cs51031 ON Semiconductor, cs51031 Datasheet - Page 7

• Manufacturer Part Number: cs51031
• Description: Hysteretic Pfet Buck Controller
• Manufacturer: ON Semiconductor

4) Inductor Selection
operation down to 0.3 Amps.
current < 0.6 A during normal operation.
Ripple current at a minimum off time is:
current, here given by:
5) Output Capacitor
filter. The output capacitor should have a low ESL and ESR.
Low impedance aluminum electrolytic, tantalum or organic
semiconductor capacitors are a good choice for an output
capacitor. Low impedance aluminum are less expensive.
Solid tantalum chip capacitors are available from a number
of suppliers and are the best choice for surface mount
applications.
CS51031 needs a maximum of 20 mV of output ripple for
the feedback comparator to change state. If we assume that
all the inductor ripple current flows through the output
capacitor and that it is an ideal capacitor (i.e. zero ESR), the
minimum capacitance needed to limit the output ripple to
50 mV peak−to−peak is given by:
ripple to 50 mV peak−to−peak is:
less than 83 mW.
and the output voltage ripple will be:
L min +
C +
The inductor value is chosen for continuous mode
The ripple current DI = 2 × I
This is the minimum value of inductor to keep the ripple
A smaller inductor will result in larger ripple current.
DI +
The core must not saturate with the maximum expected
The output capacitor and the inductor form a low pass
The output capacitor limits the output ripple voltage. The
The minimum ESR needed to limit the output voltage
The output capacitor should be chosen so that its ESR is
During the minimum off time, the ripple current is 0.4 A
I MAX + I OUT ) DI 2 + 3.0 A ) 0.4 A 2 + 3.2 A
8.0
( V OUT ) V F )
DV + ESR
( V OUT ) V D )
f SW
DI
ESR + DV
DV
L MIN
+
DI
DI
8.0
DI + 83m W
T OFF(min)
+ 50
T OFF(max)
( 200
0.6 A
OUT
10 3 Hz )
10 *3
+
0.6 A
+
min = 2 × 0.3 A = 0.6 A.
5.6 V
5.6 V
0.4 + 33 mV
+ 83 mW
( 50
28 mH
0.6 A
2.0 ms
10 *3 V )
3.0 ms
+ 0.4 A
+ 28 mH
+ 7.5mF
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CS51031
7
R1 + R2
6) V
resistor divider current should be considerably higher than
this to ensure that there is sufficient bias current. If we
choose the divider current to be at least 250 times the bias
current this permits a divider current of 1.0 mA and
simplifies the calculations.
7) Divider Bypass Capacitor C
a factor of 4.0, i.e. 5.0 V/1.25 V = 4.0, it follows that the
output ripple is also divided by four. This would require that
the output ripple be at least 60 mV (4.0 × 15 mV) to trip the
feedback comparator. We use a capacitor C
AC short.
frequency so we choose C
8) Soft−Start and Fault Timing Capacitor CS
a delay time for load transients so that the IC does not enter
a fault mode every time the load changes abruptly. Secondly
it disables the fault circuitry during startup, it also provides
Soft−Start by clamping the reference voltage during startup,
allowing it to rise slowly, and, finally it controls the hiccup
short circuit protection circuitry. This reduces the duty cycle
to approximately 0.035 during short circuit conditions.
voltage does not reach 2.5 V (the voltage at which the fault
detect circuitry is enabled) before V
otherwise the power supply will never start.
will discharge CS and the supply will not start. For the V
voltage to reach 1.15 V the output voltage must be at least
4 × 1.15 = 4.6 V.
of CS is:
The input bias current to the comparator is 4.0 mA. The
Let R2 = 1.0 K
Rearranging the divider equation gives:
Since the feedback resistors divide the output voltage by
The ripple voltage frequency is equal to the switching
CS performs several important functions. First it provides
An important consideration in calculating CS is that it’s
If the V
If we choose an arbitrary startup time of 900 ms, the value
FB
V OUT + 1.25 V R1 ) R2
CS min +
Divider
FB
V OUT
1.25
pin reaches 1.15 V, the fault timing comparator
1.0 mA
5.0 V
900 ms
* 1.0 + 1.0 kW 5.0 V
t Startup + CS
2.5 V
+ R1 ) R2 + 5.0 KW
R2
264 mA
RR
= 1.0 nF.
I Charge
RR
+ 1.25 V R1
+ 950 nF ^ 0.1 mF
2.5 V
1.25
FB
* 1.0 + 3.0 kW
reaches 1.15 V
R2
RR
) 1.0
to act as an
FB