NCP1651
http://onsemi.com
27
AC Voltage Divider
The voltage divider from the input rectifiers to ground is
a simple but important calculation. For this calculation it is
necessary to know the maximum line that the unit can
operate at. The peak input voltage will be:
Vin
peak
= 1.414 x Vrms
max
ThemaximumvoltageattheACinput(pin5)is3.75volts
(this is true for both multipliers).
If the maximum line voltage is 265 Vac, the peak input
voltage is:
Vin
peak
= 1.414 x 265 V
rms
= 375 V
pk
To keep the power dissipation reasonable for a 1/2 watt
resistor(R
ac1
),itshoulddissipatenomorethan1/4watt.The
power in this resistor is:
PR
ac1
= (375 V - - 3.75 V)
2
/ R
ac1
= 0.25 watts
so: R
ac1
= 551 k?/DIV>
Tominimizedissipation,usethenextlargeststandardvalue,
or 560 k?
Typically, two 1/4 resistors are used in series to handle the
power.
Then, R
ac2
= 3.75 V / ((375 V - - 3.75 V) / 560 k) = 5.6 k?/DIV>
Current Sense Resistor/Ramp Compensation
The combination of the voltage developed across the
current sense resistor and ramp compensation signal, will
determine the peak instantaneous current that the power
switch will be allowed to conduct before it is turned off.
Thevectorsumofthethreesignalsthatcombinetocreate
thesignalatthenon- -invertinginputtothePWMcomparator
must add up to 4.0 volts in order to terminate the switch
cycle. These signals are the error signal from the AC error
amp, the ramp compensation signal, and the instantaneous
current. For a worst case condition, the output of the AC
errorampcouldbezero(current),whichwouldrequirethat
thesumoftherampcompensationsignalandcurrentsignal
be4.0volts.Thismustbeevaluatedunder fullloadandlow
line conditions.
For proper ramp compensation, the ramp signal should
matchthefallingdi/dt(whichhasbeenconvertedtoadv/dt)
oftheinductorat50%dutycycle.50%dutycyclewilloccur
whentheinputvoltageis50%oftheoutputvoltage.Boththe
falling di/dt and output voltage need to be reflected by the
transformer turns ratio to the primary side. Thus the
following equations for R
S
and R
RC
must be satisfied:
di/dt primary = V
in
/L
P
?T/2
di/dt secondary = V
O
/L
S
?T/2
L
S
=r
N
S
N
P
u
2
L
P
di/dt reflected to the primary:
r
V
O
L
P
ur
N
P
N
S
u
2
?/DIV>
T
2
?/DIV>
N
S
N
P
Simplifies to:
V
O
/L
P
?N
P
/N
S
?T/2
di/dt primary = di/dt secondary
V
in
/L
P
?T/2 = V
O
/L
P
?N
P
/N
S
?T/2
V
in
/L
P
= V
O
/L
P
?N
P
/N
S
Equation 2)
For proper slope compensation, the relationship between
R
S
and R
RC
is:
di/dt (primary) ?T ?R
S
?High Frequency Current Gain =
V
Rcomp
V
O
/L
P
?T ?N
P
/N
S
?R
S
?16 k/3 k = 102.4 k/R
RC
R
S
= (19,200/R
RC
?T) ?(L
P
/V
O
) ?(N
S
/N
P
)
Equation 3)
t
on
= T/(N
S
/N
P
?(? ?V
LL
/V
O
)) + 1
For maximum output current, when the error amplifier is
saturated in a low state, the ramp compensation signal plus
the current signal must equal 4.0 volts (3.8 volts is used to
avoid over driving the amplifier), which is the reference
level for the PWM comparator. So:
Equation 4) Vref
PWM
= Vin
ST
+ V
Rcomp
3.8 V = I
PK
?R
S
?16 k/3 k + 102.4 k/R
RC
?t
on
/T
R
RC
=
102.4 k t
on
(3.8 5.3 I
PK
R
S
)
Combining equations 2 and 4 gives:
R
S
=
3.8
N
P
N
S
t
on
V
O
0.1875 L
P
+ 5.33 I
pk
Where:
R
S
is the current shunt resistor (Ohms)
R
RC
is the ramp compensation resistor (Ohms)
t
on
is the on time for the conditions given (ms)
T is the period for the switching frequency (ms)
L
P
is the primary inductance of the transformer (mH)
V
out
is the output voltage (VDC)
V
rms
is the rms line voltage at low line (V
rms
)
P
out
is the output power at full load (watts)
I
avg
(T) is the average current for one switching cycle (A)
I
pk
is the instantaneous peak primary side current (A)
V
(t)
is the peak line voltage (volts)
N
P
/N
S
is the transformer turns ratio (dimensionless)
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