FINAL PAPERShreesh Nanda

Peng XuAashiv Shah

ECE 445, SENIOR DESIGN PROJECT

FALL 2004

TA: Chad Carlson

Date:12/05/2004

Project No:13

Abstract

Various parts of India like Maharashtra do not have reliable power supply as a result of which people use pedal power to charge batteries and run other gadgets . Yet the main difficulty involved in this is that gadgets like LED lamps , radios and cassette players run off 6 volts and the batteries are 12 volts . Another difficulty faced is that the output voltage needs to be constant for an input voltage of 12 volts and should not vary. This idea was submitted by the Engineers Without Borders group and we had to come up with a plan to get a constant output of 3V/6V from an input of 12V . We decided to implement the circuit using a low-cost DC-DC converter which would have a feedback circuit that would change the output voltage automatically in response to any change in the input voltage which was what our design specifications required . To first test our circuit we built the open loop version of a DC-DC converter which took in an input of 12V and gave a constant output of 3V/6V . The open loop circuit consisted of 2 chips ---the TL494 which basically was a pulse-width modulation control circuit on a single chip and the IR2117 which acted like a voltage amplifier and amplified the output of the TL494 so that it could drive the MOSFET MTP50N06V . We had a potentiometer one end of which was connected to the feedback of the TL494 and the other end was connected to reference voltage(REF) of the TL494 . By varying the resistance of the potentiometer we obtained the correct output of 3V/6V for an input of 12V . The closed loop circuit was different in that we did not have to change the output manually in that we built a feedback circuit to automatically do it . We also decided to do a PCB implementation of our circuit once we had got the design down and got a working circuit on our breadboard circuit .

TABLE OF CONTENTS

1. INTRODUCTION......................................................................................................................1.1 Review and Update Material................................................................................................1.2 Specifications........................................................................................................................1.3 Subprojects...........................................................................................................................

2. DESIGN PROCEDURE.............................................................................................................2.1 Design Decisions..................................................................................................................

3. DESIGN DETAILS....................................................................................................................3.1 Components..........................................................................................................................3.2 Diagrams...............................................................................................................................

4. DESIGN VERIFICATION.........................................................................................................4.1 Testing..................................................................................................................................

5. COST..........................................................................................................................................5.1 Parts......................................................................................................................................5.2 Labor.....................................................................................................................................

6. CONCLUSIONS........................................................................................................................

INTRODUCTION

1.1 : Review and Update Material

In this project we seek to develop a low cost and efficient DC-DC converter to run various electronic gadgets (radio, cassette player, led lamps, charge 6 volt batteries) from a 12 volt lead acid battery. The converter will convert the 12 volt input to 3/6volt output . Some of the uses that this project may server are : It will run various electronic gadgets such as radio, cassette players and LED lamps It will help to use power more efficiently where meager power is available . It helps to replace the voltage generator in maintaining voltage across two terminals since it keeps the dissipation of power to a minimum . It helps to keep the voltage across two terminals constant , smoothening out the fluctuations in power . This concept is very important if someone is using a radio or cassette player and there are power fluctuations . Communities in third world countries without reliable power supply would greatly benefit from this .

We plan to implement our circuit using the closed loop feedback circuit as shown below in Fig1.

VCC12

IN 1+1

IN 1 -2

IN 2+16

IN 2 -15

C18

C211

CO MP3

DT C4

O C13

E19

E210

CT5

VRE F14 RT

6

T L494C

0.001uF

10K

IN2

HO7

V S6

CO M3

V B8

VCC1

U2

IR 2117

M TP 50N 06V

1K

1K

50 uH

1N5819LO AD

10 K

470 uF

100K 2.2u F

1N5819

3K

2K

12V

1 50 O hm

Fig 1: Closed loop feedback circuit

1.2: Specifications :

The specifications of our circuit are as follows : Input Voltage : 12 V Output Voltage : 3V/6V Output current : 0.5 A Efficiency : 80% or more

1.3: Subprojects :

Our project was divided into three parts :

a Open loop circuit : The open loop circuit consists of the TL494 , the output of which enters into the IR2117 . The TL494 is a pulse-width modulator circuit built into a chip . The output of this TL494 is then amplified by the IR2117 to drive the MTP50N06V MOSFET . The drain of this MOSFET is connected to power while a L-C circuit is connected to its source . The output is taken across a power resistor placed in parallel to the capacitance of this L-C circuit .

b Closed loop circuit : The closed loop circuit is similar to the open loop circuit and only differs in its feedback circuit . While the open loop circuit has a potentiometer , one end of which goes into the feedback of the TL494 chip and the other which is connected to the reference voltage of the TL494 , the closed loop circuit has a feedback which automatically changes the output in response to any change in the input . Thus , unlike the open loop circuit we do not need to manually control the output .

c PCB implementation of closed loop circuit : We plan to use Easytrax to layout our PCB and then submit it to the parts shop who would then implement our design on a Printed Circuit Board .

DESIGN PROCEDURE

2.1 : Design Decisions : We plan to implement our circuit using step down DC-DC buck converter . The block diagram of a step-down buck converter is given below in Fig2 :

Fig 2:Step – Down Buck Converter

This step down buck converter has two states : the on-state and the off-state . The on-state occurs when the diode is on and the off-state occurs when the diode is off . The block diagrams for the on and off states of the step-down buck converter are given below in Figs. 3 and 4 :

Fig3:On-State of a Buck Converter

During the ON stage, the diode is off and the transistor shows a small resistance of RDS. There is a voltage drop of VDS across the transistor and hence the input voltage to the inductor is VI – VDS. However there is another voltage drop of IL*RL before the inductor and the voltage that reaches the inductor is VI minus these losses. The voltage to the right side of the inductor is Vo. Hence the voltage across the inductor is VI minus all these values. VL is also given by the following equation:

..……….. <1>

The above equation shows that the inductor current increases due to the applied voltage.

The increase in inductor current is given by

…………….. <2>

Substituting the value of VL, we get

……………..<3>

Fig 4:The Off State of a Buck Converter

When the transistor is off, the diode is on and the current now goes through the diode. The voltage into the inductor becomes Vd – (IL * RL).

The voltage across the inductor is still Vo and the change in inductor current is given as follows:

……. <4>

In steady state, the change in current during the on and off stage should be the same. We use this to solve for Vo.

.<5>Using Ts = Ton + TOFF and defining the duty ratio D=TON/TS, we get

………<6>This is the equation which relates the input to the output.

DESIGN DETAILS

3.1 Components : The components that we use to build the step-down buck converter are :

1 TL 494 chip 2 IR 2117 chip3 MTP50N06V MOSFET4 1N5819 diode (2)5 Resistors : 10Kohm (2), 100Kohm(1), 1Kohm(2), 2Kohm(1), 3Kohm(1), 150ohm(1)

6> Capacitors : 0.001micro Farad(1), 2.2 micro Farad(1) , 1 micro Farad(2), 47 micro Farad(1) 7> Inductor : 50 micro Henry

3.2 Diagrams :

The diagram of the TL494 chip is given below in Fig5

Fig5: TL494

Vcc : Supply Voltage 1IN+,1IN- : input to the amplifier REF : Reference voltageC1 : Output Voltage of TL494CT : Timing Capacitor RT : Timing Resistor

The diagram of the IR2117 chip is given below in Fig6 .

Fig6:IR2117

We use the above two chips and wire them up to give us the open loop configuration as shown below in Fig7 .

VCC12

IN1+1

IN1-2

IN2+16

IN2-15

C18

C211

COMP3

DTC4

OC13

E19

E210

CT5

VREF14 RT

6

TL494C

0.001uF1

10K

IN2

HO7

VS6

COM3

VB8

VCC1

U4

IR2117

MTP50N06V

50 uH

1N5819LOAD

470 uF

1N5819

12V1

150 Ohm

pote

ntio

met

er

Fig7:Open Loop Circuit

Our closed loop feedback circuit is exactly similar to the open loop circuit except that we replace the potentiometer by a feedback circuit . The diagram of this circuit is given in Fig 1 .

DESIGN VERIFICATION

4.1 Testing

We read both the multimeter and the oscilloscope to test our ouput . For our circuit the multimeter reading showed that for an input of 12V, output is 3.1V (when the switch is turned to one side) and the output is 5.7V(when the switch is turned to other side) and the current drawn is 0.13A . The oscilloscope reading , on the other hand , gave us an output value of 3.2V and 6.23V and a current of 0.l3A .

Next , we measured the circuit efficiency for different values of output voltages and resistances for four different cases .

Case 1: Load resistance = 56 ohms Vout = 6V Vin = 12V Iin = 0.1A Pin = 1.2W Pout = Vout^2/R = 6^2/56 = 0.642W Efficiency = Pout/Pin *100 = 54%

Case 2: Load resistance = 40 ohms Vout = 3.1V Vin =12V Iin = 0.1A Pin=1.2W Pout = 3.1^2/40 = 0.171W Efficiency = 15%

Case 3 : Load resistance = 56 ohms Vout = 6.23V Vin = 12V Iin=0.139A Pin=1.66W Pout=6.23^2/56 = 0.97W Efficiency = 60%

Case 4: Load resistance = 40 ohms Vout=3.06V Vin = 12V Iin=0.106A Pin=1.2W Pout=3.06^2/40 = 0.23W Efficiency = 20%

A major problem in our circuit was noise.

For an output of 3.12V , the noise was found to have an amplitude of 510mV and frequency of 35Mhz .

Fig 8 : Circuit Noise for an output of 3.12V

For an output of 5.7V , the noise was found to have an amplitude of 531.3mV and frequency of 32Mhz as shown below in Fig. 9 .

Fig 9 : Circuit noise for an output of 5.7V

To reduce the noise in the circuit , we increased the load capacitance . For a load capacitance of 470 micro Farad, the ripple voltage decreased down to 319 mV while for load capacitances of 1000 micro Farad and 1500 micro Farads , the output ripple voltage was found to be 134mV and 98mV respectively . Thus , we managed to reduce our noise levels significantly as shown below in Figs. 10 and 11 .

Fig. 10: Circuit noise for an output of 5.7V and load capacitance of 47 microFarad

Fig 11: Circuit noise for an output of 5.7V and load capacitance of 1500micro Farad

COST

5.1 Parts

Part Part # Vendor Quanlity Cost each

Total cost

PWM Chip

TL494CN 1 $ 0.17 $ 0.17

High Side Gate Drive

IR2117 1 $ 1.04 $ 1.04

Diode 1N5820-T Parts Shop 2 $ 0.24 $ 0.48Transistor MTP50N06 Parts Shop 1 $ 0.98 $ 0.984 layers PCB Board 1 $ 30.00 $ 30.00Capacitor Parts Shop 5 $0.25 $1.25Inductor ECE Store 1 $ 1.65 $ 1.65Resistor Parts Shop 9 $1.23 $ 11.07Switch Parts Shop 2 $1.00 $2.00TOTAL COST OF CIRCUIT $48.64

5.2 LaborPeng Xu:$50/hr x 60 hrs x 2.5= $7,500.00Aashiv:$50/hr x 60 hrs x 2.5= $7,500.00Shreesh:$50/hr x 60 hrs x 2.5= $7,500.00

Total Labor:      $22,500.00 Total Parts:             $48.64

Grand Total = $22,548.64

6. Conclusion

Overall, the project was a success. The specifications required in the proposal were met. Our buck converter worked fine for different resistor values and with the increased capacitance, our voltage ripple was down to an acceptable level. The closed loop circuit accounts for any change in the input voltage and makes sure that the output voltage does not change drastically. There are some changes that can still be made to the circuit. Adding a current fuse of 0.5A would make sure that there would be no overload of current, which might damage the appliance used. Also a better feedback circuit would enable the circuit to withstand changes in input voltage better. Some of the components might also be changed to try and bring down the cost of the circuit further while maintaining an acceptable level of efficiency.

REFERENCES

1>P.T. Krein , Elements of Power Electronics , NewYork , Oxford University Press,1998, pg119-161.

2>http://focus.ti.com/docs/prod/folders/print/tl494.html

3>http://www.irf.com/product-info/datasheets/data/ir2117.pdf

4> http://www.ti.com