bcm48bf120t300a00 Vicor Corporation, bcm48bf120t300a00 Datasheet - Page 11

• Manufacturer Part Number: bcm48bf120t300a00
• Description: Bcm™ Bus Converter
• Manufacturer: Vicor Corporation

PRELIMINARY DATASHEET
9.0 SINE AMPLITUDE CONVERTER™ POINT OF LOAD CONVERSION
Figure 13 — V
The Sine Amplitude Converter (SAC™) uses a high frequency
resonant tank to move energy from input to output. (The
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 BCM48BF120T300A00 SAC can be simplified into the
preceeding model.
At no load:
K represents the “turns ratio” of the SAC.
Rearranging Eq (1):
In the presence of load, V
and I
v i c o r p o w e r. c o m
V
K =
V
I
OUT
OUT
OUT
OUT
=
V
V
= V
= V
OUT
I
IN
is represented by:
IN
IN
IN
K
– I
V
V
+
–
•
•
IN
Q
IN
•
K
K – I
I Chip
L
L
IN
OUT
IN
= 5.7 nH
TM
= 5 nH
module AC model
•
R
V•I CHIP CORP. (A VICOR COMPANY) 25 FRONTAGE RD. ANDOVER, MA 01810 800-735-6200
OUT
OUT
C
C
IN
IN
is represented by:
R
0.57 mΩ
R C
2 µF
CIN
109 mA
IN
I
I
Q
Q
1/4 • I
OUT
(1)
(2)
(3)
(4)
+
–
V•I
K
+
–
I
OUT
I
R
the R
resistance of the power transformer. I
quiescent current of the SAC control, gate drive circuitry, and
core losses.
The use of DC voltage transformation provides additional
interesting attributes. Assuming that R
Eq. (3) now becomes Eq. (1) and is essentially load
independent, resistor R is now placed in series with V
The relationship between V
V
Substituting the simplified version of Eq. (4)
(I
V
Figure 14 — K = 1/4 Sine Amplitude Converter™
OUT
Q
OUT
OUT
OUT
V
1/4 • V
Vin
is assumed = 0 A) into Eq. (5) yields:
IN
DSON
represents the impedance of the SAC, and is a function of
= (V
= V
973 pH
IN
+
3.13 Ω
–
IN
IN
of the input and output MOSFETs and the winding
•
– I
K – I
with series input resistor
IN
R
R
•
OUT
R)
9.0 mΩ
BCM
R
•
R
•
OUT
OUT
C
K
R
OUT
C
•
OUT
K
IN
2
K = 1/32
48
K = 1/4
SAC
and V
SAC™
R C
R
430 µΩ
47 µF
COUT
B
L
OUT
OUT
OUT
x 120
= 600 pH
Q
OUT
represents the
becomes:
= 0 Ω and I
V
OUT
+
–
y
V
OUT
300A00
IN
Vout
V
Page 11 of 18
Q
OUT
.
= 0 A,
Rev. 1.2
(5)
(6)
7/2011