Tutorial 14 A - Power Supplies
This is the first of two tutorials. Press the NEXT button at the bottom of the page to go to the next part.
All electronic circuits need a low voltage power supply in order to work. Almost all electronic circuits use direct current. The simplest way to provide a direct current is a battery. In the majority of circuits, batteries consist of a number of 1.5 V cells in series. (Note the rather pedantic use of cell; I am a physics teacher.)
The output voltage from this battery is 1.5 V × 4 = 6.0 V.
The 1.5 V figure comes from the most widely available disposable cells. If we use rechargeable cells, the voltage is 1.2 V. So this arrangement will give us 4.8 V. The voltage difference between 1.2 V and 1.5 V might not be significant, but when it's multiplied up, we can see that the difference becomes quite a lot. An electronic circuit requiring 6.0 V to operate may well not operate at all satisfactorily at 4.8 V.
We can wire cells in parallel as well. This enables us to get a higher current. A minibus might have two 12 V batteries in parallel to provide the massive current (1000 A) taken by the starter motor for its diesel engine.
If each cell has a voltage of 1.5 V, we see that the output voltage of this combination is 3.0 V, but the battery pack can give twice the current.
Combinations of cells shown below are electrical curiosities beloved of A-level Physics teachers, but have no use whatever in electronics.
Cells are available in a variety of sizes, e.g. AAA; AA; A; C; D. Zinc-carbon cells are the cheapest, but don't last that long. Alkaline cells can last up to three times as long, but can be quite expensive. Button cells are widely available for low voltage and low current applications.
The advantage of a battery pack is that it gives out a perfectly smooth direct current, which can be important for the stability or correct functioning of an electronic circuit. However there are a number of disadvantages:
Disposable cells are expensive;
Disposable cells can only supply a limited amount of energy before they run out;
The current is limited;
Leakage can result if the flat batteries are not removed;
There are environmental hazards in disposing such cells in landfill (a good number of shops have receptacles where you can dispose of worn-out batteries).
Rechargeable cells can be used. The most common types are:
Nickel Cadmium (NiCd - pronounced "Ni-cad");
Nickel Metal Hydride;
Lithium ion (Lion);
Lead acid (used in cars).
Lead acid batteries can hold a very large amount of energy, but are heavy and can leak out sulphuric acid (which is not good for your clothes or complexion) if tipped over.
Lion batteries can catch fire if they are punctured.
All rechargeable cells can provide a large current as they have a very low internal resistance. However, when a large current is being drawn, they will have to be recharged after a short period of time. Also the number of charge and discharge cycles is limited as there are memory effects with NiCd and NiMH. A fault in an electronic circuit could also enable a large current to flow with the possibility of a fire.
Capacity of a Rechargeable Cell
On the battery above, you can see that it has a capacity of 600 mAh. (It is pronounced 600 milliamp-hours.) This means that it can supply a current of 600 mA (0.60 A) for 1 hour when fully charged.
Time (h) = capacity (Ah) ÷ current (A)
Remember that 1 A = 1000 mA. So a cell that has a capacity of 1200 mAh can provide a current of 1.2 A for 1 hour. When using the formula, make sure that the units are consistent.
An LED lamp is powered by a 600 mAh battery and takes a current of 90 mA. How long will it remain lit for?
Time = 600 mAh ÷ 90 mA
Time = 6.67 h ( = 6 hours and 40 minutes)
A car battery has a capacity of 32 Ah. The starter motor takes 200 A and the engine will not start.
How long will the battery last for?
An 18 V cordless hedge-trimmer is powered by a 2.0 Ah battery. The manufacturer claims that the hedge-trimmer can work for 40 minutes.
(a) Work out the current.
(b) Work out the power of the motor.
(c) Comment on the power of the machine
Some electronics circuits use solar cells (or photovoltaic cells) as a power supply. The voltage given out by a solar cell varies considerably with the light level. Therefore there is usually a back-up battery. The symbol of a solar cell is this:
And this is a picture of a solar cell:
Solar cells are not efficient. The earliest cells were as little as 1 % efficient, although research has improved this. For a p-n junction solar cell, the maximum efficiency is about 34 %, although values of 25 % are often quoted and considered to be good.
The circuit below shows a simple charging circuit for a solar cell charged decorative lamp for a garden.
In the daylight the photovoltaic cell charges the rechargeable battery. The absence of light makes the transistor turn on and the LEDs glow.
This image is copyright and has been used by permission.
The average intensity of sunshine in the UK is about 500 W m-2 (Intensity (W m-2 ) = Power (W) ÷ area (m2)
The area of a solar cell is 25 cm2.
(a) What is the absolute maximum power that the cell could provide if it were 100 % efficient?
(b) In reality the cell is 10 % efficient. What is the maximum power obtained.
(c) What is the charging current if the voltage is 1.2 V?
(d) How long does it take to charge up a 600 mAh Ni-MH battery?
In reality, it would probably take longer to charge the battery fully, as the sun could be shaded by cloud, or trees in the garden. Also the actual area of electrical power generation may well be much smaller than 25 cm2.
Mains Power Supplies
While electronic circuits need DC to work, energy can only be distributed from power stations using alternating current (AC). This is because the voltage has to be stepped up to very high values to reduce energy losses due to current heating effects. Mains electricity has a voltage of 230 V at a frequency of 50 Hz. We have to convert the voltage to a low value using a transformer and from AC to DC using a rectifier. Here is a typical transformer-based power supply (for the laptop on which these notes were written).
You can get mains power supplies that plug directly into a mains socket.
You will have come across laboratory power supplies in which the voltage can be varied. Two examples are shown in the picture below:
The transformer is a machine that is simplicity itself. It consists of:
A primary coil connected to the alternating power source. This provides the changing magnetic field.
A secondary coil connected to the load (in this case, the rectifier and control equipment in the locomotive).
A laminated soft iron core.
The two coils are electrically completely different circuits. Either of the coils can act as a primary.
The laminated core is made up of layers of soft iron separated by an insulating layer of varnish or glue. This reduces losses from eddy currents. Soft iron loses its magnetism immediately the current is turned off. Therefore the magnetic field can change forwards to backwards as the current changes.
The ratio of the input voltage to the output voltage is the same as the ratio of the number of turns on the primary to the number of turns on the secondary. The turns ratio of a transformer is given by this formula:
If the voltage is reduced or stepped down, the current is increased or stepped up. If the voltage is stepped up, the current is stepped down. The formula for the current change is:
A transformer is connected to a 230 V mains supply. The output voltage is 15 V. The output current is 1.5 A.
(a) Calculate the turns ratio.
(b) Calculate the current drawn from the mains.
The circuit diagram shows the simplest way of converting AC to DC. A diode is simply placed into the secondary output. It couldn't be simpler... But it's not very good. We have a half wave rectified output, as the diode allows the current to flow one way only. So the forward (or positive) part of the wave (the forward half-cycle) is allowed to flow. However the reverse half-cycle is not. The idea is shown on the CRO on the picture on the left below:
If the diode is turned round, the reverse half cycle flows as shown on the picture above to the right. The problem is that we are losing half the cycle. Only half the energy gets to the component.
A normal diode has a forward biased voltage drop of 0.7 V
In a half wave rectified AC, the input voltage is 15 V RMS. What is the output voltage?
We need a better solution, the full-wave rectifier bridge.
We can see the result on a stylised CRO screen. Of course, the CRO does not display the rectified AC as two colours!
The two pictures below show the action of a rectifier bridge.
In reality, we get a rectified AC. This is OK to run a motor or a light bulb. For an electronic circuit, you would get a "mains background". In an audio circuit this would be an intrusive mains hum.
In a rectifier bridge, the input voltage is measured at 15 V. The output voltage is measured at 13.6 V. Explain why this occurs.
To get a smoothed DC, we add a capacitor into the circuit, which is essential for any electronic circuit to work properly. Smoothing and voltage regulation are discussed in the second part of the tutorial.