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BCM48BX120Y300A00 Datasheet, PDF (13/20 Pages) Vicor Corporation – Isolated Fixed Ratio DC-DC Converter
Sine Amplitude Converter™ Point of Load Conversion
Lin = 5.7nH
+
CCIINN
VVinIN
2µF
R0Rc.C5in7INmΩ
IIQQ
109mA
1/4 • Iout
–
IIOoUutT
973pH
V•I
++
3.13Ω
1/4 • Vin
––
K
BCM48Bx120y300A00
Rout
9R.0OmUTΩ
CCOoUuTt
47µF
Lout = 600pH
+
RRCcoOutUT
430µΩ
VVOoUuTt
–
Figure 16 — VI Chip® module AC model
The Sine Amplitude Converter (SAC™) uses a high frequency
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 BCM48Bx120y300A00 SAC can be simplified into the
preceeding model.
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.
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.
At no load:
VOUT = VIN • K
K represents the “turns ratio” of the SAC.
Rearranging Eq (1):
K = VOUT
VIN
In the presence of load, VOUT is represented by:
VOUT = VIN • K – IOUT • ROUT
and IOUT is represented by:
I OU T = IINK– IQ
R
(1)
VVinin
+
–
SSAACC™
KK==11/3/42
VVoouutt
Figure 17 — K = 1/4 Sine Amplitude Converter
(2)
with series input resistor
The relationship between VIN and VOUT becomes:
VOUT = (VIN – IIN • R) • K
(5)
(3)
Substituting the simplified version of Eq. (4)
(IQ is assumed = 0A) into Eq. (5) yields:
(4)
VOUT = VIN • K – IOUT • R • K2
(6)
BCM® Bus Converter
Page 13 of 20
Rev 1.8
08/2016
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