DS9610B Datasheet, PDF (10/12 Pages) Richtek Technology Corporation – High Voltage Synchronous Rectified Buck MOSFET Driver for Notebook Computer

RT9610B
In Figure 1, the current I_g1 and I_g2 are required to move the
gate up to 5V. The operation consists of charging C_gd1,
C_gd2 , C_gs1 and C_gs2. C_gs1 and C_gs2 are the capacitors from
gate to source of the high side and the low side power
MOSFETs, respectively. In general data sheets, the C_gs1
and C_gs2 are referred as “C_iss” which are the input
capacitors. C_gd1 and C_gd2 are the capacitors from gate to
drain of the high side and the low side power MOSFETs,
respectively and referred to the data sheets as “C_rss” the
reverse transfer capacitance. For example, t_r1 and t_r2 are
the rising time of the high side and the low side power
MOSFETs respectively, the required current I_gs1 and I_gs2,
are shown as below :
dV_g1 C_gs1 x 5
I_gs1  C_{gs1 dt}  _tr1
(1)
dV_g2 C_gs1 x 5
I_gs2  C_{gs1 dt}  _tr2
(2)
Before driving the gate of the high side MOSFET up to
5V, the low side MOSFET has to be off; and the high side
MOSFET is turned off before the low side is turned on.
From Figure 1, the body diode “D₂” had been turned on
before high side MOSFETs turned on.
I_gd1
 C_gd1
dV
dt
 C_gd1
5
t_r1
(3)
Before the low side MOSFET is turned on, the C_gd2 have
been charged to V_IN. Thus, as C_gd2 reverses its polarity
and g₂ is charged up to 5V, the required current is :
I_gd2
 C_gd2
dV
dt
 C_gd2
Vi  5
t_r2
(4)
It is helpful to calculate these currents in a typical case.
Assume a synchronous rectified buck converter, input
voltage V_IN = 12V, V_g1 = V_g2 = 5V. The high side MOSFET
is PHB83N03LT whose C_iss = 1660pF, C_rss = 380pF, and
t_r = 14ns. The low side MOSFET is PHB95N03LT whose
C_iss = 2200pF, C_rss = 500pF and t_r = 30ns, from the
equation (1) and (2) we can obtain :
I_gs1
 ^{1660 x 10-12 x 5}
14 x 10^-9
 0.593
(A)
(5)
I_gs2
 ^{2200 x 10-12 x 5}  0.367
30 x 10^-9
(A)
(6)
from equation. (3) and (4)
I_gd1

380 x 10^-12 x 5
14 x 10^-9
 0.136
(A)
(7)
500 x 10^-12 x 12+5
I_gd2 
30 x 10^-9
 0.283 (A)
(8)
the total current required from the gate driving source can
be calculated as following equations :
I_g1  I_gs1  I_gd1  0.593  0.136  0.729 (A)
(9)
I_g2  I_gs2  I_gd2  0.367  0.283  0.65 (A)
(10)
By a similar calculation, we can also get the sink current
required from the turned off MOSFET.
Select the Bootstrap Capacitor
Figure 2 shows part of the bootstrap circuit of the
RT9610B. The V_CB (the voltage difference between BOOT
and PHASE on RT9610B) provides a voltage to the gate
of the high side power MOSFET. This supply needs to be
ensured that the MOSFET can be driven. For this, the
capacitance C_B has to be selected properly. It is
determined by following constraints.
V_IN
BOOT
UGATE
PHASE
V_CC
C_B
+
V_CB
-
LGATE
GND
Figure 2. Part of Bootstrap Circuit of RT9610B
In practice, a low value capacitor C_B will lead to the over
charging that could damage the IC. Therefore, to minimize
the risk of overcharging and to reduce the ripple on V_CB,
the bootstrap capacitor should not be smaller than 0.1μF,
and the larger the better. In general design, using 1μF can
provide better performance. At least one low ESR capacitor
should be used to provide good local de-coupling. It is
recommended to adopt a ceramic or tantalum capacitor.
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DS9610B-07 August 2016