A Novel Transformer-less Adaptable Voltage Quadrupler DC Converter with Low Switch Voltage Stress

Abstract
In this paper, a novel transformer-less adjustable voltage quadrupler dc–dc converter with high-voltage transfer gain and reduced semiconductor voltage stress is proposed. The proposed topology utilizes input-parallel output-series configuration for providing a much higher voltage gain without adopting an extreme large duty cycle. The proposed converter cannot only achieve high step-up voltage gain with reduced component count but also reduce the voltage stress of both active switches and diodes. This will allow one to choose lower voltage rating MOSFETs and diodes to reduce both switching and conduction losses. In addition, due to the charge balance of the blocking capacitor, the converter features automatic uniform current sharing characteristic of the two interleaved phases for voltage boosting mode without adding extra circuitry or complex control methods. The operation principle and steady analysis as well as a comparison with other recent existing high step-up topologies are presented. Finally, some simulationand experimental results are also presented to demonstrate the effectiveness of the proposed converter.Index Terms—Automatic uniform current sharing, high step-up converter, low voltage stress, transformer-less, voltage quadrupler.

HARDWARE

SIMULATION
circuit
DESIGN
converter requirements
  • POWER =200W
  • VIN=12V 
  •  IIN=16A
  •  VOUT=200V ,
  • IOUT =1A
  • Gain of converter = 4/1-D
DUTY CYCLE CALCULATION
Gain of converter = 4/1-D
D is duty ratio
VOUT / VIN =4/1-D
 1-D = 4/ (VOUT / VIN)
D = 1- 4/ (VOUT / VIN)
D = 1- 4/ ( 200 / 12)
D = 0.77
                          
INDUCTOR VALUE CALCULATION


WHEN THE MOSFET IS ON , VOLTAGE ACROSS THE INDUCTOR WILL BE EQUAL THE VOLTAGE OF THE BATTERY
VIN =  V =12
FOR AN INDUCTOR VOLTAGE CURRENT BASIC RELATION IS
VL = L * dI / dt        
THEN,  L = VL  *  dt  / dI
HERE dt = DUTY CYCLE / FREQUENCY
Assume that operating frequency of the switch (mosfet here) =40 kHz
AND dI IS THE RIPPLE CURRENT OF INDUCTOR
Since 2 inductors of same value will take equal current from battery
IL1 =  IIN /2
IL1 =16A/2
IL1 = 8A
Assume that inductor ripple current = 20% of inductor current
dI =20%  *  IL
dI =20%  *  8A
dI =20%  *  8A
dI = 1.6A
then,
L =  VIN  *  dt   / dI 
L = VIN  *  D   / (F * di)
 L = 12V  * 0.77   / (40000 Hz * 1.6A)
L = 144uH


OUTPUT CAPACITOR C1 AND C2 VALUE

IN ABOVE MODE (MODE 1), LOAD CONSUMES POWER FROM C1 AND C2 WHICH IS IN SERIES.
IC1 = IC2 = POUT / VOUT
IC1 = IC2 = 200W / 200V
IC1 = IC2 = 1A
 FOR A CAPACITOR VOLTAGE CURRENT BASIC RELATION IS
I = C * dV / dt  
C = I * dt / dV  
 dV  is output ripple voltage. Assume that output ripple voltage is about 0.08% of output voltage
dV = 0.08% * 100V
dV =  0.08 V
C = I * dt / dV  
We have dt = duty ratio/frequency
C = I * D /  (F * dV )
C = 1A* 0.77/ (40000 Hz* 0.08 V)

C = 240 uF = 250uF (Standard value)


OUTPUT CAPACITOR C1 AND C2 VALUE
In above mode (mode2), at load side current through  c1 capacitor will be sum of load current and ica
Ic1 = ica +iout 
C1* dvC1 / / dt   =ica +iout 
C1=250uf
dv = 0.08% * 100v
dv =  0.08 v
We have dt = duty ratio/frequency
C1* dvC1 * F / D =   ica +iout 
250uF* 0.08V * 40000Hz/0.77 = ica +1A
Ica =( 250uF* 0.08V * 40000Hz/0.77)  - 1A
Ica = -1A
Ica = 0.04A
For a capacitor voltage current basic relation is
Ica = ca * dva / dt  
Ca = ica * dt / dv  
 dva  is output ripple voltage. Assume that output ripple voltage is about 0.08% of output voltage.
Ca is charged by inductor VIN / 1-D
Va = VIN / 1-D
Va = 12 / 1-0.77
Va = 52
dv = 0.08% * 52v
dv = 0.0416 v
C = ica * dt / dv  
We have dt = duty ratio/frequency
C = ica * d /  (f * dv )
C = 0.04 a* 0.77/ (40000 hz* 0.0416 v)
C = 18.50 uf = 22uf (standard value)

Hardware 
MOSFET GATE SIGNALS(S1 AND S2)

MOSFET GATE SIGNALS(S1 AND S2)
HARWARE

HARWARE

RESULT
OUTPUT GOT 163V AT 71% DUTY CYCLE FOR 1K LOAD
under work, will publish soon