NCP1380
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22
OVER POWER COMPENSATION
The over power compensation is achieved by monitoring
the signal on ZCD pin (pin 1). Indeed, a negative voltage
applied on this pin directly affects the internal voltage
reference setting the maximum peak current (Figure 40).
When the power MOSFET is turned−on, the auxiliary
winding voltage becomes a negative voltage proportional to
the input voltage. As the auxiliary winding is already
connected to ZCD pin for the valley detection, by selecting
the right values for R
opu
and R
opl
, we can easily perform
over power compensation.
ZCD/OPP
ESD
protection
Aux
Ropu
Ropl
1
Rzcd
CS
+
−
Vth
DRV
Tblank
leakage blanking
Demag
OPP
V
ILIMIT
IpFlag
Figure 40. Over Power Compensation Circuit
To ensure optimal zero−crossing detection, a diode is
needed to bypass R
opu
during the off−time.
If we apply the resistor divider law on the pin 1 during the
on−time, we obtain the following relationship:
R
ZCD
) R
opu
R
opl
+*
N
p,aux
V
in
* V
OPP
V
OPP
(eq. 2)
Where:
N
p,aux
is the auxiliary to primary turn ration: N
p,aux
= N
aux
/ N
p
V
in
is the DC input voltage
V
OPP
is the negative OPP voltage
By selecting a value for R
opl
, we can easily deduce R
opu
using Equation 2. While selecting the value for R
opl
, we
must be careful not choosing a too low value for this resistor
in order to have enough voltage for zero−crossing detection
during the off−time. We recommend having at least 8 V on
ZCD pin, the maximum voltage being 10 V.
During the off−time, ZCD pin voltage can be expressed as
follows:
V
ZCD
+
R
opl
R
ZCD
) R
opl
ǒ
V
aux
* V
d
Ǔ
(eq. 3)
We can thus deduce the relationship between R
opl
and
R
ZCD
:
R
ZCD
R
opl
+
V
aux
* V
d
* V
ZCD
V
ZCD
(eq. 4)
Design example:
V
aux
= 18 V
V
d
= 0.6 V
N
p,aux
= 0.18
If we want at least 8 V on ZCD pin, we have:
R
ZCD
R
opl
+
V
aux
* V
d
* V
ZCD
V
ZCD
(eq. 5)
+
18 * 0.6 * 8
8
[ 1.2
We can choose: R
ZCD
= 1 kW and R
opl
= 1 kW.
For the over power compensation, we need to decrease the
peak current by 37.5% at high line (370 Vdc). The
corresponding OPP voltage is:
V
OPP
+ 0.375 V
ILIM
+ −300 mV
(eq. 6)
Using Equation 2, we have:
R
ZCD
) R
opu
R
opt
+*
N
p,aux
V
lin
* V
OPP
V
OPP
(eq. 7)
+
−0.18 370 *
(
−0.3
)
(
−0.3
)
+ 221
Thus,
R
opu
+ 221
Ropl
* R
ZCD
+ 221 1k * 1k + 220 kW
(eq. 8)