LED Bargraph LM3914

Discussion in 'The Projects Forum' started by madriaanse, Mar 15, 2010.

  1. madriaanse

    Thread Starter New Member

    Mar 14, 2010
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    Dear SgtWookie,

    I found your posting on the forum regarding the use of an LM3914 to drive an LED bargraph:
    However, I am needing to adapt the above referenced circuit to monitor between 3.0 - 3.6 volts (.06V led display steps). Is it possible to do this with the LM3914? And if so, what adaptations would be required to the circuit you linked to:http://www.national.com/images/pf/LM3914/00797001.pdf?

    Thank you!!

    M.
     
  2. SgtWookie

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    Jul 17, 2007
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    Use the same diagram.
    R1:1.5k
    R2: 2.57k (use a 3k and 18k in parallel)
    50k Ohms from Rlo to ground. Use two 100k resistors in parallel, or a 20k and 30k in series.
    Or, calculate your own values:
    http://www.qsl.net/in3otd/parallr.html#

    The above values will give 10mA LED current. The 50k Ohms depends on the internal divider measuring exactly 10k Ohms.
    If your LM3914 does not measure exactly 10k Ohms from Rhi to Rlo, then you will need to adjust the 50k value correspondingly.
    Rlo to Gnd resistor = 5 times the resistance measured from Rhi to Rlo.
     
  3. madriaanse

    Thread Starter New Member

    Mar 14, 2010
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    Fantastic! You are a gentleman and a scholar! How did you gain such knowledge about this IC?

    M.
     
  4. SgtWookie

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    I started reading the datasheet a week or two before you did. :D
     
  5. madriaanse

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    Mar 14, 2010
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    Love it! :) Do the resistor values remain unchanged if power to V+/Mode is 12V instead of 9V?

    Thanks again,

    M.
     
  6. SgtWookie

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    Jul 17, 2007
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    No, but you should use individual resistors to drop some of the voltage to your LEDs, otherwise the IC will become quite hot, and it's life will be shorter.

    The resistor values I gave you will result in about 10mA current through each LED.

    If you were going to use a standard red LED 10-segment bar graph display, each LED will have a Vf of around 2v.

    So roughly, the LM3914 would have to dissipate 12v-2v *10mA * 10 = 1 Watt of power. The package is rated for 1.365W as an absolute maximum, but you really want to stay well below that.

    If you used ten 620 Ohm resistors in series with the ten LEDs, you'd cut the power dissipation in the LM3914 by a large percentage.
    10mA x 620 Ohms = 6.2v
    12v - (6.2v + 2v(the LED Vf)) = 3.8 volts remaining.
    3.8v x *10mA * 10 = 380mW; power dissipation in the LM3914 is reduced by 62%; and just under 28% of the absolute maximum rating. This is a good compromise. The resistors can be rated 1/8 Watt or higher.
     
  7. madriaanse

    Thread Starter New Member

    Mar 14, 2010
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    Thank you sir.

    One last question: I'm using this bargraph to verify the voltage of a Lithium ion battery. Could this circuit be duplicated/modified to monitor individual voltages in a series pack of, say, 20 cells (72V)? V+/Mode would still be fed from a 12V source.

    ... or am I going about this all wrong and should go with something like a PIC IC to monitor multiple cells individually?

    Thank you again for your help,

    M.
     
  8. SgtWookie

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    Now you've opened up a can of worms.

    The 12v source for the LM3914 will have to be independent of the array of batteries; IE: they cannot have a common ground or +V.

    To test an individual battery in the array, you will need to connect the LM3914 ground (pin 2) to the negative terminal of the battery being tested, and connect the positive terminal of that same battery to the LM3914 SIG input (pin 5).

    You cannot use a multiplexer/demultiplexer IC to switch the signals, as the voltage levels would exceed the limits of any mux/demux that I'm aware of. You will probably have to use a tree of double-pole single-throw normally open relays, and never have more than one relay coil energized at any given time, or you will short out all the batteries between the two relays.

    You will only be able to monitor one battery at a time.
     
  9. BillB3857

    Senior Member

    Feb 28, 2009
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    Hey, Sgt... What about supplying each cell monitor with its own DC to DC isolated converter? The input side of each converter could be a common source and the converters provide the isolation required to monitor any individual cell.
     
  10. SgtWookie

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    Sure, he could do that - if he wanted to go to that trouble and expense.

    Just a rough guess that a prefabricated DC-DC converter would cost in the neighborhood of $15-$20 each.

    But, if the OP had lots of time on their hands and knew something of winding toroidal transformers and a bit about DC-DC converters, they could probably do it for a few dollars each, or less - depending on how good of a scrounger they were.
     
  11. madriaanse

    Thread Starter New Member

    Mar 14, 2010
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    Hmmm, wowsers that got all complicated quick. I should clarify I am trying to make an easy to read indicator for individual cell voltages in a pack of 20 batteries. 10 indicator LED's per battery X 20 = a 200 led "panel".

    Is there a way that I can float the ground of each adjacent LM3914 from the previous batteries positive terminal? (and power from a +12V supply?)

    thanks!!

    M.
     
  12. BillB3857

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    Feb 28, 2009
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    See posts #9 and #10
     
  13. SgtWookie

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    That's why I indicated that you were opening up a can of worms.

    There isn't an easy way to do what you want to do. Using relays (or optoMOS relays) to basically time-share one LM3914 would be much less expensive and time-consuming than trying to build 20 DC-DC converters. If you buy off-the-shelf isolated DC-DC converters, you'll be spending around $300-$400 for them.
     
  14. BillB3857

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    Feb 28, 2009
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    Agreed, Sgt. But in Post #11, (quoted above), he still wants to know how to do it despite your earlier warning about complexity and his own admission that it is getting complicated quick.

    One other way to do it would be to use a 2 pole rotary switch with 20+ positions, NON SHORTING.Tie the commons to the LM319, one wafer for the + terminals and the other for the - terminals. (one position for each cell) May be simpler than trying to control 20 relays or optoMos devices.
     
  15. madriaanse

    Thread Starter New Member

    Mar 14, 2010
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    The rotary switch is an interesting, cost effective idea.

    Is there a (cost effective) circuit that could replace its functionality and cycle between batteries/LM3914s at a rate of 400hz or so? (20 batteries displayed 20 times per second each). This would probably be fast enough of a cycle to appear like a continuous readout to the human eye.

    Thank you for your thoughts! They have been very insightful!

    M.
     
  16. SgtWookie

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    Something different for you; DPST optical relays:
    http://www.mouser.com/ProductDetail...GAEpiMZZMtLEhJ5P/NsZ7%2bJdrkJKaXu148u3db5K1s=

    The idea here is that each LM3914 has a capacitor from to each sensing input to ground.
    You will need one of these optical relays per battery/LM3914/cap.

    You will also need a method to sequence through all of the relays, energizing just one at a time. The optical relays momentarily connect each battery - terminal to its' LM3914 ground, and the + terminal to the sense input of its' LM3914 and the + side of the input capacitor, charging it. The capacitor maintains the sample of battery voltage while the other relays cycle through their individual LM3914s.

    Since they are optical relays, they will be silent and have a long service life - no mechanical contacts to wear out.

    You would not have to cycle them very quickly. Even a clock rate of 1Hz would be fast enough.
     
    Last edited: Mar 21, 2010
  17. BillB3857

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    Feb 28, 2009
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    This is one of the many things I like about this site. SYNERGY! Ideas build on each other. If I'm thinking right, on SgtWookie's plan, it would still be imperative that the power supply for the LM3914's be isolated from the battery being measured but no longer necessary to be isolated from each other.
     
  18. SgtWookie

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    Yeah, it's like that ;) Actually, the optical relays is just a subtle shift from the DPDT/DPST relays I mentioned in the can o' worms post a ways back. And yes, the ground of the 3914 circuit would still have to be isolated from the battery circuit; the relays would momentarily connect the - terminal of each battery in turn to the 3914 supply's ground.

    I thought about the rotary switches for a while, but with DP rotary switches, the large number of connections would be a bear to solder up, and I haven't seen a DP 20-position break-before-make rotary switch. That would be a LARGE and expensive switch, if you could find one!
    [eta]
    Mouser and Digikey both carry 2-deck break-before-make 23 position switches, but they're at least $43, and have a service live of 25,000 cycles. I'd hate to have to replace one of those things.

    Actually, you'd need a 3-deck switch if you were going to have one 3914 per battery. Two decks would work for a single 3914.
    [/eta]
    Being electromechanical, it would have a relatively short life span. Something would have to keep it rotating, and after so many rotations it would be plumb worn out.

    I suppose it would be possible to use multiple DP rotary switches, but it would be imperative for only one to be in use at a time, and all others would have to be disconnected from the battery array. One mistake, and you'd have a dead short.

    Anyway, back to the optical relay idea ... something like cascaded 74HC4017 Johnson counters clocked by a 555 timer could provide the sequencing of the optical relays. A 2nd timer would need to be used to provide "dead time" between relay engagements, as you would not want more than one relay turned ON at any given time; the relays I pointed out take 5mS to turn ON and 5mS to turn OFF. A one-shot 10mS delay should provide a sufficient margin of safety.

    If the clocking 555 ran at a 50mS PRT (20Hz) then all of the 3914's will have their input caps refreshed once per second for a 35mS slice of time.
     
    Last edited: Mar 21, 2010
  19. BillB3857

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    Feb 28, 2009
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    Any bets as to when the OP will want a way to sound an alarm if any individual cell falls below (plug in a number)%?
     
  20. SgtWookie

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    They could do that by monitoring whether a certain LED in the bar graph display was lit, or just sense the cathode side of the LED to determine when it was less than Vcc.
     
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