BCM352x110y300B00
9.0 SINE AMPLITUDE CONVERTER™ POINT OF LOAD CONVERSION
108 nH
LOUT = 500 pH
R
II
ROOUUTT
LIN = 5.7 nH
OOUUT T
+
+
RRCCOOUUTT
R
R
C
CIN
480 mΩ
7.7 mΩ
9.2ImN Ω
570 µΩ
V•I
K
1/32 • IOUT
1/32 • VIN
+
COCUT
31.0 µF
C
CININ
+
0.0625 µF
OUT
VOUT
VIN
VIN
VOUT
IQ
IQ
20.0 mA
–
–
–
–
Figure 14 — BCM module AC model
The Sine Amplitude Converter (SAC™) uses a high frequency
resonant tank to move energy from input to output. (The
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 is formed by Cr and leakage inductance Lr in the
power transformer windings as shown in the BCM module
Block Diagram. See Section 8). 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 = 0 Ω and IQ = 0 A,
Eq. (3) now becomes Eq. (1) and is essentially load
independent, resistor R is now placed in series with VIN.
The BCM352F110T300B00 SAC can be simplified into the
preceeding model.
At no load:
R
SAC™
SAC
V
VoOUuTt
+
–
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 15 — K = 1/32 Sine Amplitude Converter™
VOUT
K =
with series input resistor
VIN
The relationship between VIN and VOUT becomes:
•
•
VOUT = (VIN – IIN R) K
(5)
In the presence of load, VOUT is represented by:
Substituting the simplified version of Eq. (4)
(IQ is assumed = 0 A) into Eq. (5) yields:
•
•
VOUT = VIN K – IOUT ROUT
(3)
(4)
and IOUT is represented by:
IIN – IQ
2
•
•
•
V
OUT = VIN K – IOUT R K
(6)
IOUT
=
K
BCM® Bus Converter
Page 12 of 19
Rev 1.1
07/2015
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