IEEE Solid-State Circuits Magazine - Winter 2016 - 53
Precharge Phase
VOUT = 0 V
Discharge Phase
VOUT_PEAK = (N + 1) VIN
R
G
R
G
VOUT
R
R
G
R
R
G
R
R
R
G
R
R
G
R
G
+
-
RL
R
(a)
VOUT_PEAK ~ (N + 1) VIN
TR
TF
Time
R
G
+
-
VIN
RL
(c)
VIN
(b)
Figure 2: a marx generator [2].
terminal in phase 2. Thus, a maximum voltage gain of two was realized. This simple and effective way
to generate a higher voltage became
the basis of the following innovation.
Marx Generator (1923) Figure 2
The Marx generator is composed
of N capacitors (Cs) connected in
parallel to the power supply VIN
and ground via resistors (Rs) and in
series to each other via spark gaps
(Gs) as shown in Figure 2 [2]. Initially, as shown in Figure 2(a), all the
capacitors are precharged to VIN .
When the first spark gap gets short,
the second and the following spark
gaps get short one after another.
As a result, all the capacitors are
connected in series to the load, as
shown in Figure 2(b). Because R is
made much larger than a load R L,
the output voltage VOUT can reach
the maximum of (N + 1) VIN in a
short time of period. After that, the
state gradually returns to its original
in a long period due to large R , as
shown in Figure 2(c). This technique
was originally aimed at researching lightning, but it has been also
applied to generating pulsed X-ray
for photolithography or feed grain
preservation and medical care such
as eye surgery [10], [11].
Cockcroft-Walton (CW)
Ladder (1932) Figure 3
Cockcroft and Walton needed to
have a dc voltage gain greater than
two for a particle accelerator with an
output of up to 1 MV [3]. They used
ac power, N-capacitors, and (N + 1)
diodes. Figure 3(a) depicts a 2.5-MV
generator [12]. Later, the multiplier
had been also applied for industrial
applications such as X-ray medical
equipment and an electron gun for
a cathode-ray tube [13].
Figure 3(b) shows how the CW
multiplier works in steady state
using an example of five stages.
One arrow represents an amount
of charges Q , which is output per
cycle. In steady state, each diode
transfers one Q . In phases 1 and 2,
the input power needs to transfer
three Q s to the multiplier because
it sees three diodes. As a result, the
total injected charges per cycle are
six Q s. The factor of six is considered as the number of diodes or
the number of capacitors plus one.
Thus, in general, Q IN = (N + 1) Q OUT .
Figure 3(c) calculates the relationship between the output voltage and current by proceeding from
(1) through (11). For simplicity, the
threshold voltage of the diode is
assumed to be small enough in
comparison with the input voltage
amplitude. (1) The voltage across the
first capacitor C1 (i.e., V1) in phase 1
is equal to VIN . (2) The voltage
across C1 in phase 2 is reduced by
transferred charges, which is 3Q /C,
from the voltage across C1 in phase 1.
(3) The voltage across C2 (i.e., V2)
in phase 2 is equal to V1 because
V1 is given by VIN + (V IN - 3Q/C),
V1 = 2VIN - 3Q/C. (4) The voltage
across C2 in phase 1 is reduced
by transferred charges, which is
2Q /C, from the voltage across C2 in
phase 2. (5) The voltage across C3 in
phase 1 is equal to that across C2
because both terminals of C3 are
short to those of C2. (6)-(10) In this
way, one can calculate the voltage
across each capacitor. (11) The output voltage is a sum of the voltage
at the left terminal of C1 and the
voltages across C1, C3, and C5 in
phase 2. Thus, VOUT = 6VIN - 19Q /C.
The factor of VIN is six, which is
the same as the number of diodes,
as shown in Figure 3(b). The factor
of Q/C is 19. Now we know that
such a large number results from
the fact that the voltage drops in all
the previous capacitors are added.
As a result, the total voltage drop
increases more rapidly as the number of stages increases. As for a
IEEE SOLID-STATE CIRCUITS MAGAZINE
W I N T E R 2 0 16
53
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