__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

_{v}

_{C1 /}

_{}/ dt =i

_{ca}+i

_{out}

C

_{1}=250uf

_{}

dv = 0.08% * 100v

__dv = 0.08 v__

We have dt = duty ratio/frequency

C

_{1*}d

_{v}

_{C1}* 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}= 0.04AFor 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__