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VTM 48 E x 040 y 050A00
This is similar in form to Eq. (3), where R OUT is used to
represent the characteristic impedance of the SAC TM . However,
in this case a real R on the input side of the SAC is effectively
scaled by K 2 with respect to the output.
Assuming that R = 1 Ω , the effective R as seen from the secondary
side is 0.98 m Ω , 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 V IN is added to the circuit. This is depicted
in Figure 16.
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
SAC
K = 1/32
IN
Vin
+
–
S
C
SAC TM
K = 1/32
OUT
V Vout
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 TM module are:
- No load power dissipation (P NL ): defined as the power
used to power up the module with an enabled powertrain
at no load.
Figure 16 – Sine Amplitude Converter TM with input capacitor
- Resistive loss (R OUT ): refers to the power loss across
the VTM modeled as pure resistive impedance.
A change in V IN with the switch closed would result in a
change in capacitor current according to the following
equation:
P DISSIPATED = P NL + P ROUT
Therefore,
(10)
I C (t) = C
dV IN
dt
(7)
P OUT = P IN – P DISSIPATED = P IN – P NL – P ROUT
(11)
Assume that with the capacitor charged to V IN , the switch is
opened and the capacitor is discharged through the idealized
SAC. In this case,
The above relations can be combined to calculate the overall
module efficiency:
I C = I OUT ? K
Substituting Eq. (1) and (8) into Eq. (7) reveals:
C ? dV OUT
I OUT =
K 2 dt
(8)
(9)
η =
=
P OUT = P IN – P NL – P ROUT
P IN P IN
V IN ? I IN – P NL – (I OUT ) 2 ? R OUT
V IN ? I IN
(12)
( )
The equation in terms of the output has yielded a K 2 scaling
factor for C, specified in the denominator of the equation.
= 1–
P NL + (I OUT ) 2 ? R OUT
V IN ? I IN
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.
Rev. 3.2
V?I CHIP CORP. (A VICOR COMPANY) 25 FRONTAGE RD. ANDOVER, MA 01810 800-735-6200
2/2011
Page 12 of 18
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