BCM384x120y300Azz
Sine Amplitude Converter™ Point of Load Conversion
108nH
Rout
Iout
IOUT
6.0mΩ
ROUT
Lin = 5.7nH
Lout = 500pH
+
+
R
R
RC
cout
R
cin
480mΩ
OUT
570µΩ
CIN
V•I
K
9.2mΩ
1/32 • Iout
1/32 • Vin
+
–
C
CCOUT
+
–
CIINN
out
31µF
Vout
VOUT
V
IIQQ
V
in
0.0625µF
IN
18mA
–
–
Figure 17 — VI Chip® module AC model
The Sine Amplitude Converter (SAC™) uses a high frequency
ROUT represents the impedance of the SAC, and is a function of the
RDSON of the input and output MOSFETs and the winding resistance
of the power transformer. IQ represents the quiescent current of
the SAC control, gate drive circuitry, and core losses.
resonant tank to move energy from input to output. The resonant
LC tank, operated at high frequency, is amplitude modulated as
a function of input voltage and output current. A small amount
of capacitance embedded in the input and output stages of the
module is sufficient for full functionality and is key to achieving
power density.
The use of DC voltage transformation provides additional
interesting attributes. Assuming that ROUT = 0Ω and IQ = 0A, Eq. (3)
now becomes Eq. (1) and is essentially load independent, resistor R
is now placed in series with VIN.
The BCM384x120y300A00 SAC can be simplified into the
preceeding model.
At no load:
RR
SAC™
SAC
Vout
Vout
+
–
K = 1/32
Vin
Vin
K = 1/32
VOUT = VIN • K
(1)
(2)
K represents the “turns ratio” of the SAC.
Rearranging Eq (1):
Figure 18 — K = 1/32 Sine Amplitude Converter
VOUT
K =
with series input resistor
VIN
The relationship between VIN and VOUT becomes:
In the presence of load, VOUT is represented by:
VOUT = (VIN – IIN • R) • K
(5)
VOUT = VIN • K – IOUT • ROUT
(3)
(4)
Substituting the simplified version of Eq. (4)
(IQ is assumed = 0A) into Eq. (5) yields:
and IOUT is represented by:
VOUT = VIN • K – IOUT • R • K2
(6)
IIN – IQ
IOUT
=
K
BCM® Bus Converter
Page 14 of 21
Rev 1.5
08/2016
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