RT9607 Datasheet, PDF (12/15 Pages) Richtek Technology Corporation – Dual Channel Synchronous-Rectified Buck MOSFET Driver

RT9607/A
In Figure 1, the current I_g1 and I_g2 are required to move the
gate up to 12V.The operation consists of charging C_gd
and C_gs. C_gs1 and C_gs2 are the capacitances from gate to
source of the high side and the low side power MOSFETs,
respectively. In general data sheets, the C_gs is referred as
“C_iss” which is the input capacitance. C_gd1 and C_gd2 are
the capacitances 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 showed
below
Igs1 = Cgs1 ^dVg1 = ^{Cgs1 x 12}
(1)
dt
tr1
Igs2 = Cgs2 ^dVg2 = ^{Cgs2 x 12}
(2)
dt
tr2
According to the design of RT9607/A, before driving the
gate of the high side MOSFET up to 12V (or 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
Igd1 = Cg1 ^dV = Cgd1 ^12V
(3)
dt
tr1
Before the low side MOSFET is turned on, the C_gd2 have
been charged to Vi. Thus, as C_gd2 reverses its polarity
and g₂ is charged up to 12V, the required current is
Igd2 = Cgd2 ^dV = Cgd2 ^{Vi + 12V}
(4)
dt
tr2
It is helpful to calculate these currents in a typical case.
Assume a synchronous rectified BUCK converter, input
voltage V_i = 12V, V_g1 = V_g2 = 12V. 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
Igs1
=
1660 x
14
10 ^-12
x 10-9
x 12
= 1.428A
(5)
Igs2
=
2200 x 10^-12
30 x 10-9
x 12
=
0.88A
(6)
from equation. (3) and (4)
Igs1
=
380 x 10^-12 x
14 x 10^-9
12
=
0.326A
(7)
Igs2
=
500
x
10^-12 x (12
30 x 10^-9
+
12)
=
0.4A
(8)
the total current required from the gate driving source is
Ig1 = Igs + Igd1 = (1.428 + 0.326) = 1.745A
(9)
Ig2 = Igs2 + Igd2 = (0.88 + 0.4) = 1.28A
(10)
By a similar calculation, we can also get the sink current
required from the turned off MOSFET.
Layout Consider
Figure 2. shows the schematic circuit of a two-phase
synchronous-buck converter to implement the
RT9607/A. The converter operates for the input rang from
5V to 12V.
When layout the PC board, it should be very careful. The
power-circuit section is the most critical one. If not
configured properly, it will generate a large amount of EMI.
The junction of Q1, Q2, L2 and Q3, Q4, L4 should be very
close. The connection from Q1, and Q3 drain to positive
sides of C1, C2, C3, and C4; the connection from Q2, and
Q4 source to the negative sides of C1, C2, C3, and C4
should be as short as possible.
Next, the trace from Ugate1, Ugate2, Lgate1, and Lgate2
should also be short to decrease the noise of the driver
output signals. Phase1 and phase2 signals from the
junction of the power MOSFET, carrying the large gate
drive current pulses, should be as heavy as the gate drive
trace. The bypass capacitor C7 should be connected to
PGND directly. Furthermore, the bootstrap capacitors (C_b1,
C_b2) should always be placed as close to the pins of the
IC as possible.
Select the Bootstrap Capacitor
Figure 3. shows part of the bootstrap circuit of RT9607/A.
The V_CB (the voltage difference between BOOT1 and
PHASE1 on RT9607/A) 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
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DS9607/A-07 April 2011