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Following that post here: Designing a large 7-Segment display

To give you the basic approach I did that:

  1. form the uC to shift register followed by ULN2003, then connected 8 7-segments together

  2. Put 8 P-Mosfets to control each 7-segment, turn it on, then send its digit

  3. In 2, all the 8 Mosfets are controlled by a 3 to 8 Decoder.

I managed to do everything and control the 32 7-segments. but the thing is, to control 7-segments using step 3, would require a power consumption of 8*5*120mA, considering 8 segments per row, 5 cols, and each row consumes 120mA, and that requires a very big power supply.

Is there any other idea to have a better design?

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andre_lamothe
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    Isn't this the same question as the previous one? You are basically asking for a better solution to the same question so I'm not sure why you closed the previous one by accepting an answer. – alexan_e Feb 13 '14 at 10:58
  • @alexan_e So should I close this one and doesn't mark the other one as an answer ?. As you saw I have asked several questions, but found no answer. – andre_lamothe Feb 13 '14 at 11:01
  • I'm not sure what to suggest at this point, lets see what more experienced members think about this. – alexan_e Feb 13 '14 at 11:03
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    This question is confusing, but I don't think there's any way round the problem that the max power consumption is the power of one segment times the number of segments that may be on at any one time. – pjc50 Feb 13 '14 at 11:29
  • @pjc50 so my power calculations is wrong ? I have 8 segments, 5 rows of it. each segment draws 15mA. each segment in each row is turned on/off sequentially using a decoder to Mosfets. so that total power should be 8*15*120 = 600mA ? – andre_lamothe Feb 13 '14 at 11:57
  • possible duplicate of [How can I efficiently drive an LED?](http://electronics.stackexchange.com/questions/55823/how-can-i-efficiently-drive-an-led) – Phil Frost Feb 13 '14 at 12:53
  • I thought you had 5 rows of 8 digits _each of which has 7 segments_? – pjc50 Feb 13 '14 at 13:15
  • If each segment takes 15mA average to be acceptably bright and you have n segments that could all be (visibly) on at once, you need n * 15mA power supply capacity. 40 * 8 * 0.015 = 4.8A. I don't know how to make this any simpler. If only half are actually visibly on at once, the consumption will be half, but the PS capacity should support all being visibly on at once if that is possible. – Spehro Pefhany Feb 13 '14 at 13:38

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There is no free lunch. The less average current you feed x number of display segments, the dimmer they will look, regardless (to a close approximation) of whether they're operated at 10% duty cycle or 100% duty cycle, provided only that they PWM frequency is above the human eye's flicker fusion rate. Persistence of vision does not buy you anything because the light intensity is time-averaged by the way the physiology works. Modern LEDs are more-or-less linear in that you give it more current you get incrementally more-or-less the same increment of light (less so at higher currents, so you lose).

Historically, old, badly made LEDs required a threshold current to get going and there was some improvement reported, but that's not been true for decades. Micro amperes can produce a slight visible light.

You have three choices if you want segments that appear steadily illuminated and much less current.

  1. Use a bunch of average current and use a cheap SMPS adapter to get the current (i.e. fugetaboutit).

  2. Live with very dim displays

  3. Buy more expensive displays that have more brightness at a given current (using better LED dice inside). They probably will have a bit more forward voltage and the colors may be a bit different (more orange-red than red, for example). Not all mcd claims on datasheets are reliable, in my experience, though the well known manufacturers are mostly believable.

That's about it. Incremental improvements can be had in power (not current) consumption by reducing the voltage a bit (at the expense of other things such as uniformity). It can be mitigated a bit by using more expensive drivers (eg. MOSFETs that drop mV instead of the relatively wasteful (but dirt cheap) Darlington drivers I suggested.

I must applaud you for your concern about power consumption, most companies care only about such mundane things as commercial success and short-term profits. As such, they'd rather put another $1 into a power supply capacity than put $20 into higher performance circuitry. Hard to believe, isn't it?

Spehro Pefhany
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Is there any other idea to have a better [lower-power] design ?

  1. Don't illuminate so many segments concurrently. Use persistence of vision.

  2. Use a 7-segment LCD instead of an LED display. Different drivers needed.

RedGrittyBrick
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  • I'm use POV. The point is the high 8 mosfet drivers, will consume 8*120mA. I'm not sure if that calculations is correct, consider each segment takes 15mA, then a row of 8 7-segments takes 120mA. and I'm having 8*5 segments ( Rows*Cols ). – andre_lamothe Feb 13 '14 at 11:11
  • @Ahmed: As an extreme example, if you could illuminate just one segment at a time, you'd need no more than 15 mA total (plus overheads). However it may be that to illuminate a 32-digit display without evident flicker and at sufficiently high apparent brightness you cannot get away with a much lower power need. Hence the other idea of using an inherently lower-power display technology. – RedGrittyBrick Feb 13 '14 at 11:32