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V048X320Y009A Datasheet, PDF (12/18 Pages) Vicor Corporation – High Efficiency, Sine Amplitude Converter
Not Recommended for New Designs
V048x320y009A
This is similar in form to Eq. (3), where ROUT is used to
represent the characteristic impedance of the SAC™. However,
in this case a real R on the input side of the SAC is effectively
scaled by K2 with respect to the output.
Assuming that R = 1 Ω, the effective R as seen from the secondary
side is 0.98 mΩ, with K = 1/32 as shown in Figure 15.
A similar exercise should be performed with the additon of a
capacitor or shunt impedance at the input to the SAC. A
switch in series with VIN is added to the circuit. This is depicted
in Figure 16.
S
VVINin
+
–
C
SSAACC™
KK == 11//3322
VVOUoTut
Figure 16 — Sine Amplitude Converter™ with input capacitor
A change in VIN with the switch closed would result in a
change in capacitor current according to the following
equation:
IC(t)
=
C dVIN
dt
(7)
Assume that with the capacitor charged to VIN, the switch is
opened and the capacitor is discharged through the idealized
SAC. In this case,
IC= IOUT • K
(8)
Low impedance is a key requirement for powering a high-
current, low voltage load efficiently. A switching regulation
stage should have minimal impedance while simultaneously
providing appropriate filtering for any switched current. The
use of a SAC between the regulation stage and the point of
load provides a dual benefit of scaling down series impedance
leading back to the source and scaling up shunt capacitance or
energy storage as a function of its K factor squared. However,
the benefits are not useful if the series impedance of the SAC
is too high. The impedance of the SAC must be low, i.e. well
beyond the crossover frequency of the system.
A solution for keeping the impedance of the SAC low involves
switching at a high frequency. This enables small magnetic
components because magnetizing currents remain low. Small
magnetics mean small path lengths for turns. Use of low loss
core material at high frequencies also reduces core losses.
The two main terms of power loss in the VTM module are:
- No load power dissipation (PNL ): defined as the power
used to power up the module with an enabled powertrain
at no load.
- Resistive loss (ROUT): refers to the power loss across
the VTM modeled as pure resistive impedance.
PDISSIPATED = PNL + PROUT
(10)
Therefore,
POUT = PIN – PDISSIPATED = PIN – PNL – PROUT
(11)
The above relations can be combined to calculate the overall
module efficiency:
h
=
POUT
PIN
=
PIN – PNL – PROUT
PIN
(12)
Substituting Eq. (1) and (8) into Eq. (7) reveals:
IOUT
=
C
K2
•
dVOUT
dt
(9)
The equation in terms of the output has yielded a K2 scaling
factor for C, specified in the denominator of the equation.
A K factor less than unity, results in an effectively larger
capacitance on the output when expressed in terms of the
input. With a K=1/32 as shown in Figure 16,
C =1 µF would appear as C=1024 µF when viewed
from the output.
=
VIN • IIN – PNL – (IOUT)2 • ROUT
VIN • IIN
( ) =
1–
PNL + (IOUT)2 • ROUT
VIN • IIN
VTM® Current Multiplier
Page 12 of 18
Rev 1.0
01/2014
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