Driving 3" Seven Segment Display

Discussion in 'General Electronics Chat' started by elec_mech, Nov 12, 2008.

  1. elec_mech

    Thread Starter Senior Member

    Nov 12, 2008
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    Hello All,

    I'm getting my butt kicked and could really use some help. I'm building a sports scoring device using very bright 3" seven segment displays. Because I can do nothing simple, these are also dual colored seven segments with a common anode for each color. I'm going to use a PIC and multiplex a total of four digits so only one is on at any given moment which I'm figuring will reduce the overall current consumption. The caveat is this is for a battery powered device that I'm trying to make as small as possible. The red color uses a typical forward voltage of 12.0V while the green calls for a 12.6V, therefore I'm planning on using eight AA batteries to get my 12V.

    The biggest problem I've run into so far is driving the segments. I originally planned on using ULN2803's to drive the cathodes of the segments and an Allegro 2982 source driver for the anodes. However, I failed to realize these transistors have a 1-2V drop across the collector and emitter and the segments come out dim at best. I then tried playing with an IRL520 MOSFET, just driving the cathode of one segment to see how it looked, but the current on the power supply goes well beyond 400mA and doesn't turn on the segment at all. I've included a PDF of the test circuit - I'm not 100% sure I did it right. I've also tried opto-isolators, specifically the LTV-847. Those seem to work - the segment under test appears reasonably lit, but still not as bright as when I hook a segment directly up to the 12V source (with a 10Ω resistor for safety).

    Anyone have any suggestions on how to drive an LED segment, with rapid on/off, with a 12V battery without dropping the voltage or consuming a great deal of current? Or would you recommend sticking with the ULN2803 and 2981 and increasing the batteries to 9 or 10 to get 13.5-15V?
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    You said that each red led needs 12V to turn on, i think its too much, whats the part number of the display?

    Also, if each led needs 12 volts to turn on, in your circuit you connected them in series so they cant turn on, you have to connect them in parallel with a limiting resistor for each one.
     
  3. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    You need to limit the current to the LEDs, otherwise they will burn out. I see that you know the forward voltages, but do not specify a current. Yes, it is best to run the LEDs directly with their intended forward voltage for the most efficiency, but I have always given overhead for supply changes, etc., then limited the current via a resistor. Your batteries will start out at a much higher voltage when fresh, then dip significantly. This isn't going to work well in your setup.

    Either you give enough voltage overhead to run at the min battery voltage, or you need to further complex things by using a switching regulator. Or, perhaps a constant-current switching regulator.

    Something along the lines of:
    http://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1094,C1766,P38340

    You can do a linear current control by the following:
    http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm

    You can replace the transistor with a mosfet.

    Steve
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    896
    Are you planning to throw away many AA alkaline batteries?
    Eight alkaline batteries are 12V to 13V only when they are brand new. As they are used the voltage quickly driops to 9.6V and your LEDs will be extremely dim. Then the voltage drops slower to 8V.

    Your LEDs might be typically 12V and 12.6V but they are all different and have a range of voltages. Maybe you can buy many extra LEDs and use only the "typical" ones.

    Your circuit should allow for the battery voltage to drop to 8V or 9V and for the max voltage that is spec'd for the LEDs.

    Your circuit has nothing limiting the current of the LEDs so when the battery is new and the LEDs are typical or less then the LEDs will instantly burn out. When the current was 400mA then the LEDs got fried.

    You cannot save current by multiplexing the digits. Then each digit is lighted for 1/4 of the time and will look dimmed. You must increase the current 4 times for the brightness to look 'normal". Re-calculate how long the battery will last.
     
  5. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
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    Typical battery has 15390 Joules of energy, you have ~123KJ of energy.

    Assuming three seven segment displays half lit, at 12V @ 100mA, 3X12*0.1A = 3.6W = 3.6J/S = 9.5hours for the 123KJ of energy. This is largely dependent on what type of forward currents you have, which I guessed at 200mA max. Also, I am assuming no losses.. If you were using a switching converter to maintain the brightness, then you'd need to multiply by 0.8-0.9.

    Steve
     
  6. elec_mech

    Thread Starter Senior Member

    Nov 12, 2008
    1,513
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    Here is the datasheet for the display. I'm using part number:
    NFD-30011BUEUG-11. This is an assembled seven segment display which shows six LEDs hooked up in series - I just showed six in the previous circuit to follow true to the datasheet. Tentatively, I was using 10Ω resistors to limit current since I was planning on having 12V. I've always used: (V_batt - V_f_led)/I_led = R, but since I was expecting the forward voltage to be about equal to the battery/supply voltage, I kept the resistor value low. Is there another calculation to follow to find the proper resistor value for an LED when V_supply ≈ V_f_led?

    Ugh, you are right, I had forgotten that the voltage drops rapidly in AA alkaline batteries - those will make this pretty useless pretty quick. So, a switching regulator? I was planning on using one to drop the 12V to 5V to power the PIC since the power consumed would be low and the loss would be small. I would really prefer to use a wallwart to supply power, but it doesn't look like there is power available where this will be, thus the battery issue.

    Okay, so I can either use more batteries and hope the voltage drop is slow enough over time so batteries don't have to be changed every few hours or use a switching regulator which will guarantee a longer, steady power supply at the cost of efficiency? It sounds like the switching regulator is the way to go.

    The datasheet calls for 30mA per segment. Assuming worse case, all seven segments are lit giving us 0.03A x 7 segments x 4 digits = 840mA. I also have some additional LED bars, 10 total rated also for 30mA, but at 10V. Sticking to current: 840mA + 10 * 0.03A = 1.14A. I apologize, I used the term multiplexing to mean PWM. If I turn on one digit for a quarter of the time, then off, then turn on the next digit for a quarter of the time and so on, aren't I only powering each digit for a quarter of the time and, in effect, consuming a quarter of the current that I would be if I just ran all four continuously? This is assuming I can run them fast enough so the human eye can't see the shift (~+60Hz for one, I need four, so 60 * 4 = 240Hz ≈ 4.1ms). Am I correct or am I making a fatal flaw somewhere?

    Did I set up the MOSFET circuit right? I still can't figure out why the current draw went up by an order of magnitude. I'll try it again with some spare individual LEDs.

    Thank you everyone for your advice!
     
  7. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    It's about your average current draw, not your peaks. It is usually better to multiplex, since you do not need as much decoding logic. Nonetheless, you still should be drawing 30mA per segment, otherwise it will not be lit properly. If you want to run them at a duty cycle, then scale up the current to a point as indicated in the datasheet. It illudes that you cannot go below 1/10 duty cycle, so be aware of this.

    You can go to 30Hz if you need to. Movies go down to 25fps, without much flicker. Lowering the frequency helps minimize switching losses, but they're insignificant anyways, because of all the power being use.

    Following from my previous calculation, 1.14A X 12V = 13.68W, so 123KW/13.68 = 2.5 hours of operation, assuming 100% conversion efficiency.

    Your MOSFET should be working.. I am guessing the current wasn't limited and it burned out the segment.

    Steve
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    Movies blink each frame twice so the flicker rate is 48Hz and not too bad.
    Car LED tail-lights flicker at a few hundred Hz but can still be seen as a trailing line of dashes in your slower vision.
     
  9. k7elp60

    Senior Member

    Nov 4, 2008
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    I have a digital clock I built several years ago that uses 4" displays very similar to what are described. I am using ULN2004A to drive each display. I use a resistor between each segment and the ULN2004A. The +Vcc is supplied by a 3 terminal regulator that I adjust the brightness for day and night by changing the adjust resistor. The + voltage to the regulator is about 13.6V so the output from the regulator is about 11.6 volts. I am not sure where the displays were made, but I have had them for over 10 years and and they work real well.

    I assume that my displays are not multiplexed as the clock IC I am using has individual segment drive for each digit, and does not have digit drive.

    Any help I can be let me know.

    I just found the data sheet on my displays. They are AND4107 and the data sheet shows 4 LED's in series for each segment and a typical Vf of 8.4 to 12V.
     
    Last edited: Nov 12, 2008
  10. elec_mech

    Thread Starter Senior Member

    Nov 12, 2008
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    Ahh, alright, I think I see now. Using PWM can be used to increase an LED's brightness by applying a high current for a short period of time in bursts which is about equivalent in average power as running the same LED as its rated current continuously. Alternately, if I attempted to PWM an LED at its typical current instead of a higher one, I'd end up dimming the LED.

    At a little over 1A*hr in the worst case and not even factoring in efficiencies or discharge rates, I don't see how I can practically run this off batteries unless I use a car battery, but who wants to lug that back and forth between the charger? :) I'm assuming this device will get used an average of 4-6 hrs a day, maybe a lot more so to even last a week, the battery would have to be 1A*5hr*7days = 35A*hr! Ugh. :(

    k7elp60 - thank you for the part number of your displays. Unfortunately, I didn't have any luck on a Google search. Do you know who the manufacturer is? I've done a lot of research on 7-segment displays and the China manufacturer I'm using is the only one I found who produces bright digits - on the order of 200-360mcd. All the US manufacturers offer little in the way of digits bigger than 0.80" and those that do only offer a brightness well under 100 mcd. They call out the brightness spec in ucd instead of mcd if that's any indication.

    Thank you everyone for your help and insight!
     
  11. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    This is a totally untested design, it was something I was working on for my robotics club before it went defunct. Their may be some ideas you could use.
     
  12. k7elp60

    Senior Member

    Nov 4, 2008
    478
    69
    The ones I am using were made by a company called AND and I was unable to find any info also. My displays I have had in stock well over 10 years. I orinally had to buy 10 each and to day I have used 8 so I still have 2 that have only been used on a very limited basis.
    Jameco has some 4" in there current catalogue. P/N LSD40054-20. There website is www.jameco.com.
    I have had good luck in building my on displays. I layout a seven segment display the size I want, then drill individual holes for 3MM or 5MM LED's. I wire each segment in series and drive them with conventional circuits. I have used opaque plastic panels as a base. I usually drill the holes slightly smaller than the diameter of the LED's and then they pressfit into the panel.
     
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