What it shows
Phone chargers, LED drivers and solar chargers use switching converters because linear regulators waste power as heat. The project tests a ready-made LM2596 buck module and compares its efficiency with an LM7805 linear regulator at the same load.
The science
A buck converter switches the input on and off quickly through an inductor and diode, and the average voltage at the output equals the duty cycle times the input. The switch is either fully on or fully off, so little power is lost, and efficiency often exceeds 80 percent. A linear regulator drops the excess voltage as heat, so for 12 V to 5 V its best efficiency is only 5 divided by 12, which is about 42 percent.
Materials
Cost: under Rs 500. Many items come from home or from waste; spare parts such as LEDs, motors and wire are in the store.
- An LM2596 buck converter module with a display or preset
- An LM7805 regulator with a heat sink and two 100 nF capacitors
- A 12 V battery or 12 V, 2 A sealed adapter
- Power resistors or a rheostat as loads
- Two multimeters (current and voltage)
- A thermometer or IR thermometer
Safety
For this project: Use a sealed ready-made 12 V adapter or a battery. Resistors and regulators get very hot: an adult must supervise, and nothing flammable may be near the table.
Never use mains electricity: power models with cells, batteries or a low-voltage adapter. Sharp and hot tools, flames and lab chemicals need an adult or your teacher beside you. Keep magnets, button cells and small parts away from young children, and wash your hands after making.
All the safety rulesStep by step
10 steps. Do them in order, and test each part before you go on to the next. Write what you notice in a notebook as you go.
Set the LM2596 output to 5.0 V using the preset and a multimeter, with no load.
Connect the input through the ammeter to the 12 V source, and the output to the load through a second ammeter, with a voltmeter on each side.
Switch the load to draw 100 mA and record Vin, Iin, Vout and Iout.
Repeat for 200, 400, 600, 800 and 1000 mA by changing the load resistors.
Calculate input power, output power and efficiency for each reading.
Replace the converter with the LM7805 on its heat sink and repeat the same loads.
Measure the temperature of each regulator after 5 minutes at 500 mA.
Plot both efficiency curves on one graph against load current.
Calculate the heat lost by each regulator at 500 mA and discuss the effect on a battery runtime.
Write the report with a table, graph and recommendation for which to use in a solar charger.
What you should see
The buck module shows efficiency of about 75 to 90 percent over most loads, the 7805 about 40 percent from 12 V, and its heat sink is clearly hotter. The graph makes the difference obvious.
Viva questions
A teacher or a judge will ask how it works. Try to answer each question before you open it, then compare.
1Why is a buck converter more efficient?
Its switch is either fully on or fully off, so almost no power is lost across it.
2What does the inductor do?
It stores energy while the switch is on and releases it while the switch is off, smoothing the current.
3What is duty cycle?
The fraction of each switching period during which the switch is on.
4What is the drawback of switching regulators?
They make electrical noise that can affect sensitive circuits, so filtering may be needed.
Take it further
Measure the output ripple with an oscilloscope or sound-card scope, and add an LC filter to reduce it.
Change one thing, measure what happens, and add a table or chart of your readings: that is what turns a model into your own investigation.
Free ebooks for this project
Hardware & repair
Using a Digital Multimeter
Measure voltage, current, resistance and continuity safely
Robotics
Electronic Components Handbook
Resistors, capacitors, LEDs, diodes, transistors and more, with how to use each