Hey guys this is my first post, I'm from South Africa. Basically just started with with electronics and the arduino. For my first project i want to build a lap timer for our motocross track. The input signal will come from a laser diode shining into a photodiode which would be connected to an analog pin which will act like a laser switch for when the bikes ride past. i only plan to time one bike at a time.
My trouble is with the display. Our track is about 1 minute and 20 seconds or that's at least the time we get. So i am busy building 3 seven segment displays. 1 segment will contain 12 LED's and the point will use 4. So in total ill be using 264 5mm green LED's. ill be using multiplexing because the arduino uno r3 does'nt have enough pins.
Ok so im new to all this so if i have done something stupid understand why. What i would think i must do is have the arduino switching transistors which then switch relays which then power the display. So i will have 3 power circuits? or can i do away with the relays and just use transistors? or am i missing something completely?
The power i will be using to power the display will be a 12v 2amp battery, its like a small sealed motorbike battery. im using this cause i have a few of them and i have a charger for them so its the most convenient thing for me to use. The way i understand it is that i will need to divide the voltage to the LED's, can i do this by putting 4 LED's in series and multiple parallel lines of series lines. the series lines will divide the voltage, so if i use 4 LED's in a line it will be 12/4 = 3v.
Now for the current, i would think it will be divided by how many parallel lines i have? for example 3 lines of (5 LED's in series) will divide the current in 3 equal portions so 2A/3 = 666mA. ok so this is clearly tooooooo much current for the led's so do i need to add resisters in each parallel line or do i add a single resister before in series before the wires split out to parallel?
My other question is this. when you display a "1" on the display, you are only lighting up 2 segments. Now when you display an 8 you lighting up all the segments. So will the current drawn for the "1" not be different for the current drawn for the "8"? This might be a dumb question but if my power supply is a 2A supply will the 2A not be too much when the "1" is displayed? compared to when the "8" is displayed? Or will the current that is drawn dependent on how many segments are lit up? like the power supply only delivers what the circuit asks for?
You're reasoning on the current is incorrect. The current is determined by the voltage and the resistance. Your 2A rating on the battery is the MAXIMUM current it will supply. But, with 12V and say, a 1K resistor, it would actually be only supplying 12mA. Think of it the 2A rating as max allowable current, not actual current. V=IR is your friend.
See (1) above. Yes, there will be less current drawn when only displaying 2 instead of 7 segments. But, each segment should have a current limiting resistor, normally limiting to around 15mA. In your case, worst case is with all 264 LEDS on, which gives max power per led of 7.5mA with your 2A supply. You won't have them all on at the same time, and certainly not for very long, so you could probably go with about 10mA per LED. Therefore, for your 12V/2A supply I would suggest a 1.2KOhm resistor in series with each of the LEDS. That's the basic theory, but it also depends on the driver chip you end up using and the way you connect up the LEDS.
You need to read up a bit.
This is basic electronics so you need to get this right.
Sure, lighting up 2 segments will draw less current than lighting up 7 segments.
You do not need to use relays, just use transistors that can handle voltage and current you need.
If you put 4 LEDs in series, then you will have to add their forward voltages.
This will result in something like this: (1.6 * 4 ==) 6.4 volt.
You have 12 volts (or so, check it first, depends on battery type) and your LEDs need this 6.4 volts.
So you've got 5.6 volts to "burn".
Use that in your calculations for the resistor you'll need to limit the current to the value your LEDs can handle.
In other words: find the datasheets of all components you're going to use.
They contain information you need.
Our track is about 1 minute and 20 seconds or that's at least the time we get. So i am busy building 3 seven segment displays.
Aren't you interested in tenths and hundredths?
Yes i am actually interested in the split seconds but for now i'm just building a basic one to learn. Later on i will upgrade it to where it displays split seconds and learns times so more than one guy can ride at a time. but for now simple is king
You're reasoning on the current is incorrect. The current is determined by the voltage and the resistance. Your 2A rating on the battery is the MAXIMUM current it will supply. But, with 12V and say, a 1K resistor, it would actually be only supplying 12mA. Think of it the 2A rating as max allowable current, not actual current. V=IR is your friend.
See (1) above. Yes, there will be less current drawn when only displaying 2 instead of 7 segments. But, each segment should have a current limiting resistor, normally limiting to around 15mA. In your case, worst case is with all 264 LEDS on, which gives max power per led of 7.5mA with your 2A supply. You won't have them all on at the same time, and certainly not for very long, so you could probably go with about 10mA per LED. Therefore, for your 12V/2A supply I would suggest a 1.2KOhm resistor in series with each of the LEDS. That's the basic theory, but it also depends on the driver chip you end up using and the way you connect up the LEDS.
Thanx for your input. Im confused why u say my reasoning on the current is wrong? If i have 4 LED's in series, and 4 parallel lines each containing the 4 LED's in series, each series line will have equal resistance cause they each have the identical amount of LED's in them, so will the current not be equally divided? do LED's not act like resistors by themselves?
MAS3:
You need to read up a bit.
This is basic electronics so you need to get this right.
Sure, lighting up 2 segments will draw less current than lighting up 7 segments.
You do not need to use relays, just use transistors that can handle voltage and current you need.
If you put 4 LEDs in series, then you will have to add their forward voltages.
This will result in something like this: (1.6 * 4 ==) 6.4 volt.
You have 12 volts (or so, check it first, depends on battery type) and your LEDs need this 6.4 volts.
So you've got 5.6 volts to "burn".
Use that in your calculations for the resistor you'll need to limit the current to the value your LEDs can handle.
In other words: find the datasheets of all components you're going to use.
They contain information you need.
Use 1 resistor per parallel line.
Thanx man, this helps alot. I know my knowledge on electronics is very bad but i am in the process of learning, the learning curve is just so steep haha!
In your mind, you are thinking of the Voltage and Current ratings of your battery in the same way. Voltage is the force that WILL be supplied. Current is a maximum amount that MAY be supplied at a given moment in time. A 12V/2A battery supplies 12V, and some current value up to 2A depending on the attached circuit.
LEDs are Diodes (PN junctions). There do not behave like resistors. They have a voltage drop across them, but other than that, they are essentially a piece of wire (for calculation purposes). The current through them is determined by the rest of the circuit, until you put too much through them and they die. They are current driven and need around 15mA. More current, brighter. Less current, duller. You generally never want to go more than about 20mA though. Check your LED datasheets. In your case though, your battery can't source this much current per led due to the sheer number of LEDs you're using.
arduinodlb:
In your mind, you are thinking of the Voltage and Current ratings of your battery in the same way. Voltage is the force that WILL be supplied. Current is a maximum amount that MAY be supplied at a given moment in time. A 12V/2A battery supplies 12V, and some current value up to 2A depending on the attached circuit.
LEDs are Diodes (PN junctions). There do not behave like resistors. They have a voltage drop across them, but other than that, they are essentially a piece of wire (for calculation purposes). The current through them is determined by the rest of the circuit, until you put too much through them and they die. They are current driven and need around 15mA. More current, brighter. Less current, duller. You generally never want to go more than about 20mA though. Check your LED datasheets. In your case though, your battery can't source this much current per led due to the sheer number of LEDs you're using.
Thank you for clearing that up for me, I see how I was confusing it. I would like to ask a few questions if you don't mind.
You say LED's are current driven so if I build a basic circuit. As an example battery, resister, LED. If the voltage of the battery is 2v and the current is 20mA then the LED will shine xy bright. If we take the same LED put it in a circuit with battery voltage of 3v but we ad a different value resister so the current remains 20mA, will the LED still be xy bright? Basically what I'm asking is will the brightness of the LED stay the same irrelevant of the voltage as long as the current is the same?
Basically what I'm asking is will the brightness of the LED stay the same irrelevant of the voltage as long as the current is the same?
Yes, provided you take into account the forward voltage of the LED. Around 1.6VDC. For 12V, the 1.6V is small enough to not matter. For 2V or 3V, you need to base your current limiting resistor calculation on (2-1.6) = 0.4V.
Basically what I'm asking is will the brightness of the LED stay the same irrelevant of the voltage as long as the current is the same?
Yes, provided you take into account the forward voltage of the LED. Around 1.6VDC. For 12V, the 1.6V is small enough to not matter. For 2V or 3V, you need to base your current limiting resistor calculation on (2-1.6) = 0.4V.
Not strictly correct.
Different manufactures and even different batches of the same led type will not always yield the same brightness for the same current.
Normally this does not matter as few projects have so many leds right next to each other.
This is one of the reasons I asked the question several posts back as to the required size of the segments the OP wants to achieve by using discrete leds as it may be better to use the "jumbo" 7 segments that are available as the leds in them will most certainly be matched in terms of brightness and colour temperature.
The other reason is of course it's a heck of a lot easier to use complete 7 segment displays then making them up from individual leds.
Just 1 point. The battery is 12V, 2 AH - Ampere-hour, which means it could supply two Amps for one hour (although you really should not make it do this as it would not be good for battery life - you should limit it to one amp.)
As you will be using multiplexing, only one display will be on at any one time, so your current draw will correspond to this.
Green LEDs do not have a voltage drop of 1.6V; more like 3.3, so you will not be able to use four in series, only three. You need to test this with a breadboard and your battery with various resistors. I suggest you start with a 100 ohm resistor and see what current you measure. It will however be different as the multiplexing drivers (transistors) introduce further voltage drops. You may well need to use drivers which themselves control the current.
If all LEDs are from the same batch, you can wire them in direct series-parallel connection without separate resistors for each series chain.
Let's see - four chains of three, 20 mA per chain is 80 mA, seven segments gives 560 mA total if showing "all 8s". That sounds entirely do-able with that battery. You do not need to use more current as it will be bright enough with a one-in-three multiplex and if the program were to fail with a certain digit lit, it will be perfectly safe.
Basically what I'm asking is will the brightness of the LED stay the same irrelevant of the voltage as long as the current is the same?
Yes, provided you take into account the forward voltage of the LED. Around 1.6VDC. For 12V, the 1.6V is small enough to not matter. For 2V or 3V, you need to base your current limiting resistor calculation on (2-
Not strictly correct.
Different manufactures and even different batches of the same led type will not always yield the same brightness for the same current.
Normally this does not matter as few projects have so many leds right next to each other.
This is one of the reasons I asked the question several posts back as to the required size of the segments the OP wants to achieve by using discrete leds as it may be better to use the "jumbo" 7 segments that are available as the leds in them will most certainly be matched in terms of brightness and colour temperature.
The other reason is of course it's a heck of a lot easier to use complete 7 segment displays then making them up from individual leds.
But hypothetically speaking if the LED's were identical then what Karma is saying is correct right?
I'm building it myself with individual LED's cause big SSD's are super expensive in South Africa or any electronic components for that matter
Paul__B:
Just 1 point. The battery is 12V, 2 AH - Ampere-hour, which means it could supply two Amps for one hour (although you really should not make it do this as it would not be good for battery life - you should limit it to one amp.)
Good point. Thanks for pointing that out Paul. I was going on the original poster's comments and should have checked. Especially for Batteries which are only ever in AH.
calvingloster:
... the learning curve is just so steep haha!
The curve is as steep as you set it.
Lower the bar a bit so you can keep an overview if necessary.
@Paul__B:
I know different color LEDs have different forward voltages.
And different LEDs have optimum performance at different currents.
That's exactly why i told OP to check datasheets.
calvingloster:
What do you mean by "You may well need to use drivers which themselves control the current" ?
Since you are requiring currents greater than the ATmega chip can provide, you are going to need transistors (or one or more driver ICs designed for the purpose) to drive these LED arrays. I was meaning constant current drive circuits in which the transistor rather than simply a series resistor is used to regulate the current.
You can use TPIC6B595 shift registers to sink current thru the strings of LEDs that make up your segments.
Each 12V powered string, generally 5 LEDs, maybe 6, will require up to 20mA.
Each output of the TPIC6B595 can sink 150mA, so you can have 2 or 3 strings arranged to be as long as you want the segment, or strings next to each other for fatter/wider segments, and each shift register output can still handle 1 segment.
You'd then have a shift register per digit.
This board I make has 12 shift registers, fully populated it could do 12 digits. No multiplexing required, so the digits are nice & bright all the time. http://www.crossroadsfencing.com/BobuinoRev17/
Here's another that supports just 4 digits - you don't have to add all the extra parts (3.3V regulator, SD card socket, RS232 driver, or even the screw terminals).
