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BCD384F120T300A00 Datasheet, PDF (14/21 Pages) Vicor Corporation – BCM® Bus Converter
Sine Amplitude Converter™ Point of Load Conversion
Lin = 5.7nH
+
R9Rc.C2inmINΩ
CCIINN
VVinIN
0.0625µF
IIQQ
18mA
1/32 • Iout
–
IIOoUutT
108nH
V•I
++
480mΩ
1/32 • Vin
––
K
BCM384x120y300Azz
Rout
6R.0OmUTΩ
CCOoUuTt
31µF
Lout = 500pH
+
RRCcoOutUT
570µΩ
VVOoUuTt
–
Figure 17 — 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 BCM384x120y300A00 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 = IIN
–
K
IQ
R
(1)
VVinin
+
–
SSAACC™
KK == 11//3322
VVoouutt
Figure 18 — K = 1/32 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 14 of 21
Rev 1.5
08/2016
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