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
• V_IN=12V 
•  I_IN=16A
•  V_OUT=200V ,
• I_OUT =1A
• Gain of converter = 4/1-D

DUTY CYCLE CALCULATION
Gain of converter = 4/1-D
D is duty ratio
V_OUT / V_IN =4/1-D
 1-D = 4/ (V_OUT / V_IN)
D = 1- 4/ (V_OUT / V_IN)
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
V_IN =  V_L  =12
FOR AN INDUCTOR VOLTAGE CURRENT BASIC RELATION IS
V_L = L * dI / dt        
THEN,  L = V_L  *  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
I_L1 =  I_IN /2
I_L1 =16A/2
I_L1 = 8A
Assume that inductor ripple current = 20% of inductor current
dI =20%  *  I_L
dI =20%  *  8A
dI =20%  *  8A
dI = 1.6A
then,
L =  V_IN  *  dt   / dI 
L = V_IN  *  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.
I_C1 = I_C2 = P_OUT / V_OUT
I_C1 = I_C2 = 200W / 200V
I_C1 = I_C2 = 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 i_ca
I_c1 = i_ca +i_out 
C_1* d_{vC1 /} / dt   =i_ca +i_out 
C₁=250uf
dv = 0.08% * 100v
dv =  0.08 v
We have dt = duty ratio/frequency
C_1* d_vC1 * F / D =   i_ca +i_out 
250uF_* 0.08V * 40000Hz/0.77 = i_ca +1A
I_ca =( 250uF_* 0.08V * 40000Hz/0.77)  – 1A
I_ca = -1A
I_ca = 0.04A
For a capacitor voltage current basic relation is
I_ca = c_a * dv_a / dt  
C_a = i_ca * dt / dv  
 dv_a  is output ripple voltage. Assume that output ripple voltage is about 0.08% of output voltage.
C_a is charged by inductor V_IN / 1-D
V_{a =} V_IN / 1-D
V_{a =} 12 / 1-0.77
V_{a =} 52
dv = 0.08% * 52v
dv = 0.0416 v
C = i_ca * dt / dv  
We have dt = duty ratio/frequency
C = i_ca * 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