參數(shù)資料
型號(hào): VJ1206Y224KXXAT
廠商: ON SEMICONDUCTOR
英文描述: Low-Cost 100 mA High-Voltage Buck and Buck-Boost Using NCP1052
中文描述: 低成本百毫安高電壓降壓和降壓升壓使用NCP1052
文件頁(yè)數(shù): 2/10頁(yè)
文件大小: 93K
代理商: VJ1206Y224KXXAT
AND8098/D
http://onsemi.com
2
through the output and a low-frequency ripple will be found
in the output voltage. Hence, the value of C
2
is needed to be
small enough to increase this charging frequency f
VCC
in
order to reduce output voltage ripple because some
efficiency is lost due to this low-frequency ripple.
Input
Output
D
3
C
3
D
L
C
1
C
Z
1
C
2
V
CC
FB
D
S
D
2
Z
2
D
1
R
1
(a) Buck
Input
Output
D
3
C
3
D
L
C
1
C
Z
1
C
2
V
CC
FB
D
S
D
2
Z
2
D
1
R
1
(b) Buck-boost
I
start
I
start
Figure 2. Charging Current of C
2
In Figure 2b it is noted that in the buck-boost topology the
charging current path is blocked by diode D and hence the
charging of C
2
does not affect the output voltage directly.
However, it still affects the output voltage indirectly and
slightly by adding some low-frequency noise on the
inductor. Hence, small value of C
2
is also wanted.
C
1
V
out
R
1
(a) Buck
D
1
C
1
V
out
R
1
(b) Buck-boost
D
1
Figure 3. Output Voltage Couples to C
1
with a
Charging Current
The function of diode D
1
, capacitor C
1
and resistor R
1
are
to transfer the magnitude of output voltage to a voltage
across C
1
so that the IC can regulate the output voltage. In
Figure 3, when the main switch inside the IC is opened and
the diode D is closed. In buck, the potential of the IC
reference ground (pin S) becomes almost 0 V in this
moment. In buck-boost, the potential of the IC reference
ground (pin S) becomes -V
out
in this moment. The voltage
in C
1
will be charged to the output voltage. On the other
hand, when main switch is closed and the diode D is opened,
diode D
1
is reverse biased by a voltage with magnitude V
in
and V
in
+V
out
respectively. Hence, D
1
does not affect the
normal operation of the buck and buck-boost converter.
It is noted that the instantaneous voltage in C
1
can be
possibly greater than the output voltage especially when
output current or output ripple is too large. It directly affects
the load regulation of the circuit since the IC regulates the
output voltage based on the voltage in C
1
. In order to solve
it, larger values of L and R
1
can help to slow down the
charging speed of C
1
. It reduces the maximum instantaneous
voltage in C
1
so that output voltage at high output current
can be pulled up and a good regulation is made.
Larger value of L can help the load regulation but it
usually unwanted because it is bulky. Hence, resistor R
1
is
recommended. Larger value of R
1
makes higher output
voltage. Hence, it is called as a “pull-up resistor” and it can
help to pull up the output voltage slightly.
The voltage in C
1
representing the output voltage is
feedback to the feedback (FB) pin of the NCP1052 through
a diode D
2
and zener diode Z
2
. When output voltage is too
high, there will be a greater-than-50 A current inserting
into the feedback pin of the NCP1052. The NCP1052 will
stop switching when it happens. When output voltage is not
high enough, the current inserting into the feedback is
smaller than 50 A. The NCP1052 enables switching and
power is delivered to the output until the output voltage is
too high again.
The purpose of the diode D
2
is to ensure the current is
inserting into the feedback pin because the switching of
NCP1052 can also be stopped when there is a
greater-than-50 A current sinking from the FB pin. The
purpose of the zener diode Z
2
is to set the output voltage
threshold. The FB pin of NCP1052 with a condition of
50 A sourcing current is about 4.3 V. The volt-drop of the
diode D
2
is loosely about 0.7 V at 50 A. Hence, the output
voltage can be loosely set as follows:
Vout
zener
zener
According to (1), the possible minimum output voltage of
the circuit is 5.0 V when there is no zener diode Z
2
.
If there is no load, the IC will automatically minimize its
duty cycle to the minimum value but the output voltage is
still possible to be very high because there is no passive
component in the circuit try to absorb the energy. As a result,
4.3 V
5 V
0.7 V
(eq. 1)
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