When the LM3405 starts up, internal circuitry from VIN supplies a 20mA current to the BOOST pin, flowing out of the
BOOST pin into C3. This current charges C3 to a voltage sufficient to turn the switch on. The BOOST pin will continue to
source current to C3 until the voltage at the feedback pin is
greater than 123mV.
There are various methods to derive V
BOOST
:
1.
From the input voltage (VIN)
2.
From the output voltage (V
OUT
)
3.
From a shunt or series zener diode
4.
From an external distributed voltage rail (V
EXT
)
The first method is shown in the Simplified Block Diagram of
Figure 1. Capacitor C3 is charged via diode D2 by VIN. During
a normal switching cycle, when the internal NMOS power
switch is off (T
OFF
) (refer to Figure 2), V
BOOST
equals VIN minus the forward voltage of D2 (VD2), during which the current
in the inductor (L1) forward biases the catch diode D1 (VD1).
Therefore the gate drive voltage stored across C3 is:
V
BOOST
- VSW = VIN - VD2 + V
D1
When the NMOS switch turns on (TON), the switch pin rises
to:
VSW = VIN – (R
DS(ON)
x IL)
Since the voltage across C3 remains unchanged, V
BOOST
is
forced to rise thus reverse biasing D2. The voltage at
V
BOOST
is then:
V
BOOST
= 2VIN – (R
DS(ON)
x IL) – VD2 + V
D1
Depending on the quality of the diodes D1 and D2, the gate
drive voltage in this method can be slightly less or larger than
the input voltage VIN. For best performance, ensure that the
variation of the input supply does not cause the gate drive
voltage to fall outside the recommended range:
2.5V < VIN - VD2 + VD1 < 5.5V
The second method for deriving the boost voltage is to connect D2 to the output as shown in Figure 3. The gate drive
voltage in this configuration is:
V
BOOST
- VSW = V
OUT
– VD2 + V
D1
Since the gate drive voltage needs to be in the range of 2.5V
to 5.5V, the output voltage V
OUT
should be limited to a certain
range. For the calculation of V
OUT
, see OUTPUT VOLTAGE
section.
20178993
FIGURE 3. V
BOOST
derived from V
OUT
The third method can be used in the applications where both
VIN and V
OUT
are greater than 5.5V. In these cases, C3 cannot
be charged directly from these voltages; instead C3 can be
charged from VIN or V
OUT
minus a zener voltage (VD3) by
placing a zener diode D3 in series with D2 as shown in Figure
4. When using a series zener diode from the input, the gate
drive voltage is VIN - VD3 - VD2 + VD1.
20178999
FIGURE 4. V
BOOST
derived from VIN through a Series
Zener
An alternate method is to place the zener diode D3 in a shunt
configuration as shown in Figure 5. A small 350mW to
500mW, 5.1V zener in a SOT-23 or SOD package can be
used for this purpose. A small ceramic capacitor such as a
6.3V, 0.1µF capacitor (C5) should be placed in parallel with
the zener diode. When the internal NMOS switch turns on, a
pulse of current is drawn to charge the internal NMOS gate
capacitance. The 0.1µF parallel shunt capacitor ensures that
the V
BOOST
voltage is maintained during this time. Resistor R2
should be chosen to provide enough RMS current to the zener
diode and to the BOOST pin. A recommended choice for the
zener current (I
ZENER
) is 1mA. The current I
BOOST
into the
BOOST pin supplies the gate current of the NMOS power
switch. It reaches a maximum of around 3.6mA at the highest
gate drive voltage of 5.5V over the LM3405 operating range.
For the worst case I
BOOST
, increase the current by 50%. In
that case, the maximum boost current will be:
I
BOOST-MAX
= 1.5 x 3.6mA = 5.4mA
R2 will then be given by:
R2 = (VIN - V
ZENER
) / (I
BOOST_MAX
+ I
ZENER
)
For example, let VIN = 12V, V
ZENER
= 5V, I
ZENER
= 1mA, then:
R2 = (12V - 5V) / (5.4mA + 1mA) = 1.09kΩ
20178994
FIGURE 5. V
BOOST
derived from VIN through a Shunt Zener
7 www.national.com
LM3405