AIC AN023 An useful model for charge pump converter Datasheet

AN023
An Useful Model for Charge Pump Converter
Analysis
Introduction
This application note focuses on the analysis of
VIN
ION
VOU T
Q1
the charge pump and its equivalent circuit. The
AIC1845 is a micro-power charge pump DC/DC
C IN
RDS-ON
Q2
COUT
ESR
converter that produces a regulated 5V output. This
Q3
C1
kind of converter uses capacitors to store and transfer
R DS-ON
energy. Since the capacitors can not change their
voltage level abruptly, the voltage ratio of VOUT over VIN
Q4
Fig.2 The on State of Charge Pump Circuit
is limited to some range. Capacitive voltage conversion
is obtained by switching a capacitor periodically. It first
VIN
charges the capacitor by connecting it across a voltage
source and then connects it to the output. Referring to
RDS-ON
Q1
CIN
Q2
ESR
Fig.1, During the on state of internal clock, Q1 and Q4
Q3
are closed, which charges C1 to VIN level. During the off
RDS-ON
state, Q3 and Q2 are closed. The output voltage is VIN
IOFF
C1
C OUT
VOU T
Q4
Fig.3 The OFF State of Charge Pump Circuit
plus VC1, that is, 2 VIN.
VIN
Q1
CIN
Q2
C1
Q3
Referring to Fig.2 and Fig.3, here shows the circuit of
VOUT
COUT
Q4
Fig.1 The Circuit of Charge Pump
charge pump at different states of operation. RDS-ON is
the resistance of the switching element at conduction.
ESR is the equivalent series resistance of the flying
capacitor C1. ION-AVE and IOFF-AVE are the average
current during on state and off state, respectively. D is
the duty cycle, which means the proportion the on state
takes. Let’s take advantage of conversation of charge
for capacitor C1. Assume that the capacitor C1 has
reached its steady state. The amount of charge flowing
into C1 during on state is equal to that flowing out of C1
at off state.
ION− AVE × DT = IOFF− AVE × (1 − D)T
(1)
ION- AVE × D = IOFF- AVE × (1 − D)
(2)
February, 2002
1
AN023
2
2
PESR ≅ ION
× ESR × D + IOFF
× ESR × (1 − D)
− AVE
− AVE
I
I
= ( IN ) 2 × ESR × D + ( OUT ) 2 × ESR × (1 − D)
2D
1− D
1 2
1
2
= IOUT
× ESR × + IOUT
× ESR ×
D
1- D
1
2
= IOUT
× ESR ×
D(1 - D)
IIN = ION- AVE × D + IOFF- AVE × (1 − D)
= 2 × ION- AVE × D
(3)
= 2 × IOFF- AVE × (1 - D)
IOUT = IOFF- AVE × (1 − D)
IIN = 2IOUT
(4)
When the duty cycle is 0.5, the power loss of switching
element is
For AIC1845, the controller takes the PSM (Pulse
Skipping Modulation) control strategy. When the duty
2
PRDS −ON ≅ IOUT
×
2
× RDS -ON
0.5(1 − 0.5)
2
= IOUT
× 8RDS −ON
cycle is limited to 0.5, there will be:
… … … … ..(8)
ION- AVE × 0.5 × T = IOFF- AVE × (1 − 0.5) × T
ION- AVE = IOFF- AVE … … … … … … … ..(5)
2
PESR ≅ IOUT
× ESR ×
According to the equation (4), we know that as long as
1
0.5(1 − 0.5)
2
= IOUT
× 4ESR
the flying capacitor C1 is at steady state, the input
In fact, no matter the current is at on state or off state, it
current is twice the output current. The efficiency of
decays exponentially rather than flows steadily. And the
charge pump is given below:
root mean square value of exponential decay is not
? =
VOUT × IOUT
V
×I
V
= OUT OUT = OUT
VIN × IIN
VIN × 2I OUT
2VIN
(6)
equal to that of steady flow. That is why the
approximation comes from.
Let’s consider the power dissipation of RDS-ON and ESR.
Let’s treat the charge pump circuit in another approach
Assume that the RDS-ON of each switching element is
and lay the focus on the flying capacitor C1. Referring to
equal. The approximation of the power loss of RDS-ON
Fig.2, when the circuit is at the on state, the voltage
and ESR are given below:
across C1 is:
PRDS −ON
2
2
≅ ION
× 2R DS −ON × D + IOFF
× 2R DS −ON × (1 − D)
- AVE
- AVE
I
IIN 2
) × 2R DS-ON × D + ( OUT ) 2 × 2R DS-ON × (1 - D)
2D
1- D
2I
I
= ( OUT ) 2 × 2R DS-ON × D + ( OUT ) 2 × 2R DS-ON × (1 - D)
2D
1- D
2
2
2
2
= IOUT
× ( R DS-ON ) + IOUT
×(
R DS-ON )
D
1- D
2
2
= IOUT
×
× R DS-ON
D(1 - D)
=(
… … … … ..(7)
2
AN023
VC-ON (t) = VIN − 2R DS−ON × ION (t) - (ESR × ION (t)) (9)
The average of VC1 during the on state is:
VC−ON− AVE = VIN − (2R DS −ON × ION− AVE ) − (ESR × ION− AVE ) .… … … ..(10)
Similarly, referring to Fig.3, when the circuit is at the off state, the voltage of C1 is:
VC-OFF (t) = VOUT − VIN + 2R DS-ON × IOFF (t) + ESR × IOFF (t)
(11)
The average of VC1 during the off state is:
VC−OFF− AVE = VOUT − VIN + 2R DS−ON × IOFF− AVE + ESR × IOFF− AVE
(12)
The difference of charge stored in C1 between on state
and off state is the net charge transferred to the output
in one cycle.
∆Q = Q ON - Q OFF
= C1 × (VC1−ON− AVE − VC1−OFF− AVE )
= C1 × (2VIN - VOUT - 2R DS-ON × ION- AVE - 2R DS-ON × IOFF- AVE - ESR × ION− AVE - ESR × IOFF- AVE ) … … ..(13)
I
I
I
I
= C1 × (2VIN − VOUT − 2R DS −ON × OUT − 2R DS −ON × OUT - ESR × OUT - ESR × OUT )
D
1− D
D
1- D
1
= C1 × [2VIN − VOUT − (2R DS −ON + ESR) × IOUT ×
]
D(1 − D)
Thus the output current can be written as
IOUT = f × ∆Q = f × (Q ON − Q OFF )
= f × C1 × [2VIN − VOUT - (2R DS - ON + ESR ) × IOUT ×
… … … ..(14)
1
]
D(1 - D)
When the duty cycle is 0.5, the output current can be
IOUT = f × C1 × [2VIN − VOUT − (2R DS − ON + ESR) × IOUT ×
1
]
0.5(1 − 0.5) … … … (15)
= fC1 × [2VIN − VOUT − (8R DS − ON + 4ESR) × IOUT ]
And equation (15) can be re-written as:
2VIN − VOUT =
1
× IOUT + (8R DS−ON + 4ESR) × IOUT … … … … … … ..(16)
fC1
3
AN023
According the equation (16), when the duty cycle is 0.5,
chosen to make the duty cycle of AIC1580 to be 0.5.
the equivalent circuit of charge pump is shown in Fig.4.
The experiment data and the corresponding RDS-ON
The term 8 RDS-ON is the total effect of switching
(evaluated from the equivalent circuit in Fig.4) are listed
resistance, 1/fC1 is the effect of flying capacitor and
in Table.1. The value of RDS-ON is then assigned to all
4ESR is its equivalent resistance.
switching resistors of the simulation circuit, Fig.6. For
IOUT
2VIN
the given loading RLOAD, the output voltage and output
VOUT
current of experiment results and those from simulation
1/fC1 8RDS-ON 4ESR
are shown in Fig.7. From Fig.7, it is easily seen that the
LOAD
COUT
derived equivalent circuit in Fig.4 is quite corresponding
to the original charge pump circuit.
Fig.4 The Euqivalent Circuit of Charge Pump
Experimental and Simulation Results
For the reason of simplicity, the output loading is
VOUT
VIN
1
C1
10µF
2
3
U1
VOUT
GND
CVIN
C+
SHDN
AIC1845
C3
6
5
C2
4
1µF
10µF
RL
Fig. 5 The Test Circuit of AIC1845
4
AN023
Fig.6 The Simulation Circuit of AIC1845
Table. 1 The experiment results and the evaluated RDS-ON.
VIN (V)
RLOAD (Ω)
ESR (mΩ)
VOUT (V)
IOUT (mA)
Frequency(KHz)
RDS-ON(Ω)
2.7
99.4
20
4.77
48.20
640
1.43
2.7
109.4
20
4.83
44.30
640
1.40
2.7
121.2
20
4.89
40.32
640
1.38
2.7
131.3
20
4.92
37.49
640
1.40
2.7
141.2
20
4.95
35.07
640
1.40
Table. 2 The simulation results. (RDS-ON is from Table.1)
VIN(V)
RLOAD(Ω)
ESR(mΩ)
RDS-ON(Ω)
2.7
99.4
20
1.43
2.7
109.4
20
2.7
121.2
2.7
2.7
Frequency
VOUT(V)
IOUT(V)
640
4.83
48.62
1.40
640
4.89
44.68
20
1.38
640
4.94
40.77
131.3
20
1.40
640
4.97
37.84
141.2
20
1.40
640
5.00
35.38
(kHz)
5
IOUT(mA)
AN023
55
53
51
49
47
45
43
41
39
37
35
33
31
29
4.5
4.6
4.7
4.8
4.9
5
5.1
VOUT(V)
5.2
5.3
5.4
5.5
Experiment results
Simulation results-new solution
Simulation results-previous solution
Fig. 7 The Distribution of Experiment Results and Simulation Results
The previous solution means that the equivalent
seen that the terms 1/fC1, 4ESR and 8RDS-ON
resistance includes only the 1/fC term.
should be as small as possible to get large output
current. However, for users, since the RDS-ON is
fixed and manufactured in IC, what we can do is to
lower 1/fC1 and ESR. However even the effect of
Summary:
1/fC1 and ESR can be kept as small as possible,
1. For the capacitive switched converter, the function of
the term 8RDS-ON still dominates the role that limits
the flying capacitor C1 is storing and transferring charge.
the maximum output current.
Due to the law of conservation of charge for capacitor,
we can only obtain the relationship between input
From Fig.4, the equivalent circuit shows a one-pole
current
system. Therefore there is no need to worry about
and
output
current.
Contrarily,
in
the
inductor-used application, owing to the voltage-second
oscillation problem for charge pump converter.
balance of inductor, various voltage conversions are
easily obtained. Therefore for the capacitor switched
converter, to have an arbitrary voltage conversion is
more difficult than inductor-used converter.
2.
From the equivalent circuit shown in Fig.4, it is
6
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