10W LED 555 Sweeper Circuit

Discussion in 'General Electronics Chat' started by mtpalymore, May 18, 2017.

  1. mtpalymore

    Thread Starter New Member

    May 18, 2017
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    I'm building a 555 sweeper circuit with 4017's in series triggering various colored 10W LED's for signage. Using a 12V 1A power supply. I thought the 555 and 4017 could handle the 12V no problem, but the LED's are drawing anywhere from 500mA to 900mA through those chips and the 555 is getting very hot. Not only that, but the oscillation seems to either go insanely fast or just freezes no matter what I set the duty cycle. So I'm thinking I should use transistors to handle the higher current draw of the LED's and keep a lower current for the timer circuit part of it. Running at 12V and 500mA to 900mA what would be a good transistor for this application?
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Yes, you are seriously abusing the IC's.
    Look at the data sheets and see what the maximum current ratings are.
    A MOSFET would be good to handle the current.
    Post your circuit and we can give more detail.
     
  3. mtpalymore

    Thread Starter New Member

    May 18, 2017
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    Thank you, I'm basing it off this schem attached only the LED's are rated 9-15V and draw 750mA. So I saw other schematics where they were adding a MOSFET to each LED, but they were for even lower power LED's and I was trying to figure out a good match for this 10W one. Unfortunately, the only datasheet I had on the LED's is no longer available on the web, so I'm just getting these ratings off the variable power supply.
     
  4. crutschow

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    Just about any P-MOSFET should do the job with an ON resistance of less than an ohm.
    Connect the MOSFET source to V+, the gate to the respective 4017 output, and the drain to the LED going to ground.

    What supplier do you use for parts?

    But I don't understand why the 555 is overheating.
    What part values did you use for the 555 which aren't on the schematic?
     
  5. dl324

    AAC Fanatic!

    Mar 30, 2015
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    CD4017 won't be sourcing anywhere near 750mA.
     
  6. AnalogKid

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    If you don't read the chip data sheets, everything will fail. You are overcurrenting the 4017 output stage by 125x to 225x.

    To select a driver transistor, it should be rated for at least twice the peak current your circuit will require, and twice the power supply voltage.

    ak
     
  7. mtpalymore

    Thread Starter New Member

    May 18, 2017
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    I used Mouser for the IC's and resistors and everything since they're really close to Austin and it's like next day shipping, but I got the LED's from an Ebay shop in the UK. I'm using all the same rated components as in the schematic, though before looking at the IC current ratings I got 2W and 1W resistors for the power draw of the LED's which are really excessive for those IC's and I should probably get some 1/8th watt ones for the timing circuit. I wasn't sure why the 555 was getting hot either, maybe it's from the resistors being to high or because the 4017 is so over current it's coming back out or something. Thankfully i got several extra's for testing/spares so next time I'll get it right since I am too impatient to read everything before testing it out.
     
  8. mtpalymore

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    May 18, 2017
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    That's what I needed to know.
    Awesome, Thanks everyone.
     
  9. ian field

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    Oct 27, 2012
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    One of the standard wire ended LEDs pushes a CMOS output to the limit.

    There are various Darlinton driver arrays that save PCB real estate, but 10W LEDs may need discrete BJT or MOSFET drivers to handle the current.
     
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  10. crutschow

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    Post a schematic of the 555 with the resistor values, as I previously requested, and we can likely tell you why it's getting hot.
     
  11. Tonyr1084

    Well-Known Member

    Sep 24, 2015
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    Here's how I did mine:

    Rather than use a 555, since all I wanted was a clock pulse I chose to use 1/2 of a Dual D Flip-Flop for the clock pulse and the other half of the DDFF to switch between 1-10 and 11-20.

    Here's how the clock works: PNP transistor (2N3906) base is drawn to ground via the capacitor. This holds set and reset of the DDFF (CD4013) at ground, giving /Q (not Q) the chance to go and stay low. When the 0.1 µF cap charges up the base of the 3906 opens and allows Set and Reset to oscillate. Set and Reset oscillate providing the clock pulse, as they are constantly toggling (or flip-flopping) back and forth. This clock pulse is then fed to the CD4017 (Decade counter) which outputs one high per clock pulse (as you know). When it gets to output #9, the next pulse drives the chip back to the #0 output and the "Carry Out" output clocks a signal clock pulse to the clock input of the CD4013 B, flipping Q to low and /Q to high. Each transistor is on only one at a time, so that as the 4017 sweeps through 10 outputs, the DDFF switches ground via the transistors for either the first set of 10 or the second set to ground so that only one LED is lit at any given time.

    The cap is given as a 104, meaning it's a 0.0000000100000 farad cap (AKA 100,000 Pico, or 100 Nano, or 0.1 Micro) capacitor, and my resistors are coded according to standard markings on a numerically identified resistor, be it SMT or PTH type. A 473 is 47000 (or 47KΩ) The 471 is 470Ω

    The larger ground resistor protects all the LED's AND the two NPN transistors (2N3904). Since only one line is active at any given moment there is no large amount of current running through the circuit.

    To modify my circuit to accommodate what you wish to build you'd need a MOSFET in place of each LED, and two MOSFETs (with my setup) toggling the ground.

    This isn't your project completely engineered for you, I'll leave that up to you. After all, I don't have immediate access to your data sheets, and only you know how you intend to build.

    Just one comment on the circuit you provided: The first 4017 strobes ten outputs one at a time in sequence. The second 4017 strobes ten outputs, stepping one step with each TEN clock pulses. Meaning the first will sweep across the 10 LED's while LED #11 is constantly lit. Then with the next sweep through the first 10 LEDs, #12 will be lit. At any given instant you will have two led's lit. The first 10 counts the clock pulses one for each pulse and the second counts each TENTH clock pulse as a single clock pulse.
     
    Last edited: May 18, 2017
  12. AnalogKid

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    4017 outputs are active-high. For one-at-a-time LEDs, shouldn't the driver be n-channel?

    ak
     
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  13. crutschow

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    Your are certainly correct. :oops:
    That, of course, means that the LEDs will be low-side driven rather than high-side.
     
  14. mtpalymore

    Thread Starter New Member

    May 18, 2017
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    I was looking up various ones, I think STQ3N45K3-AP should handle them fine in this application.
     
  15. crutschow

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    Not quite.
    That has an on-resistance of 3.3Ω which would dissipate 2.7W when carrying 900mA, which means it will be operating very close to its maximum junction temperate at an ambient of 25°C.
    You should find one with an on-resistance below 1Ω.
    You also don't need one with a 450V rating.
     
  16. ian field

    AAC Fanatic!

    Oct 27, 2012
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    The type number looked sort of like a high voltage part - a high RDSon wouldn't be a big surprise.

    I'd use the SMD power MOSFETs off scrap motherboards because I have some laying around. With drain current rating in the general direction of 90A, its probably overkill, but the RDSon is a few tens of mV - at less than 1A; you might even get away with no heatsinking.

    Most types can only handle 30V, some only 20V - but VGSthr is very much in the same ball-park as logic level types.
     
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