Home > Tutorials > Designing Large 7-Segment Displays
Modified: 22:37, 22 October 2013

Seven segment displays have been around for ages but are a very effective and cheap way of displaying numbers and words to a limited extent. All letters except "k m v w x" can be formed in either lower or upper case. The main problem is the size - I have not found any larger than 100mm high.

An alternative is to wire some LED's in series to form each segment. This has the following advantages:

  • The segments can be almost any size - literally!
  • Any colour of LED's can be used giving a wide range of options - even multicolour.
  • Much more cost effective - 100mm displays cost around £6 each.
  • Wide range of brightness available with hyper-bright LEd's clearly visible in daylight.

Common type available and how its segments are labelled

The SpeedTrap project uses 4 LED's in series for each segment. The holes for the LED's were made with a laser cutter but you can drill them with a pillar drill. Use a template to keep all LED's in line. The digits look terrible if they are not straight and aligned properly.
A rear view of the display showing the series connections of the LED's using the component legs. Extra wiring was kept to a minimum.

This is a Go-Kart timer and has been running for about 10 years and is very bright. There is a sheet of red semi-transparent acrylic to help improve contrast.

A view of a giant scoreboard for a school gym using 8mm LED's, 6 LED's form each segment and they are arranged 2 x 3 in series.

LED's mounted directly to a PCB makes it easy to form the display accurately using PCB software - unfortunately photo-etch PCB is not cheap!

Careful consideration should be given when designing displays in order to not waste too much power.

A quick explanation:

LED's have a fixed operating voltage (forward voltage) which if exceeded will destroy the LED. You could use exactly the correct voltage to drive an LED - true this would work fine - BUT if the voltage rises for whatever reason, more than a few tenths of a volt, the LED will start to degrade, colour will alter and the LED will ultimately fail.

Details of some standard diffused 5mm LED types.
Fwd Voltage
Therefore generally it is wiser and safer to use a series resistor - the only problem is that power is wasted in the resistor.

For example suppose you are driving a single red LED from a 12 volt supply at 25mA. The calculation for the series resistor would be:

R = V/I = (12v - LED) / 0.02 = 10 / 0.025 = 400 ohms

No problem - but when you examine how much power is lost in the resistor compared to what the LED uses, it's not so good.

Power in LED:

P = IR = 0.025A x 2v = 50mW

Power in resistor:

P= IR = 0.025A x 10v = 250mW!!!


5 times what you actually need is lost. And if you connect 4 LED's in parallel its even worse. Normally if only one or two LED's are being used it's not a worry but if you are using tens or hundreds for display, then another method must be used.


The trick is this. Try to wire LED's in series until their total forward voltage comes close to your supply (within 2 to 3v). If you need more LED's in the segment - parallel up another series of LED's. It's best to have at least 2 volts across the resistor to give you control over the current more easily.

Let's assume you have segments of 8 green LED's running from a 12v supply. The forward voltage is 2.2 volts and a current of 30mA. There are 4 possible combinations:

8 LED's in series
All LED's in series would give a total voltage drop of 8 x 2.2v = 17.6v which is higher than the supply so this would not work at 12v.

8 LED's in parallel
Each LED will require 25mA so 8 x 25 = 200mA will flow through the resistor with 12 - 2.2 = 9.8 volts across it. The resistor required can be calculated:

R = V/I = 9.8v / 0.2A = 49ohms

Now calculate the power dissipated:

Power dissipated in LED's = P = IV = 0.2A x 2.2v = 0.44W
Power wasted in resistor = P = IV = 0.2A x 9.8v = 1.96W

A 2Watt resistor gets hot and clearly we are wasting a lot of power. This is a very inefficient option.

2 LED's in series x 4
This option has rows of 2 LED's in series giving 12 - 4.4 = 7.6 volts across the resistor. Note that in series 25mA will flow through both LED's meaning we need only 4 x 25mA = 100mA this time. Calculating the values gives:

R = V/I = 7.6v / 0.1A = 76ohms
Power dissipated in LED's = P = IV = 0.1A x 4.4v = 0.44W
Power wasted in resistor = P = IV = 0.1A x 7.6v = 0.76W

A better ratio but still wasting nearly twice what we are using for the LED's

4 LED's in series x 2
This option has rows of 4 LED's in series giving 12 - 8.8 = 3.2 volts across the resistor. Since that in series 25mA will flow through all 4 LED's, we need only 2 x 25mA = 50mA this time. Calculating the values gives:

R = V/I = 3.2v / 0.05A = 64ohms
Power dissipated in LED's = P = IV = 0.05A x 8.8v = 0.44W
Power wasted in resistor = P = IV = 0.05A x 3.2v = 0.16W

Now we are actually only wasting a third of the power used by the LED's so this is by far the best option to use. Not only has the current dropped 50mA, the voltage across the resistor has also dropped to only 3.2v.


Have the least possible voltage across the resistor and the most LED's in series.

Written by Phil Townshend 2008
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