Final Year ProjectFinal Presentation

Title: Energy Conversion for low

voltage sources.Supervisor: Dr.Maeve DuffySupervisor: Dr.Maeve Duffy

Aim of Project

The aim of this project was to develop circuits

to demonstrate the performance of bio fuel

cells which are being developed by the Energy cells which are being developed by the Energy

research centre in NUI Galway.

The ideal end goal would have been where a

Microbial Fuel Cell arrangement has the ability

to charge a mobile phone battery.

Outline of Presentation

This presentation will deal with the following topics:

1. Overview of Project

2. Work Completed2. Work Completed

3. Outlook for future development

4. Conclusions

5. Questions

1) Overview of project:

Demonstration Circuit:

2) Work Completed:

• Thévenin Equivalent circuit:

• LED Demonstration

• Demonstration of fuel cell powering low

power devicespower devices

• Storage Capacitor

• Knowledge of charging algorithms

• Demonstration of fuel cell powering a DC Fan

Thévenin Equivalent circuit:Power Density curve:

0.4

0.5

0.6

Power density (mW/m2)

800

1000

1200

Voltage (V)

0

0.1

0.2

0.3

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

Current density (mA/cm2)

Power density (mW/m

0

200

400

600

Voltage (V)

Blue line represents the power density Vs current density.

White line represents Voltage Vs current density .

Area across which power density is measured is 5.4cm^2.

1cm^2 = 0.0001m^2

The point at which we have maximum power output is the

second from right so we take this point.

When worked out the following outputs result:

Power ~ 0.486 milli-WattsPower ~ 0.486 milli-Watts

Voltage ~ 0.42 volts

Current ~ 1.215 milli-Amps

Internal Resistance of Fuel Cell ~ 345 ohms

Thévenin Equivalent circuit:

LED Demonstration:

On testing the LED’s found in the electronics labs it was found

that the lowest power LED needed a minimum of 3.8 milli-

Amps and a minimum of 1.83 volts to light.

This meant the voltage & current output from the fuel cell

needed to be stepped up.

There is three solutions to this problem:There is three solutions to this problem:

1) Cascade a number of fuel cells in parallel, this way increasing

the current output and then use a DC-DC boost converter to

step up the voltage.

2) Use an RC circuit to boost the current using a mosfet for

switching and then use a DC-DC boost converter to step the

voltage up.

3) Use a low power LED (1 milli-Amp LED can be obtained)

Low power devices identified:

Voltage needed:

1.5 Volts DC

Power needed:

0.0001 Watts

Current needed:

66.66 micro-

Amps

Voltage needed:

2.7 volts DC

Power needed:

1.4 Watts

Current needed:

0.42 Amps

Voltage needed:

~3.3 volts DC

Power needed:

unknown

Current needed:

unknown

Demonstration of fuel cell powering low power

devices:

To demonstrate these devices a DC-DC boost converter needed

to be designed.

This caused problems as most common DC-DC boost converters

use either diodes or BJT’s which have a diode between the base

and emitter. The BJT is used due to its fast switching speeds. Theand emitter. The BJT is used due to its fast switching speeds. The

diodes cause a minimum of 0.3 voltage drop. As the output

voltage from the fuel cell is so low already we can not afford to

use BJT’s.

Demonstration of fuel cell powering low power

devices:

Using a boost converter obtained from Texas instruments called

the TPS61200 the output voltage could be boosted.

This converter gets around the problem of using BJT’s by using

MOSFET’s instead.

The TPS61200 can needs 0.8 volts to startup, after which it can The TPS61200 can needs 0.8 volts to startup, after which it can

operate at a voltage as low as 0.3 volts.

As the TPS61200 was to small to fit on a board I needed to order

the evaluation module.

Demonstration of fuel cell powering low power

devices:

Demonstration of fuel cell powering low power

devices:

Demonstration of fuel cell powering low power

devices:

Demonstration of fuel cell powering low power

devices:

From using the formula to work out the minimum inductance

needed (Vin = L * DI/DT) ,I found that the minimum

inductance required was 2.1333 micro-Henry’s.

So the 2.2 micro-Henry should be satisfactory to induct the

input current from the fuel cell.input current from the fuel cell.

Storage capacitor:

• As the DC-DC boost converter needed more power at start up

than the MFC could provide a Capacitor needed to integrated

into the system to output enough power.

• It was found through using the equation:

that the Energy which could be obtained from a 0.1 Farad

capacitor would be enough to get over the start up power

requirements.

• 3.3 Farad and 10 Farad Capacitors were also obtained to

enable the powering of high load devices for longer and

devices which require more Energy than the 0.1 Farad

capacitor can store.

2

21 CVE =

Storage capacitor:

Research of battery chemistries, charging

algorithms:

Example of type of voltage and current used to charge a

phone:

My phone (Sony Ericsson) is a lithium-polymer battery

which supplies 3.6 volts to the phone. And has 780 milli-

Amp hours.Amp hours.

The charger for the phone supplies 5 volts and a current

of 1Amp. This is probably implementing a charging

algorithm known as constant charge where a constant

charge is applied to the battery.

The type of charging algorithm that could be implemented

for this project would be either pulse charging or trickle

charging.

Demonstration of fuel cell powering a DC Fan:

Demonstration of fuel cell powering a DC Fan:

3) Outlook for future

development• Continuous powering of low power device

– How many capacitors are needed.

– Possible switching control devices.

– How many Microbial Fuel Cells are needed

• Research on more efficient DC-DC boost converters.

• Further Research on battery charging profiles.

• There are various ways in which a continuous powering of

devices using this circuit can be implemented.

• Obviously as the load attached to the DC-DC boost converter

changes so too does the rate of current discharge from the

capacitors. For this reason a system has to be devised for each

specific device.

Continuous powering of a low power device

specific device.

• An example chosen for this presentation is the continuous

powering of a 1.5 volt DC calculator

• Through testing it has been found that the calculator draws a

constant current of 9 μAmps from output of the Boost

converter and a constant voltage of 3.3 volts is applied across

it.

Continuous powering of a low power device

• A 0.1 Farad capacitor took approximately 194 seconds to

charge fully.

• When tested a fully charged capacitor could power the

calculator for 100 seconds.

• This meant that if the system was going to be implemented by

allowing the capacitors to fully charge then three 0.1 Farad allowing the capacitors to fully charge then three 0.1 Farad

Capacitors would have to be used as the charge rate does not

equal the discharge rate.

• This also meant using two extra Microbial Fuel Cells as each

capacitor would need to be charged separately .

Continuous powering of a low power device

• The alternative to this is to charge the capacitor for about 40

seconds. If you do this the calculator can be powered for

nearly 40 seconds meaning you will only need two capacitors

to continuously power the calculator. This in turn means

there is less MFC’s needed to charge the capacitors. there is less MFC’s needed to charge the capacitors.

Continuous powering of a low power device

555 TimerMSP430C1101

Possible switching devices:

Voltage Comparator

Continuous powering of a low power device

555 Timer:

Advantages:

– Inexpensive (47 cent)

– Easy to implement

Disadvantages:Disadvantages:

– Sync issues may arise

– Inflexible

– Higher operational voltage & input current (More MFC’s used to

power it)

• Minimum of 3 milli-Amps & 4.5 volts

Continuous powering of a low power device

Voltage Comparator:

Advantages:

– Inexpensive (€1.65)

– Easy to implement– Easy to implement

– Low voltage and current input (Less MFC’s used to power it)

• 1.8 volts & 15 μAmps

Disadvantages:

– Value of voltage across capacitor has to be very precise

Continuous powering of a low power device

MSP430C1101:

Advantages:

– More Flexible

– MP could also be used if implementing a smart battery charger– MP could also be used if implementing a smart battery charger

– Low voltage and current input (Less MFC’s used to power it)

• 2.2 volts and 150 μAmps

Disadvantages:

– Expensive($49.49 – Evaluation module & chip)

– More complex to implement

Research on more efficient DC-DC boost

converters.

The following graph shows the efficiency of the DC-DC boost

converter at an input voltage of 0.8 volts:

Research on more efficient DC-DC boost

converters.

An example of this lack of efficiency was observed when

powering the calculator. At the start there was from 0.9milli-

Amps being drawn from the capacitor into the Boost

converter which had 0.8 volts applied across the input.

At the end there was 0.3 volts applied across the DC-DCAt the end there was 0.3 volts applied across the DC-DC

boost converter and 2.2 milli-Amps being drawn from it.

This means the input power was between 0.72 milli-Watts

down to 0.66 milli-Watts yet the output power was only

0.0297 milli-Watts. This is equal to 4.5 % efficiency.

Further Research on battery charging profiles.

• As the 0.1 Farad capacitor takes so long to charge and outputs

relatively so little power it is hard to know if a system like the

system proposed for the calculator can be altered to trickle

charge a even a 1 milli-Amps hour battery without drastically

increasing the number of Microbial Fuel Cells on available.increasing the number of Microbial Fuel Cells on available.

• For a trickle charge algorithm usually the rate at which the

battery is Charged is 15 % of the rate at which constant charging

Algorithms are implemented.

• If implemented it is hard to know whether the current being

drawn into the battery would be enough to compensate for

the idle discharge of the battery through air.

Further Research on battery charging profiles.

• Through testing of the discharge rate of a 10 Farad capacitor

a possible way to implement a constant voltage\current

charging algorithm was identified.

• The 10 Farad capacitor powered an Led in series with a 1k

resistor for 5 minutes supplying the Led with a constant resistor for 5 minutes supplying the Led with a constant

current of 1.6 milli-Amps .

• This means that if twelve 10 Farad capacitors where charged a

1.5 milli-Amp Hour battery could be charged.

4) Conclusion:

• Better understanding of Electronic circuit design & MFC’s

• LED Demonstration

• Demonstration of fuel cell powering low power devices• Demonstration of fuel cell powering low power devices

• Knowledge of charging algorithms

5) Questions!!