NCP1650_05 ONSEMI [ON Semiconductor], NCP1650_05 Datasheet - Page 26

• Manufacturer Part Number: NCP1650_05
• Description: Power Factor Controller
• Manufacturer: ONSEMI [ON Semiconductor]

be advisable to increase the input power to assure operation
at maximum power over production tolerance variations.
it’s pole about a factor of 10 below the switching frequency.
Where:
C
f = pole frequency (kHz)
so, for a 100 kHz switching frequency, a 10 kHz pole is
desirable, and C
Maximum Power Circuit
and a scale factor, to output a value that is proportional to the
input power. This voltage is filtered to remove the line
frequency components. The resulting output is compared to
the 2.5 volt reference on the power error amplifier. When the
output of the multiplier reaches 2.5 volts the power loop
takes control and will reduce the output voltage as necessary,
but can not reduce it to less than the peak of the line voltage.
that the unit will power limit at a value slightly greater than
the maximum power desired. R
formula:
Where:
V
R
AC
Pin = rated input power (w)
R
current shunt and a 1% AC divider, the RSS error will be 7%
maximum, or a worst case error of 14%. In order to assure
maximum power output the reference voltage (V
be reduced by the error factor.
close to a DC level, so a filter cap needs to be added with a
high frequency pole relative to the line frequency. For a
60 Hz line, a 0.6 Hz pole would allow 40 dB of attenuation,
or .01 which would reduce a 5.0 volt p−p signal to a DC level
of 2.5 volts, with 50 mv of ripple. The chosen frequency will
be a tradeoff of response time vs. ripple. For a pole of 0.6 Hz:
Where:
C
R
11
10
S
9
9
9
This equation does not allow for tolerances, and it would
The current sense filter capacitor should be selected to set
The power multiplier multiplies the input voltage, current
For proper operation, resistor R
The NCP1650 has been designed such that with a 2%
The output signal from the power multiplier should be
= Shunt resistance (W)
= Pin 9 capacitance (F)
= Pin 9 resistance (W)
= Power reference voltage (2.5 v nom)
ratio
= Pin 11 capacitance (nF)
= Current scaling resistor (W)
= AC attenuation factor at pin 5
R 9 +
11
C 9 +
would be 1.0 nF.
AC ratio Pin R S 3.75
2 @ p @ R 9 @ 0.6
C 11 + 10.6
V 9 R 10
1
9
f
can be calculated by the
9
should be chosen such
+ 0.265
R 9
9
) should
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NCP1650
26
Reference Multiplier
modulated representation of the analog input. The multiplier
is internally loaded with a resistor to ground which will set
the DC gain. An external capacitor is required to filter the
signal back into one that resembles the input fullwave
rectified sinewave. The pole for this circuit should be greater
than the line frequency and lower than the switching
frequency.
starting value for a 60 Hz line frequency. The filter capacitor
for pin 4 can be determined by the following equation:
C
f
AC Error Amplifier
that is terminated with a series RC impedance. This creates
a pole−zero pair.
look at the two signals that reach the PWM inputs. The
non−inverting input is a slow loop using the averaged
current signal. It’s gain is:
amplifier averaging circuit. The next term is the gain of the
transconductance amplifier and the constant is the gain of
the AC Reference Buffer.
current signal to the PWM non−inverting input. This gain is
simply 16, since the input signal is converted to a current
through a 1 k resistor, and then terminated by the 16 k
resistor at the PWM input.
less than the gain of the high frequency path. This can be
written as:
and for a zero at 1/10
Where:
R
g
C
f
pole
sw
m
4
3
3
The output of the reference multiplier is a pulse width
1/15
The AC error amplifier is a transconductance amplifier
To determine the values of R
Where the first two terms are the gains in the current sense
The high frequency path is that of the instantaneous
For stability, the gain of the low frequency path must be
The suggested resistor and capacitor values are:
= Pin 4 capacitance (F)
and R
is in Farads
is in units of mhos
is in Hz
= Ref gain pole freq (Hz)
th
10
of the switching frequency is a recommended
C 4 +
are in units of W
A lf + 15 k
517, 500 @ g m @ R 3
2 @ p @ 25 k @ f pole
1 k
R 3 +
th
C 3 + 1.59
@ 15 k
of the switching frequency
R 10
R 10
1
56, 000 g m
f sw R 3
@ (g m @ R 3 ) @ 2.3
R 10
3
and C
t 16
+ 6.366E * 6
3
, it is necessary to
f pole