The post details the construction of a 200 to 600 LED project, using series parallel LEDs for creating an alphabetical display sign board. The idea was requested by Mr. Mubarak Idris.
Let's Make a Blinking 200 String LED Circuit
I need a blinking LED light that show"WELCOME TO" blink and then "COLLEGE OF ENGINEERING" base on my rough estimation I'm going to use about 696 LEDs e.g for"WELCOME TO" = 216 LEDS"COLLEGE OF ENGINEERING" 480 LEDS the name welcome and college of engineering is going to flip flop and I'm thinking of connecting them to AC and only use relay to toggle the
"welcome to" and "college of engineering" alternately. hope to hear from you sir very soon and thanks in advance.
I have already discussed one related article where we learned how to calculate connect LEDs in series and parallel, in this post we are going to incorporate the same concept and formulas for estimating the connection details of the proposed 200 to 600 LED project for making the specified display sign board.
Since the LEDs are supposed to be operated from 220V mains, after rectification and filtration this would end up being at a 310V DC level.
Therefore we'll have to configure the LED groups as per the above mentioned DC level.To do this we'll first have to evaluate the total forward drop of the LED series that would fit comfortably within the 310 V limit.
Let's assume the LEDs are rated at 20mA / 3.3V, if we divide the 3.3v value with 310V, we get:
310/3.3 = 93nos.
It implies that 93 LEDs can be connected in series with the 310 input comfortably for getting an optimal illumination, however considering a possible low voltage situation and to ensure that the LEDs continue to glow even at low voltages we can go for 50% less LEDs in series, that is may be around 46 LEDs.
As per the request the welcome sign needs to have 216 LEDs, dividing this 216 with 46 gives us approximately 5 strings, in which 4 strings having around 46 LEDs in series, while the 5th could have 32 LEDs.
Therefore now we have 4 strings of 46 series LEDs and 1 string having 32 LEDs, all these strings now needs to be connected in parallel.
But as we know, in order to allow proper current distribution across the strings and allow uniform illumination, these LED strings need to have calculated resistors in series with them.
Calculating LED Current Limiter Resistor
This can be calculated with the help of the following formula:
R = Supply - Total LED FWD voltage / LED Current
= 310 - (46 x 3.3) / 0.02
here 310 is the DC supply voltage after rectification of the 220V AC supply, 46 is the total number of LEDs, 3.3 is the forward operating voltage of each LED, 0.02 is the current in amps for each LED (20mA), and 4 is number of strings.
Solving the above gives us: 7910 ohms or 7.9K, or simply a standard will 8k2 resistor will do.
wattage will be = 310 - (46 x 3.3) x 0.02 = 3.164 watts or simply a standard 5 watts resistor will do the job
the above 8k2 5 watt resistor will need to be connected with each of the strings having 46 LEDs
Now for the single 32 LEDs, we may have to follow the above procedures separately, as shown below:
R = 310 - (32 x 3.3) / 0.02 = 10220 ohms or 10.2 k or simply a standard 10K will do the job
wattage will be 310 - (32 x 3.3) x 0.02 = 4.088 or again a 5 watts will do.
Through the above formulas we calculated the series parallel connections with resistor for configuring a 216 LED display, however, the above strings will now need to be arranged appropriately in the shape of the alphabets, corresponding to the word "WELCOME". This might require some effort and could be a little time consuming, and might require some patience and skill.
For the second group of LEDs consisting of 696 LEDs, the process will be quite similar. We first divide the 696 with 46 which gives us around 15.13, meaning 14 strings can be configured with a series of 46 LeDs and one string having 52 LEDs...all these strings will likewise need to be connected in parallel and physically arranged to represent the phrase " COLLEGE OF ENGINEERING".
The resistor values for the 46 LED strings can be as calculated in the above sections, while for the 52 LED, it may done as given below:
R = 310 - (52 x 3.3) / 0.02 = 6920 ohms or simply a 6k9 standard resistor may be used.
wattage will be = R = 310 - (52 x 3.3) x 0.02 = 2.76 watts or 3 watts
The above explanation provides us the information regarding how to build any 200 to 400 LED based project for boards or display sign boards using mains voltage without the need of a transformer.
Now, to enable the two sets of LED groups flash alternately using a relay, the following simple IC 555 flasher could be used:
The LED Flasher Circuit
R1, R2, and C can be suitably adjusted for getting the desired blinking rate over the connected 200 to 400 LED strings. The relay does not need to be a 15amp as indicated in the diagram it could be any ordinary 12v 400 ohm 5 amp type of relay