FA5331P(M)/FA5332P(M)
■ Design advice
1. Start circuit
DB1
Figure 9 shows a sample start circuit. Since the IC current
while the Vcc pin voltage rises from 0V to VTHON is as small as
90µA (typ.), the power loss in resistor RA is small. If an
additional winding is prepared in the voltage step-up inductor
(L), power to the control circuit can be supplied from this
circuit. However, the voltage must be stabilized by a regulator
circuit (REG) to prevent an excess rise of the IC supply voltage
(Vcc). Use fast or ultra-fast rectifier diodes for the rectifier circuit
(DB1) of the winding for high-frequency operation.
L
Io
RA
AC input
Vcc
REG
16
10
C
CA
RS
FA5331/FA5332
7
2. Current sensing resistor
The current sensing resistor (Rs) detects the current in the
inductor. Rs is used to make the input current sinusoidal. The
current in the inductor produces a negative voltage across Rs.
The voltage is input to IC pin 16 (IDET). Determine the value
of Rs so that the peak voltage of the IDET pin is –1V.
Fig. 9 Start circuit
Vin
√2 • Pin
Rs =
.................................................. (11)
Example: FA5332
When Vin is 85V and Pin is 300W, the formulas of (11)
and (12) can be calculated as:
Vin: Minimum AC input voltage (effective value) [V]
Pin: Maximum input power [W]
Since the threshold voltage of the overcurrent limiting circuit
(pin 16) is –1.15V for FA5311 for and –1.10V for FA5332, the
peak input current limit (ip) is determined by:
85
√ 2 • 300
Rs =
= 0.2 [ Ω ]
1.10
0.2
ip =
= 5.5 [ A ]
1.15
FA5331: ip=
.............................................................................(12)
Rs
And,
1.10
Rs
FA5332: ip=
R6
R6 + R7
√ 2 • 85 •
= 0.65 [ V ]
3. Voltage step-up type converter
Figure 9 shows the basic circuit of a voltage step-up type
converter which is used as a power factor correction.
If R6 is set to 2.7kΩ to satisfy these formulas, R7 becomes
480kΩ.
(a) Output voltage
Example:
For stable operation, set the output voltage to be 10V or more
over the peak value of the maximum input voltage. When
using this IC for an active filter, set the output voltage (Vo) as
follows:
When Vin is 85V, Vo is 385V, and γ is 0.2, the formula of (14)
can be calculated as:
2.48 ✕ 104
fs • Pin
......................................... (15)
[ H ]
L ≥
............................................ (13)
Vo ≥ √ 2 • Vin + 10V
Vin: Maximum AC input voltage [V]
(effective value of sinusoidal wave)
(c) Smoothing capacitor
When a voltage step-up converter is used in a power factor
(b) Voltage step-up inductor
correction circuit, the input current waveform is regulated to be
in-phase with the input voltage waveform. Therefore, ripple
noise of twice the input line frequency appears at the output.
The output voltage (υo) is represented as:
When using a voltage step-up converter in continuous current
mode, the ratio of inductor current ripple to the input peak
current is set to about 20%. Determine the inductance as
follows:
Vin2 ( Vo – √ 2 • Vin )
Io
L ≥
υo = Vo –
• Sin 2 ωo t
................................ (14)
................... (16)
γ • fs • Pin • Vo
2 • ωo •C
Vin: Minimum AC input voltage (effective value) [V]
γ : Ratio of inductor current ripple (peak to peak value) to the
input peak current (about 0.2)
fs: Switching frequency [Hz]
Pin: Converter’s maximum input power [W]
Vo:Average output voltage
Io: Output current
ωo: 2π fo (fo: Input power frequency, 50 or 60Hz)
C: Smoothing capacitor value
Therefore, the peak-to-peak value of the output ripple voltage
Vrp is given by:
As the characteristic curves on page 66 show, the peak
voltage at pin 3 should be at least 0.65V, even when the AC
input voltage is minimal. Considering this, determine R6 and
R7 shown in Fig. 6.
Io
Vrp =
..................................................... (17)
ωoC
Using formula (17), determine the necessary C value.
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