer charge from a source to a load. Fig. 1 shows
the basic model of a charge pump. It can represent several
topologies, such as regulated stepdown charge pumps, inverting regulated charge pumps and inverting unregulated
charge pumps. For inverting charge pumps, the output
voltage is negative, but the magnitude of that voltage is the
same as that predicted by the model. Stepup or boost charge
pumps operate in a very similar fashion; however, there are a
number of key differences that are not taken into account in
the model described in this article. Thus, the model cannot
readily be applied to describe their operation.
For unregulated charge pumps, R1 represents the total
resistance of the internal switches, which are usually MOSFETs. For regulated charge pumps, the resistor R1 is a variable
resistance that can be implemented by varying the bias of
the MOSFET switch in the on state.
This resistor regulates the charge pump’s output voltage. Unlike inductive dc-dc converters, in which the output
voltage is regulated according to the duty cycle of the switch,
2
1
R1
C2
C1
V1 u(t)
VG
R2
VC1I (t)
+
–
+
–
+
VC2I (t)
–
V2 u(t)
Fig. 2. This circuit includes the initial conditions for position 1 in steady-state operation.
30
+
–
2
i1
i2
R1
C2
C1
V1 u(t)
R2
VC1II (t)
+
–
+
–
i3
+
VC2II (t)
–
V2 u(t)
Fig. 3. This circuit includes the initial conditions for position 2 in steady-state operation.
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POWER MANAGEMENT
output voltage for the charge pump is regulated by changing
the value of this resistance.
The switch is controlled by an oscillator that alternates
between position 1 and position 2 (Fig. 1). We assume for
this analysis that the circuit is operating in the steady-state
mode. That is the condition used to derive an expression for
output voltage as a function of all the variables.
voltages by placing an initial condition on each capacitor
(i.e., a voltage source connected in series with the respective capacitor), which we will call V1 and V2. The resulting
circuit is shown in Fig. 2.
Having established the initial conditions, let’s solve
for the two capacitor voltages during the first interval (I),
when the switch is connected to position 1. (For interval
two [II], the switch is in position 2.) The voltage across C2
is always the output voltage of the charge pump. Solving
for the voltage across C1 during interval one, i.e., VCI 1(t),
we obtain the following expressions:
Interval One
Because of the steady-state assumption, each capacitor
will be charged to some initial voltage. We can provide those
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t
t
−
−


VCI 1 = VG  1 − e C1R1  + V1 × e C1R1 .


(Eq. 1)
Eq. 1 gives an expression for voltage
across the flying capacitor (C1) during interval one. Note that this voltage
is a function not only of the element
values, but also of the initial condition
V1. By inspection, the output voltage
VCI 2 (t) across output capacitor C2 during interval one is:
t
(Eq. 2)
−
VCI 2 (t) = V 2 × e C 2 R 2 .
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Interval Two
Next, we evaluate what happens
when the switch flips to position 2.
It goes to position 2 at t=DT, where
D is the duty cycle of the switching
frequency and T is its period. Since
we evaluate this interval separately, we
must set the appropriate initial conditions for the two capacitors. We can call
these initial conditions V3 and V4, but
they are actually the final states of the
capacitors at the end of interval one,
and therefore can be derived from the
expressions obtained in Eqs. 1 and 2.
These conditions are shown in the schematic of Fig. 3.
Before analyzing the circuit in
Fig. 3, we need the initial conditions
V3 and V4. They can be obtained by
recognizing that V3 is the voltage across
C1 at time t = DT, and V4 is the voltage
across C2 at t=DT. Therefore, we write
the following expressions:
DT
DT
−
−


V 3 = VG  1 − e C1R1  + V1
V1 × e C1R1


(Eq. 3)
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−
DT
(Eq. 4)
V 4 = V2 × e C2R 2 .
Having obtained the initial conditions for interval two, we can find an
Power Electronics Technology July 2006
32
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POWER MANAGEMENT
Unifying Initial and Final Conditions
expression for voltages across the flying and output capacitors as we did for the first interval. To preserve clarity, we
refer to the initial conditions in this interval as V3 and V4,
even though they are actually functions of the initial conditions at interval one. Solving for the voltage across output
capacitor C2 ( VCII2 ), we obtain the following equation (see
derivations 1, 2 and 3 online):
So far, we have equations that describe the voltages across
the two capacitors in each of the intervals. Next, we need to
equate the initial and final conditions in a way that forces
the equations to represent steady-state operation.
To discuss this point further, we first examine the voltage
across the output capacitor, which is the output voltage of
the charge pump. During interval one, the output capacitor
is connected to the load by itself, and its voltage drops from
the initial condition V2. During that interval, the flying capacitor charges to a voltage higher than its initial condition
V1. During the second interval, the flying capacitor connects
to the output capacitor through R1, so the output voltage
begins to increase.
For the system to be in equilibrium, note that the average
output voltage has to be constant. The only way to achieve
that condition is for the output voltage at the end of interval
two to equal its initial voltage at the beginning of interval
one. The same consideration applies to the conditions for the
flying capacitor, because the average charge supplied to the
output must be constant. We can write these two conditions
in the following way:
VCI 1 (t=0)=V1
(Eq. 7)
VCII1 (t=0)=V3
(Eq. 8)
VCI 1 (t=DT)=V3
(Eq. 9)
VCII1 (t = DT) = V1,
(Eq. 10)
where D equals (1-D).
VCI 2 (t=0)=V2
(Eq. 11)
VCII2 (t=0)=V4
(Eq. 12)
VCI 2 (t=DT)=V4
(Eq. 13)
VCII2 (t = DT) = V 2.
(Eq. 14)
VCII1 (t = DT) =is a function of V1 and V2, and
Since V1=V
II
V2 = VC 2 (t = DT) =is also a function of V1 and V2, we can
form a system of two equations in two unknowns (V1 and
V2), and solve for the equilibrium (steady-state) initial
conditions for interval one. When we obtain V1 and V2, we
can solve for every other parameter of interest, including
the average output voltage, efficiency and output ripple, as
a function of capacitor size, duty cycle, switching frequency,
series resistance and load.
We now write and solve the system of equations. Eqs. 10
and 14 reduce to:
VCII2 (t) = (V 3 × A
A′′ + V 4 × C
C′′)e − s1 × t +
(Eq. 5)
(V 3 × B
B′′ + V 4 × D′
D ′)e − s2 × t ,
wheree s1 , s2 , A ′, B ′ , C ′ and D ′ are constants that depend
on the values of C1, C2, R1, and R2. Because the equations
for these constants are rather complex and require a lot of
space, they are presented in an appendix at the end of the
online version of this article. (See Eqs. A1 through A6 in
the appendix.)
The voltage across the flying capacitor during interval two is
expressed as a function of the output voltage during that period,
and simplified as follows (see derivations 4 and 5 online):
V II (t) = N 2 (V 3 × A ′ + V 4 × C ′)e − s1 × t +
N 4 (V 3 × B ′ + V 4 × D ′)e − s2 × t .
(Eq. 6)
As for Eq. 5, the terms N2 and N4 are constants that depend
on the values of C1, C2, R1 and R2. Because of the complexity
of these equations, they are presented in the online appendix.
(See Eqs. A7, A8, A9 and A10 in the appendix.)
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V1 = N 2 (V 3 × A
A′′ + V 4 × C
C′′)e − s1 × DDT +
− s2 × D
N4(V 3 × B ′ + V4
V 4 × D ′)e DT
V 2 = (V 3 × A ′ + V4
V 4 × C ′)e − s1 × DDT +
(Eq. 16)
(V 3 × B′
B ′ + V 4 × D′
D ′)e − s2 DDT .
Plugging Eqs. 3 and 4 into Eqs. 15 and 16, and rearranging
the terms slightly, we obtain the final system of two equations
in the two unknowns V1 and V2:
DT
DT
 

−
−

V 2 =  VG  1 − e C1R1  + V1
V1 × e C1R1  ×

 

(A ′ × e − s1 × DDTT + B ′ × e − s2 × DDTT ) +
DT
−


− s1 × D
DT
T
C2R 2
V
2
×
e
+ D ′ × e − s 2 × DT )

 (C ′ × e
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Power Electronics Technology July 2006
(Eq. 15)
34
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POWER MANAGEMENT
DT
DT
 

−
−

V1 =  VG  1 − e C1R1  + V1
V1 × e C1R1 

 

VC 2 AVERAGE =
− s × DT
(N2 × A′
A ′ × e − s × DT + N 4 × B ′ × e −s
)+
1
2
(Eq. 17)
(1 − e − s × DDTT ) +
1
DT
−


− s1 × DT
C2R 2
V
2
×
e
+ N 4 × D ′ × e − s2 × DT ).

 (N 2 × C ′ × e
=
1
T
DT
∫V
I
C1
(t)dt +
0
1
T
DT
∫V
II
C1
(t)dt.
(Eq. 23)
0
Evaluating the integrals in Eq. 23, we obtain (see derivation 7 online):
DT
−


VG × D × T + C1
C1R1(V
V11 − VG ) 1 − e C1R1 


VC1 AVERAGE =
+
T
1 N 2 ( V 3 × A ′ + V 4 × C ′)
× (1 − e − s1 × DT ) +
T
s1
N 4 ( V 3 × B ′ + V 4 × D ′)
(Eq. 24)
× (1 − e − s2 × DT ) .
s2
Finally, we calculate the charge-pump circuit’s efficiency. By
definition, efficiency equals power delivered to the output
divided by power consumed at the input. Since we have a
device that is switching and whose voltage waveforms vary
with time, we are interested in the ratio of the average power
at input and output, as opposed to the ratio of instantaneous powers. We define the average power per period as
the integral of instantaneous power divided by the period
interval: (V (tt))2
POUT (t) = C 2
(Eq. 25)
R2
T
1
POUT = POUT (t) = ∫ POUT (t)dt =
T0
(
)
(Eq. 20)
T1 × T3
.
(1 − T4 × T5 )(1 − T2 × T6 ) − T2 × T3 × T4 × T7
Eqs. 19 and 20 represent the result for which we have
been working. They give the precise initial conditions for the
charge-pump model in equilibrium, as a function of all of
the parameters we specified at the beginning of this article:
capacitor size, load, series resistance, frequency or period,
and duty cycle. Knowing the values of V1 and V2, we can
easily solve for other parameters of interest. Knowing V1
and V2 also allows us to plot waveforms for the capacitor
voltages. Because the equations for V1 and V2 are rather
complicated, we can use mathematical software such as
Matlab or Excel to give precise numerical values and plot
the results.
The latter part of this article presents several such plots,
but first we derive some useful and important expressions.
One such is the average output voltage.
T
DT
DT

1 (VC 2 (t))2
1 
I
2
dt
=
(
V
(
t
)
)
d
t
+
(VCII2 (t))2 d
dtt .
C2

∫
∫
∫
T0
R2
T × R2  0

0
(Eq. 26)
Evaluating the integrals in Eq. 26, we obtain (see derivation 8 online):
1
POUT =
T × R2
2 DT
−
 (V 3 × A
V 2 2 C 2R 2 
A′′ + V 4 × C
C′′)2
C2R 2
1
−
e
+
×


2
2s1
Average Output Voltage
We find the average output voltage by integrating the
output capacitor waveform over one period and dividing the
result by the period interval T. Since this waveform consists
of two separate waveforms for each interval, we can split the
integral into two separate integrals as follows:
T
1
VC 2 AVERAGE = VC 2 (t) = ∫ VC 2 (t)dt =
T0
(Eq. 21)
DT
DT
D
T
1
1
I
II
VC 2 (t)dt + ∫ VC 2 (t)dt.
T ∫0
T 0
Evaluating the integrals, we obtain (see derivation 6
online):
Power Electronics Technology July 2006
(V 3 × B
B′′ + V 4 × D
D′′)
× (1 − e − s2 × DT ).
s2 × T
(Eq. 22)
Similarly, we obtain the average flying-capacitor voltage
as follows:
T
1
VC1 AVERAGE = VC1 (t) = ∫ VC1 (t)dt
T0
To solve these equations, it is convenient to reduce their
size by adopting a short notation for some of the constant
terms.
We must also bear in mind that a final numerical solution is feasible only with a computer, which further justifies
the change in notation. Making substitutions for all constant terms except V1 and V2, and rearranging them (see
Eqs. A11 through A17 in the online appendix), we can rewrite
the system of equations as follows:
-(T2  T3)V1+(1-T4  T 5)V2=T1T3
(1-T2  T 6)V1-(T4  T7)V2= T1T6.
(Eq. 18)
We’ve now reduced the system of equations to a manageable form that can be solved for V1 and V2:
(Eq. 19)
T × T6 + (T4 × T7 )V2
V1 = 1
1 − T2 × T6
V2 =
DT
−
 (V 3 × A
C 2R 2 V 2 
A′′ + V 4 × C
C′′)
C2R 2
1
−
e
+
×


T
s1 × T


(
(1 − e −2s × DDT ) +
1
2(V 3 × A
A′′ + V 4 × C
C′′) × (V 3 × B
B′′ + V 4 × D
D′′)
×
s1 + s2
(1 − e −(s + s )DT ) +
1
2
)
(V 3 × B ′ + V 4 × D ′)2
(1 − e −2 × s2 × DT ) .
2s2
(Eq. 27)
Next, we evaluate input power
power
the

the power that is delivered from the source VG to the charge pump. Note that input
36
www.powerelectronics.com
POWER MANAGEMENT
power is consumed only during the first interval, when the
flying capacitor is connected. Because no constant load is
connected to the input (or to the output), we can find the
instantaneous input power by multiplying the input voltage VG by the input current. Thus, it can be shown that the
instantaneous input power is:
−
series resistors for the capacitors. However, the subsequent
analysis is fundamentally the same.
We can now plug the expressions obtained previously
into our numerical-computation software and run some
simulations to see how the circuit works. More importantly,
we can discover whether our model predicts the operation
of a real circuit with accuracy. A calculation spreadsheet in
Excel allows you to input all circuit parameters discussed
in this article, and obtain numerical values for the model’s
average output voltage and efficiency using the equations
derived in this article.
t
V (V − V1)e C1R1
PIN (t) = G G
.
R1
(Eq. 28)
From Eq. 28, we find the average input power is:
DT
1
PIN = PIN (t) = ∫ PIN (t)dt =
T 0
1
T
DT
∫
0
VG (VG − V1 )e
R1
−
t
C 1R 1
dt =
DT
−

C1 × VG (VG − V1) 
C 1R 1
1
−
e

 .
T

(Eq. 29)
Having obtained the average input
power and average output power, we
can now calculate the charge-pump
efficiency as:
P
(Eq. 30)
η = OUT .
PIN
Applying the Results
So far, we’ve analyzed the operation
of the charge-pump model presented
at the beginning of this article. We
obtained a number of equations that express circuit parameters in steady-state
mode as a function of the capacitors
used in the model, the series resistor
with the flying capacitor, switching
frequency, duty cycle, input voltage and
load. These expressions were obtained
by evaluating the circuit in each interval
separately, and then forcing the initialcondition values in each interval to
support steady-state operation in the
circuit.
Many of the equations are complex,
which hampers our intuitive understanding of how the circuit works.
However, you can compensate for this
drawback with mathematical software
such as Matlab or Excel. With the expressions derived in this article, we can
accurately simulate numerous scenarios
by varying the circuit parameters, and
thereby reveal a much more comprehensive picture of the circuit operation.
You can add refinements to the simplified model presented previously, such
as series inductances for the traces and
www.powerelectronics.com
37
Power Electronics Technology July 2006
POWER MANAGEMENT
5
5.5
4.5
4
Average absolute output voltage (V)
Average absolute output voltage (V)
5
4.5
4
Theoretical
model
MAX870
3.5
C2 = 0.22 �F
C1 = 0.22 �F
R1 = 5 �
VG = 5 V
3
2.5
10
Theoretical
model
3.5
3
2.5
2
MAX870
1.5
C2 = 0.204 �F
R2 = 1.0188 k�
R1 = 5 �
VG = 5 V
1
0.5
0
100
1000
10000
100000
0.001
1000000
0.01
0.1
1
10
100
1000
C1(�F)
Load resistance (�)
Fig. 4. A plot of the average output voltage versus load resistance
(R2) for the theoretical model agrees closely with the actual results
obtained with a switched-capacitor voltage inverter (MAX870).
Fig. 5. Values of average output voltage versus flying capacitance (C1)
calculated using the theoretical model are close to the actual results
obtained with the MAX870.
The spreadsheet, which is available in the online version
of this article, provides a quick and easy way to visualize results and use them for designing basic charge-pump circuits.
Also available online is an extended version of this article,
featuring mathematical derivations for all the equations
presented here.
To test the validity of our model, a bench test was performed using a switched-capacitor voltage inverter from
Maxim (MAX870). The internal structure of this device
(an unregulated inverting charge pump) is close to that of
the model, except the output voltage is inverted (i.e., the
output ground connection is flipped during interval two).
That action has no effect on the validity of the equations,
but to compare with results from the model, we must ignore
the output sign and consider only the absolute value of the
output voltage.
A series of tests was performed in which the MAX870’s
load resistor (R2), flying capacitance (C1) and output capacitance (C2) were varied while measuring the absolute value of
the output voltage. The results were then compared with those
predicted by the calculation spreadsheet (Figs. 4-6).
Other parameters for the circuit were measured experimentally: f = 134 kHz, D = 50% and R1 = 5 Ω. As can be
seen, the experimental results agree very closely with results
obtained using the MAX870, thereby validating the model
as a convenient tool for quick calculations and first-order
approximations when designing simple charge pumps.
Note that the charge pump’s average output voltage is
strongly influenced by the value of R1 (Fig. 7). Higher values
Power Electronics Technology July 2006
38
www.powerelectronics.com
POWER MANAGEMENT
5
Theoretical
model
4
4
3.5
MAX870
3
C1 = 0.212 �F
R2 = 1.0188 k�
R1 = 5 �
VG = 5 V
2.5
2
0.0001 0.001
C1 = 0.22 �F
C2 = 0.22 �F
R2 = 1.0188 k�
VG = 5 V
4.5
Average output voltage (V)
Average absolute output voltage (V)
4.5
5
0.01
0.1
1
10
100
3.5
3
2.5
2
1.5
1
0.5
0
1000
0
100 200 300 400 500 600 700 800 900 1000
C2(�F)
R1(�)
Fig. 6. Average output voltage versus output capacitance (C2) for the
theoretical model is compared with actual results for the MAX870.
Fig. 7. The theoretical model reveals the strong influence of resistance
R1 on average output voltage.
of R1 cause the output voltage to drop in magnitude, and a
proper choice of switching frequency, load and flying capacitor make the relationship nearly linear. This is important
because it allows us to understand how to regulate the output
voltage. The mechanism for regulation is almost analogous
to that of a linear regulator.
For regulated charge pumps, R1 is a MOSFET whose resistance is regulated according to a feedback signal that senses
the output voltage. You should obtain the range of values that
R1 can assume from the data sheet for that particular charge
pump. When working with unregulated charge pumps, on
the other hand, R1 represents the sum of switch resistances
seen by the flying capacitor.
That parameter, usually called internal switch resistance,
should be specified in the data sheet for the charge-pump
IC. Note that the capacitor may see more than one switch,
in which case the resistance of individual switches should
be added. If you include a flying capacitor with substantial
series resistance, you should also account for that value in
the parameter R1.
Finally, charge-pump efficiency is an important consideration. An increase of R1 causes efficiency to go down
almost the same way as the output voltage. For relatively high
switching frequencies, it can be shown that charge-pump
efficiency is identical to that of an LDO (i.e., it equals the
ratio of output voltage to input voltage).
Many other conclusions can be drawn from the model.
To examine a parameter of interest while varying the input
variables, be sure to use the calculation spreadsheet, appendix and derivations in the online version of this article at
www.powerelectronics.com.
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Power Electronics Technology July 2006
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Appendix
s1 =
(C1(R1 + R 2) + C2R 2) + (C1R1 − C2R 2)2 + C12 (2 × R1R
R22 + R 22 ) + 2 × C1C2R 22
2 × C1C2R1R 2
s2 =
A′ =
(C1(R1 + R 2) + C2R 2) − (C1R1 − C2R 2)2 + C12 (2 × R1R
R22 + R 22 ) + 2 × C1C2R 22
2 × C1C2R1R 2
−C1R 2
(C1R1 − C2R 2) + C1 (2R1R 2 + R
R22 ) + 2C1C2R 2
2
D′ =
N1 =
N2 =
N3 =
N4 =
2
(Eq. A.2)
(Eq. A.3)
2
C1R 2
B′ =
C′ =
2
(Eq. A.1)
(C1R1 − C2R 2) + C1 (2R1R 2 + R 22 ) + 2C1C2R 22
2
2
C2R 2 − C1(R1 + R 2) − (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R2
R 22
−2 (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R 22
C2R 2 − C1(R1 + R 2) + (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R2
R 22
2 (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R
R222
2 × C2R2
R
C1(R1 + R 2) − C2R 2 + (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R 22
2 × C2R2
R
C2R 2 − C1(R1 + R 2) − (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R 22
2 × C2R2
R
C1(R1 + R 2) − C2R 2 − (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R 22
2 × C2R2
R
C2R 2 − C1(R1 + R 2) + (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R 22
DT
−


T1 = VG  1 − e C1R1 


T2 = e
−
DT
C 1R 1
T3 = (A ′ × e − s1 × DT + B ′ × e − s2 × DT )
T4 = e
−
DT
C2R 2
T5 = (C
C′′ × e − s1 × DDTT + D ′ × e − s2 × DT )
(Eq. A.4)
(Eq. A.5)
(Eq. A.6)
(Eq. A.7)
(Eq. A.8)
(Eq. A.9)
(Eq. A.10)
(Eq. A.11)
(Eq. A.12)
(Eq. A.13)
(Eq. A.14)
(Eq. A.15)
T6 = (N 2 × A ′ × e − s1 × DDTT + N 4 × B ′ × e − s2 × DT ) (Eq. A.16)
T7 = (N 2 × C ′ × e − s1 × DDTT + N 4 × D ′ × e − s2 × DT ) (Eq. A.17)
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Derivations
Derivation 1
V3
s
i1 (s) =
1
R1 +
s × C1
VCII2 (s) −
i 2 (s) = (VCII2 (s) −
i 3 (s) =
V4
)s × C2
s
VCII2 (s)
R2
i1 (s) + i 2 (s) + i 3 (s) = 0
VCII2 (s)
V3
V IIII (s)
−
+ VCII2 (s) × s × C2
C2 − V 4 × C2
C2 + C 2 = 0
1
1 
R2

R1 +
s  R1 +


s × C1
s × C1


1
1
V3


VCII2 (s) × 
+ s × C2 +  =
+ V4
V 4 × C2
1
1 
R2

 R1 +
 s  R1 +


s × C1
s × C1

1
1 
1
 2  1

 C2R 2  s + s  C2R1 + C1R1 + C2R 2  + C1C2R1R 2  
=
VCII2 (s) × 
1 



R
2
s
+




C1R1
V3
1 

+ V 4 × C2  s +
 C1R1
R1
II
VC 2 (s) =
1
1 
1


 1
C2  s 2 + s 
+
+
+

 C2R1 C1R1 C2R 2  C1C2R1R 2 

 V3 
 
R1
+ V 4 × C2
1 

s
+


C1R1
Derivation 2
We now factor the denominator of the final equation in Derivation 1. As a quadratic equation we know it has two roots,
but we must show that the roots are real, regardless of the component values selected for C1, C2, R1 and R2. Imaginary
roots in our time-domain equation would indicate the presence of sinusoidal signals, which (intuition tells us) should
not happen with RC circuits.
1
1 
1
 1
s2 + s 
+
+
+
=0
 C2R1 C1R1 C2R 2  C1C2R1R 2
1
 C1R 2 + C2R 2 + C1R1
s2 + s 
=0
 +

C1C2R1R 2
C1C2R1R 2
2
s1,2
4
 C1(R1 + R 2) + C2R 2 
 C1(R1 + R 2) + C2R 2 
−
 ± 
 −

C1C2R1R 2
C1C2R1R 2
C1C2R1R 2
=
2
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Next, we concentrate on the term inside the square root sign, and make sure that it is a positive number, which in turn
guarantees real roots and no sinusoids at the output:
2
4
C12 (R1 + R 2)2 + 2 × C
C1(R1 + R 2)C2R 2 + C22 R 22 − 4C1C2R1R 2
 C1(R1 + R 2) + C2R 2 
=
=

 −
C1C2R1R 2
C1C2R1R 2
(C1C2R1R 2)2
C12 (R12 + 2R1R 2 + R 22 ) + 2C1C2R1R 2 + 2C1C2R2
R 22 + C22 R2
R 22 − 4 × C1C2R1R 2
=
2
(C1C2R1R 2)
C12 R12 + 2 × R1R 2C12 + C1
C12 R 22 + 2 × C
C11C2R1R 2 + 2 × C1C2R 22 + C22 R 22 − 4 × C1C2
C R
R11R 2
2
(C1C2R1R 2)
C12 R12 + 2 × R1R 2C12 + C1
C12 R 22 − 2C1C2R1R 2 + 2 × C1C2R 22 + C22 R 22
(C1C2R1R 2)2
Rearranging the terms of the numerator we get:
(C12 R12 − 2 × C1C2R1R 2 + C22 R 22 ) + 2 × R1R 2C12 + C
C112 R 22 + 2 × C1C2R 22
=
(C1C2R
R11R 2)2
(C1R1 − C2R 2)2 + C12 (2 × R1R 2 + R 22 ) + 2 × C1C2R 22
(C1C2R1R 2)2
We see from the last equation that the quantity inside the square root is positive, so the roots are indeed real, and correspond to exponential functions in the time domain. We can see this more clearly after factoring them. Going back to
the equation for roots s1 and s2, we get:
s1,2 =
s1,2 =
s1 =
(C1R1 − C2R 2)2 + C12 (2 × R
R11R 2 + R 22 ) + 2 × C1C2R2
R 22
 C1(R1 + R 2) + C2R 2 
−
±

2


C1C2R1R 2
(C1C2R1R 2)
2
−(C1(R1 + R 2) + C2R 2) ± (C1R1 − C2R 2)2 + C12 (2 × R1R
R22 + R 22 ) + 2 × C1C2R 22
2 × C1C2R1R 2
(C1(R1 + R 2) + C2R 2) + (C1R1 − C2R 2)2 + C12 (2 × R1R
R22 + R 22 ) + 2 × C1C2R 22
2 × C1C2R1R 2
(C1(R1 + R 2) + C2R 2) − (C1R1 − C2R 2)2 + C12 (2 × R1R
R22 + R 22 ) + 2 × C1C2R 22
2 × C1C2R1R 2
Note that the minus sign has been omitted from s1 and s2, because we want to represent this quadratic equation in
terms of its roots, in the form (s + s1)(s + s2).
s2 =
Derivation 3
1 
 V3 

V4  s +




C2R1
C1R1
VCII2 (s) =
+
(s + s1 )(s + s2 ) (s + s1 )(s + s2 )
 A′
 C′
B′ 
D′ 
VCII2 (s) = V 3 
+
+ V4 
+

 s + s1 s + s2 
 s + s1 s + s2 
VCII2 (s) =
(V 3 × A
A′′ + V 4 × C
C′′) (V 3 × B′
B ′ + V 4 × D′
D ′)
+
(s + s1 )
(s + s2 )
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where:
 1 


C2R1
A′ =
s + s2
=
−C1R 2
(C1R1 − C2R 2) + C1 (2 × R
R11R 2 + R 2)2 + 2 × C1C2R 22
2
2
s = − s1
 1 


C2R1
B′ =
s + s1
=
C1R 2
(C1R1 − C2R 2) + C1 (2 × R
R11R 2 + R 22 ) + 2 × C1C2R 22
2
2
s = − s2
1
s+
C
1
R1
C′ =
s + s2
=
C2R 2 − C1(R1 + R 2) − (C1R1 − C2R 2)2 + C12 (2 × R1R2
R + R 22 ) + 2 × C1C2R2
R 22
−2 (C1R1 − C2R2
R 2)2 + C12 (2 × R1R 2 + R 22 ) + 2 × C1C2
C 2R 2 2
s = − s1
D′ =
1
C1R1
s + s1
s+
=
C2R 2 − C1(R1 + R 2) + (C1R1 − C2R 2)2 + C12 (2R1R 2 + R 22 ) + 2C1C2R2
R 22
2 (C1R1
R1 − C2R
R22)2 + C12 (2R1R 2 + R
R222 ) + 2C1C2R 22
s = − s2
VII (t) = (V 3 × A
A′′ + V 4 × C
C′′))ee − s1 ×(t ) + (V 3 × B ′ + V 4 × D ′)e − s2 ×(t )
Derivation 4
VCII1 (s) = i1 (s) ×
1
V3
+
s × C1 s

 V3  
  
II

VC 2 (s)
s
1
V3
R1
VCII1 (s) =  C2
×
−
+
×
1
1
R
1
s
×
C
1
s


s+
s+

C1R1
C1R1 
VCII1 (s) =
II
VC2
C 2 (s)
×
C1R1
1
V3
1
V3
−
×
+
1
1
C1R1 
s

ss +
 C1R1
C1R1
Derivation 5
  1  
 1 


  C1R1  V 3


(
V
3
×
A
+
V
4
×
C
C′
)
(
V
3
×
B
+
V
4
×
D′
D
)
′
′
′
′
C
1
R
1
VCIII (s) = 
+
×
−
V
3

+

1
(s + s1 )
(s + s2 )
s

 s  s + 1  
s+
  C1R1 
C1R1

 1 
 1 


(V 3 × A′
A ′ + V 4 × C′
C ′) 
( V 3 × B ′ + V 4 × D ′) 


 C1R1
 C1R1 
Z2  V 3
 Z1
II
VC1 (s) = 
+
+
 − V3  +
1 
1 
s s+ 1 
s






(s + s1 )  s +
(s + s2 )  s +


 C1R1
 C1R1
C1R1 
VCII1 (s) =
(







N1
N2
N3
N4 
(V 3 × A ′ + V 4 × C
C′′)
+
+
 + ( V 3 × B ′ + V 4 × D ′) 
−
  s + 1  (s + s1 )
  s + 1  (s + s2 )
  C1R1

  C1R1



Z2  V 3
 Z1
V3  +
+
s s+ 1 
s


C1R1
)
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VCII1 (s) =
(
(N1(V 3 × A ′ + V4
V 4 × C ′) + N3(V 3 × B ′ + V 4 × D′)
D ) − V 3 × Z2) N2(V 3 × A ′ + V 4 × C ′)
+
+
1 
(s + s1 )

s
+


C1R1
)
N4(V 3 × B ′ + V 4 × D ′) V 3(1 − Z1
Z1)
+
(s + s1 )
s
 1 


2 × C2R2
R
C1R1
N1 =
=
2
(s + s1 )
C1(R1 + R 2) − C2R 2 + (C1R1 − C2R 2) + C12 (2 × R1R2
R 2 + R 22 ) + 2 × C1C2R 22
s= −
 1 


C1R1
N2 =
1 

 s +

C1R1
1
C 1R 1
=
2 × C2R2
R
C2R 2 − C1(R1 + R 2) − (C1R1 − C2R 2)2 + C12 (2 × R11R
R 2 + R 22 ) + 2 × C1C2R 22
s = − s1
 1 


C1R1
N3 =
(s + s2 )
=
2 × C2R2
R
C1(R1 + R 2) − C2R 2 − (C1R1 − C2R 2)2 + C12 (2 × R1R2
R 2 + R 22 ) + 2 × C1C2R 22
1
s= −
C 1R 1
 1 


C1R1
N4 =
1 

 s +

C1R1
=
2 × C2R2
R
C2R 2 − C1(R1 + R 2) + (C1R1 − C2R 2)2 + C12 (2 × R11R
R 2 + R 22 ) + 2 × C1C2R 22
s = − s2
 1 


C1R1
Z1 =
=1
1
s+
C1R1 s = 0
 1 


C1R1
Z2 =
s
= −1
s= −
VCII1 (s) =
1
C 1R 1
(N1(V 3 × A ′ + V4
V 4 × C ′) + N3(V 3 × B ′ + V 4 × D′)
D ) + V3
V 3) (N2(V 3 × A ′ + V 4 × C ′) N4(V 3 × B ′ + V 4 × D ′)
+
+
1
(s + s1 )
(s + s2 )


s
+


C1R1
However, the numerator of the first term in the above equation is equal to zero and thus the voltage of the fly capacitor becomes:
VCII1 (s) =
N2(V 3 × A ′ + V4
V 4 × C ′) N4(V 3 × B ′ + V 4 × D ′)
+
(s + s1 )
(s + s2 )
VCII1 (t) = N2(V 3 × A ′ + V4
V 4 × C ′)e − s1 ×(t ) + N4(V
V33 × B ′ + V 4 × D ′)e − s2 ×(t )
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Derivation 6
1
T
DT
∫
VCI 2 (t)dt =
0
DT
t
DT
−
−

1
C2R 2V
2V2 
V 2 × e C 2 R 2 dt =
1 − e C2R 2 
∫

T0
T


DT
DT
1
1
VII (t)dt = ∫ ((V 3 × A ′ + V 4 × C ′)e − s1 ×(t ) + (V 3 × B ′ + V 4 × D ′)e − s2 (tt ) )dt =
∫
T 0
T 0
( V 3 × A ′ + V 4 × C ′)
(1 − e −s1 × DT ) + (V3 × B ′ + V 4 × D′)(1 − e −s2 × DT )
s1 + T
s2 × T
Derivation 7
T
VC1AVERAGE = ⟨ v C1 (t)⟩ =
VC1AVERAGE =
1
T
1
T
DT
∫
0
1
1
VCI 1 (t)dt =
T ∫0
T
DT
∫V
II
C1
(t)dt
0
t
t
 

−
−

C 1R 1
C 1R 1
+
V
1
×
e
 VG  1 − e
 dt +



DT
∫ (N2(V3 × A ′ + VV44 × C ′)e
− s2 × ( t )
+ N44((V 3 × B ′ + V 4 × D ′)e − s2 ×(t ) )dt =
0
DT
−


VG DT
DT + C1R1(V
V11 − VG ) 1 − e C1R1 


+
T
1  N2(V 3 × A ′ + V 4 × C ′)
N4(V 3 × B ′ + V 4 × D ′)
(1 − e − s1 × DDTT ) +
(1 − e − s2 × DT )]

T
s1
s2
Derivation 8
DT
DT
0
0
I
2
∫ (VC2 (t)) dt =
DT
∫ (V
II
C2
(t))2 dt =
0
∫
2
DT
t
2× t
2 × DT
−
−
−



V 2 2 C 2R 2 
2
C2R 2
C2R 2
C2R 2
V
2
×
e
dt
=
V
2
×
e
dt
d
t
=
1
−
e
∫




2

DT
0
C )e
∫ ((V3 × A ′ + V 4 × C′
− s1 × ( t )
2
+ (V 3 × B ′ + V 4 × D ′)e − s2 ×(t ) ) dt =
0
(V 3 × A ′ + V 4 × C ′)2
2(V 3 × A ′ + V 4 × C ′))((V 3 × B′
B + V 4 × D′
D)
(1 − e −2 × s1 × DT ) +
(1 − e −(s1 + s2 )DT ) +
2 × s1
(s1 + s2 )
(V 3 × B ′ + V 4 × D ′)2
(1 − e −2 × s2 × DT )
2 × s2
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