ANP001

ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
Contents
1.
Introduction
2.
Principle of Operation
3.
General Description
4.
Functional Description
5.
Step-down Switching Regulator Design Example
6.
Step-up Switching Regulator Design Example
7.
Voltage-inverting Switching Regulator Design Example
This application note contains new product information. Diodes, Inc. reserves the right to modify the product specification without notice. No
liability is assumed as a result of the use of this product. No rights under any patent accompany the sale of the product.
1/16
ANP001 – App. Note 1
Jul 2005
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© Diodes Incorporated
ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
This paper describes the principle of the AP34063 switching regulator subsystems. Three converter
design examples and application circuits with test data are included.
1.0
Introduction
The AP34063 is monolithic switching regulator subsystems intended for as dc to dc converters.
The device has highly efficient and simple switching power supplies. The use of switching regulator is
becoming more pronounced over that of linear regulators because the size reductions in new
equipment designs require greater conversion efficiency. Another major use of the switching regulator
is that it has flexibility of output voltage. The output can be less than, greater than, or of opposite
polarity to that of the input voltage.
2.0
Principle of Operation
The switching regulator consists of a static reference and a high gain error amplifier identical to
that of the linear regulator. This system added a free running oscillator and a gated latch. The error
amplifier again monitors the output voltage, compares it to the reference level and generates a control
signal. If the output voltage is below nominal, the control signal will go to a high state and turn on the
gate, thus allowing the oscillator clock pulses to drive the series-pass element alternately from cutoff to
saturation. This will continue until the output voltage is pumped up slightly above its nominal value. At
this time, the control signal will go low and turn off the gate, terminating any further switching of the
series-pass element. The output voltage will eventually decrease to below nominal due to the presence
of an external load, and will initiate the switching process again. The increase in conversion efficiency
is primarily due to the operation of the series-pass element only in the saturated or cutoff state. When
saturated, the voltage drop across the element is as small as the dissipation. When in cutoff, the
current through the element and likewise the power dissipation are also small. The most common are
the fixed frequency pulse width modulator and the fixed on-time variable off-time types, where the
on-off switching is uninterrupted and regulation is achieved by duty cycle control. Generally speaking,
the example given in Figure 1 does apply to AP34063.
2/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
VI
N
VOUT
ERROR
AMP
7
V REF
3
+
6
2
4
A. Linear Regulator
VOU
T
VIN
INDUCTO
R
RESISTOR
CAPACITOR
ERROR AMP
+
6
-
3
V REF
2
4
GATED LATCH
7
DIODE
RESISTOR
OSCILLATOR
B. Switching Regulator
Figure 1. Step-Down Regulators
3/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
3.0
General Description
The AP34063 is a monolithic control circuit containing all the active functions required for dc to dc
converters. This device contains an internal temperature compensated reference, comparator,
controlled duty cycle oscillator with an active peak current limit circuit, driver, and a high current output
switch. This series was specifically designed to be incorporated in step-up, step-down and
voltage-inverting converter applications. These functions are contained in an 8 pin dual in-line package
shown in Figure 2.
8
1
A
S
B
Q
Q1
AND
GATE
Q2
R
7
R
R-S LATCH
2
Ipk
Ct
OSCILLATOR
6
3
+
1.25V
REFERENCE
VOLTAGE
COMPARATOR
5
4
Figure 2. AP34063
4/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
4.0
Functional Description
The oscillator is composed of a current source and sink which charges and discharges the
external timing capacitor CT. The typical charge and discharge current are 35μA and 200uA
respectively, yielding about 1 to 6 ratio. The ramp-up period is 6 times longer than ramp-down period
as shown in Figure 3. The upper threshold is equal to the internal reference voltage 1.25V and the
lower is close to 0.75V. The oscillator runs continuously at a rate controlled by the selected value of
capacitor CT.
During the ramp-up portion of the cycle, a “HIGH” present at the ‘A’ input of the AND gate. If the
output voltage of the switching regulator is below nominal, a “HIGH” will also be present at the ‘B’ input.
This condition will set the latch and cause the ‘Q’ output to go “HIGH”, enabling the driver and output
switch to conduct. When the oscillator reaches its upper threshold, CT will start to discharge and
“LOW” will be present at the ‘A’ input of the AND gate. This logic level is also connected to an inverter
whose output presents a “HIGH” to the reset input of the latch. This condition will cause ‘Q’ to go
“LOW”, disabling the driver and output switch.
The output of the comparator can set the latch only during the ram-up of CT and can initiate a
partial or full on-cycle of output switch conduction. The comparator sets the latch, but it cannot reset
the latch. The latch will remain set until CT begins ramping down. Thus the comparator can initiate
output switch conduction, but cannot terminate it and the latch is always reset when CT begins ramping
down. The comparator’s output will be “LOW” when the output voltage of the switching regulator is
above nominal.
V
6t
T
t
charge
discharge
Figure 3. Ct Voltage Charge-Discharge Waveform
5/16
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Application Note
Application of the AP34063 Switching Regulator Control Circuits
5.0
Step-Down Switching Regulator Design Example
A schematic of the basic step-down regulator is shown in Figure 4. The frequency chosen is a
comprisal between switching losses and inductor size. Given are the following conditions:
Vout = 5.0V
Iout = 500mA
fmin = 50KHz
Vin = 12V~16V
Vripple(p-p) = 50mVp-p
5.0.1
Determine the ratio of switch conductor ton versus diode conduction toff time.
ton
toff
=
=
Vout + VF
Vin - Vsat - Vout
5 + 0.8
............................................. (1)
12 - 1.4 - 5
= 1.036
5.0.2 The cycle of the LC network is equal to ton(max) + toff.
t on + t off =
1
=
f min
1
50 × 10
3
..................... (2)
= 20 μ s per cycle
5.0.3
Using equation (1) and equation (2), we can obtain Ton and Toff time respectively.
t on + t off
t off =
t on
+1
t off
=
20 × 10
-6
1.036 + 1
.................................................... (3)
= 9.8 μs
Since t on + t off = 20 μ s
t on = 20 - 9.8 = 10.2 μ s ..................................... (4)
6/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
5.0.4
The ton is set by selecting a value for CT
C T = 4.0 × 10
−5
× t on
−5
−6
= 4.0 × 10 (10.2 × 10 ) = 408pF ................... (5)
Use a standard 470pF capacitor.
5.0.5 The circuit works in Continuous-Conduction Mode (CCM), so the peak switch current
in inductance is:
Ipk = 2Iout
= 2(500mA) ..................................................... (6)
= 1A
5.0.6 By using the peak switch current and on time, a minimum value of inductance can be
calculated.
L (min) =
=(
Vin(min) − Vsat − Vout
Ipk
12 - 1.4 - 5
1
× t on
.................... (7)
) × 10.2 × 10
−6
=57μH
Since the minimum value of inductance is 57μH, we use 100μH for the inductance.
5.0.7 The current limit resistor RSC can be determined by using the peak switch current
when Vin = 24V.
I'pk =
Vin − Vsat − Vout
L
× t on
⎛ 16 - 1.4 - 5 ⎞
−6
× 10.2 × 10
-6 ⎟
⎝ 100 × 10 ⎠
=⎜
= 0.98A
0.33
R SC =
Ipk
=
0.33
..................... (8)
0.98
= 0.33Ω
7/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
5.0.8 With knowledge of the peak switch current, minimum operation frequency, and the
peak to peak voltage of the ripple, an ideal output filter capacitor can be obtained.
Cout =
=
Ipk
8Vripple(p-p) f(min)
1
8(50 × 10
−3
3
)(50 × 10 )
............................. (9)
= 50 μ F
In order to suppress the output voltage of ripple under 40mV, we choose 470μF for this capacitor. If
you want the lowest output voltage of the ripple, consider using the lowest ESR of capacitor you
used now.
5.0.9 The output voltage is programmed by the R1, R2 resistors divider. The output voltage
is:
⎛
⎝
Vout = 1.25⎜ 1 +
R2 ⎞
⎟ .............................................. (10)
R1 ⎠
Since the divider current can go as low as 100μA without affecting system performance, a
minimum current divider R1 is equal to:
R1 =
1.25
100 μ A
= 12500Ω = 12.5KΩ
We choose R1 = 12KΩ.
Using equation (10), R2 can be solved:
⎛ Vout ⎞
− 1⎟
⎝ 1.25 ⎠
R2 = R1⎜
= 12 × 10
3⎛
5
⎞
− 1⎟
⎜
⎝ 1.25 ⎠
= 36KΩ
Using the above derivation, the design circuit is optimized to meet the assumed conditions.
8/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
ANALOG TECHNOLOGY, INC.
8
Rsc
7
0.33
Vi
n
12V~16V
U1
6
VDC
ISW
C
IPK
ISWE
VCC
5
+
C
T
COM
GND
P
AP3406
3
1
2
3
4
470
p
C
t
R2
Cin
470
u
100
u
D1
1N581
9
L1
Vout
1% 36K
5V/500mA
R1
+
1% 12K
Cou
470
t
u
STEP-DOWN CONVERTER
Figure 4
6.0
Step-Up Switching Regulator Design Example
A schematic of the basic step-down regulator is shown in Figure 5. Given are the following
conditions:
Vout = 28V
Iout = 200mA
fmin = 50KHz
Vin = 9.0V~12.0V
Vripple(p-p) = 40mVp-p
6.0.1
Determine the ratio of switch conductor ton versus diode conduction toff time.
ton
toff
=
=
Vout + VF − Vin(min)
Vin(min) - Vsat
28 + 0.8 - 9
9 - 0.8
........... (11)
= 2.41
9/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
6.0.2
The cycle of the LC network is equal to ton(max) + toff.
ton + toff =
1
=
fmin
1
50 × 10
3
..................... (12)
= 20 μ s per cycle
6.0.3
Using equation (11) and equation (12), we can obtain Ton and Toff time respectively.
t off =
t on + t off
t on
t off
=
+1
20 × 10
-6
..........................................…..... (13)
2.41 + 1
= 5.87 μ s
Since t on + t off = 20 μ s
t on = 20 - 5.87 = 14.13 μ s ............................... (14)
6.0.4 The ton is set by selecting a value for CT
C T = 4.0 × 10
= 4.0 × 10
Use a standard 680pF capacitor.
−5
(14.13 × 10
−5
−6
× t on
) = 565.2pF ....…... (15)
6.0.5 The circuit works in Continuous-Conduction Mode (CCM), so the peak switch current
in inductance is:
⎛ t on ⎞
⎟
⎜ t + 1⎟
⎝ off
⎠
Ipk = 2 Iout ⎜
= 2 (200mA)(2. 41 + 1) .................................... (16)
= 1.364A
6.0.6 By using the peak switch current and on time, a minimum value of inductance can be
calculated.
L (min) =
=(
Vin(min) − Vsat
Ipk
12 - 0.8
1.364
× t on
) × 14.13 × 10
−6
.......................... (17)
= 116 μ H
Since the minimum value of inductance is 116μH, we use 120μH for the inductance.
10/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
6.0.7 The current limit resistor RSC can be determined by using the peak switch current when
Vin = 12V.
Ipk =
Vin − Vsat
L
× t on
⎛ 12 - 0.8 ⎞
−6
× 14.13 × 10
-6 ⎟
⎝ 120 × 10 ⎠
=⎜
= 1.32A
0.3
R SC =
Ipk
=
0.3
1.32
................................ (18)
= 0.22Ω
6.0.8 With knowledge of the peak switch current, minimum operation frequency, and the
peak to peak voltage of the ripple, an ideal output filter capacitor can be obtained.
Cout =
=
Iout
Vripple(p-p) f(min)
200 × 10
(40 × 10
−3
-3
3
)(50 × 10 )
..........(19)
= 100 μ F
In order to suppress the output voltage of the ripple under 40mV, we choose 470μF for this
capacitor. If you want the lowest output voltage of the ripple, consider using the lowest ESR of
capacitor you used now.
11/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
6.0.9
is:
The output voltage is programmed by the R1, R2 resistors divider. The output voltage
⎛
⎝
R2 ⎞
Vout = 1.25⎜ 1 +
⎟ .............................................. (20)
R1 ⎠
Since the divider current can go as low as 500μA without affecting system performance, a
minimum current divider R1 is equal to:
R1 =
1.25
= 2500Ω = 2.5KΩ
500 μ A
We choose R1 = 2.5KΩ.
Using equation (20), R2 can be solved:
⎛ Vout ⎞
− 1⎟
⎝ 1.25 ⎠
⎞
3 ⎛ 28
= 2.5 × 10 ⎜
− 1⎟
⎝ 1.25 ⎠
R2 = R1⎜
= 53.5KΩ (use 53KΩ or 54KΩ )
Using the above derivation, the design circuit is optimized to meet the assumed conditions.
ANALOG TECHNOLOGY, INC.
L1
R
3
Rsc
Vin
9.0V ~12V
0.2
1
+
180
Vin
9.0V~12V
8
7
6
5
Ci
n
470u
U
1
VD
C
IP
K
VC
C
120u
GN
D
C
T
ISW
E
COMP
ISWC
4
3
2
1
D
1
AP34063
R
2
1N5819
1% 56K
R
1
10K
Ct
Cout +
680p
470u
Vout
28V/200mA
R
5
5.1K
D202
LED
STEP-UP CONVERTER
Figure 5
12/16
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Application Note
Application of the AP34063 Switching Regulator Control Circuits
7.0
Voltage-Inverting Switching Regulator Design Example
A schematic of the basic voltage-inverting regulator is shown in Figure 6. The frequency chosen
is a comprisal between switching losses and inductor size. Given are the following conditions:
Vout
Iou t
fmin
Vin
=
=
=
=
=
Vripple(p-p)
-12V
100mA
50KHz
4.5V~6.0V
40mVp-p
7.0.1 Determine the ratio of switch conductor ton versus diode conduction toff time.
ton
=
toff
=
| Vout | + VF
Vin - Vsat
12 + 0.8
........... (21)
4.5 - 0.8
= 3.46
7.0.2
The cycle of the LC network is equal to ton(max) + toff
ton + toff =
1
=
fmin
1
50 × 10
3
.......... (22)
= 20 μ s per cycle
7.0.3
Using equation (21) and equation (22), we can obtain Ton and Toff time respectively.
t off =
t on + t off
t on
t off
=
+1
20 × 10
-6
3.46 + 1
...................................... (23)
= 4.48 μ s
Since t on + t off = 20 μ s
t on = 20 - 4.48 = 15.52 μ s ..............(24)
13/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
7.0.4
The ton is set by selecting a value for CT
C T = 4.0 × 10
−5
× t on
−5
(15.52 × 10
= 4.0 × 10
Use a standard 680pF capacitor.
−6
) = 620.8pF ............... (25)
7.0.5 The circuit works in Continuous-Conduction Mode (CCM), so the peak switch current
in inductance is:
⎛ t on ⎞
+ 1⎟
⎝ t off ⎠
Ipk = 2Iout ⎜
= 2(100mA)(3 .46 + 1) .................................... (26)
= 0.892A
7.0.6 By using the peak switch current and on time, a minimum value of inductance can be
calculated.
Vin(min) − Vsat
L (min) =
Ipk
=(
4.5 - 0.8
0.892
× t on
) × 15.52 × 10
−6
........................ (27)
= 64 μ H
Since the minimum value of inductance is 64μH , we use 100μH for the inductance.
7.0.7 The current limit resistor RSC can be determined by using the peak switch current
when Vin = 6.0V.
Ipk =
Vin − Vsat
L min
× t on
⎛ 6.0 - 0.8 ⎞
−6
× 15.52 × 10
-6 ⎟
⎝ 64 × 10 ⎠
=⎜
= 1.261
0.33
R SC =
Ipk
=
0.33
1.261
............................. (28)
= 0.26Ω
14/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
7.0.8 With knowledge of the peak switch current, minimum operation frequency, and the
peak to peak voltage of the ripple, an ideal output filter capacitor can be obtained.
Cout =
=
Iout
Vripple(p-p) f(min)
100 × 10
(40 × 10
−3
-3
3
)(50 × 10 )
...................... (29)
= 50 μ F
In order to suppress the output voltage of ripple under 40mV, we choose 470μF for this
capacitor. If you want the lowest output voltage of ripple, consider using the lowest ESR of
capacitor you used now.
7.0.9 The output voltage is programmed by the R1, R2 resistors divider. The output voltage
is:
⎛
⎝
Vout = 1.25⎜ 1 +
R2 ⎞
⎟ ............................(30)
R1 ⎠
Since the divider current can go as low as 400μA without affecting system performance, a
minimum current divider R1 is equal to:
R1 =
1.25
400 μ A
= 31250Ω = 3.125KΩ
We choose R1 = 3KΩ.
Using equation (30), R2 can be solved:
⎛ Vout
R2 = R1⎜⎜
⎞
− 1⎟⎟
⎝ 1.25 ⎠
⎞
3 ⎛ 12
= 3 × 10 ⎜
− 1⎟
⎝ 1.25 ⎠
= 25.8KΩ (use 26KΩ )
Using the above derivation, the design circuit is optimized to meet the assumed conditions.
15/16
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ANP001
Application Note
Application of the AP34063 Switching Regulator Control Circuits
ANALOG TECHNOLOGY,INC.
8
Vin
4.5V to 6.0V
Rsc
7
0.26
6
5
U1
VDC
ISWC
IPK
ISWE
VCC
1
2
3
CT
COMP
4
GND
Ct
680p
L1
D1
1N5819
100u
AP34063
1% 3K
+ Cin
Vout
-12V/100mA
R1
470u
D2
LED
Cout
+
R2
50K
470u
R3
2.4K
VOLTAGE INVERTING CONVERTER
Figure 6
16/16
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