Shunt regulator/Zener diode - current regulation for LEDs

Discussion in 'General Electronics Chat' started by akirby, Oct 4, 2007.

  1. akirby

    Thread Starter New Member

    Oct 4, 2007
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    To make a long story short, I'm trying to replace a small LED built in to a circuit board with an external LED indicator light. It works fine with the LED receiving 3.8VDC when it's supposed to be on. The problem is when the LEDs are off they still receive 1.8VDC. I've tried both 2.2Vf and 5Ff 5mm bare LEDs with the same results - they still glow dimly when they're supposed to be off.

    I understand basic electricity but not a lot of the IC stuff. It sounds like I either need to find a LED that will turn on with 3.8V but not with 1.8V, or use a resistor to lower the voltage to achieve the same affect. However, the descriptions I've read of zener diodes and/or shunt regulators sound like they are exactly what I want.

    If I get a zener diode/shunt regulator with a rating of 3.3V, will it turn off the LED completely with the 1.8V current and turn it on with 3.3V or 3.8V when it sees the 3.8V current?

    Is there an easier way to do this? What's the difference between a shunt regulator and a zener diode? I have 5 LEDs (4 green, 1 amber). If you can provide part numbers and sources plus wiring instructions I would be grateful.

    Allen
     
  2. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
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    If you use a resistor, you will not completely eliminate the effect. You can use a normal diode to drop the voltage about 0.6V. Most LEDs won't lit at 1.2V, but if yours does lit, add another diode. You can also use a Zener diode, but keep in mind that the minimal Zener diode has a voltage breakdown of 2.4V. In that case you will need to change your protection resistor to a lower value.

    If you are willing to use another LED colour, try a green one. These LEDs need a slightly bigger voltage to lit than the red ones.
     
  3. niftydog

    Active Member

    Jun 13, 2007
    95
    0
    A Zener won't do what you want it to do. Best bet here is to use a LED with a higher forward voltage drop, or to reduce the voltage by using a conventional diode in series with the LED.
     
  4. arthur92710

    Active Member

    Jun 25, 2007
    307
    1
    can he do it like this?
    [​IMG]

    one of these should work
    The zener voltage Vz is the output voltage required. So if you want 3.8v out you would get a 3.8v zener diode.
     
  5. niftydog

    Active Member

    Jun 13, 2007
    95
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    akirby, the voltages you have measured, where they measured with the LED in circuit or out of circuit?

    arthur92710, the trouble is not the "on" state of 3.8V, it's the "off" state which still gives 1.8V. A 3.8V Zener won't prevent the dim glow from the 1.8V.
     
  6. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    What about this: ledoff.jpg
     
  7. niftydog

    Active Member

    Jun 13, 2007
    95
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    The LED is a diode, give it sufficient current and it's going to have a forward voltage drop - in this case I'm guessing around 1.2V. So, with 1.8V supplied, 1.2V across the LED, that only leaves 0.6V across the zener - not enough to make it break down.

    My vote is for a series diode - it should drop the 1.8V enough to go very close to being below the Vfd of the LED.
     
  8. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    Not actually. He should put the zener and the resistor in series with the LED. The series zener will keep the LED from being lit when 1.8V are supplied to the series. It could be a zener or a normal diode. It depends if the voltage breakdown of a normal diode will be enough.

    Also, keep in mind that a current limiting resistor is included in the circuit. If you use the zener, you might have to replace it by a lower value. Hence my preference for the normal diode.

    That is more like it. By the way, can you find a 1.8V zener? I think the minimum is 2.4V (or at least in the BZX series)! Indeed a 1.8V zener would me more suitable than a 2.4V one.
     
  9. akirby

    Thread Starter New Member

    Oct 4, 2007
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    What I have are 5mm Linrose LEDs - 4 green and 1 yellow. 140 MCD and a Vf of 2.2V which I assume is the voltage drop.
     
  10. akirby

    Thread Starter New Member

    Oct 4, 2007
    7
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    I believe I measured them with the LED out of the circuit, but I'll go back and measure them both ways to be safe.
     
  11. akirby

    Thread Starter New Member

    Oct 4, 2007
    7
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    Unless I measured it with the LED in the loop, that's the part that's confusing. The LED Vf is 2.2V. I even tried it with a 5Vf LED (with a built-in resistor I assume) and it still lights up.

    Let me recheck the voltages and report back. I'll also try to post a picture of the circuit board - there are several IC components in the circuit already on the board. Maybe you guys can explain what they are and what they're doing and see if they can be bypassed or not.
     
  12. akirby

    Thread Starter New Member

    Oct 4, 2007
    7
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    So the zener voltage needs to be based on the residual current WITH the LED in the loop. That makes sense.

    Is this similar to or the same as a shunt? From what I read it sounded like a 2.4V shunt would send any current less than 2.4V straight to ground but would pass a current higher than 2.4V to the LED. Then again I could be dead wrong.

    And thanks very much for all the help so far.
     
  13. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Digikey has three different flavors of 1.8V Zener in stock, and five more in their catalog. Other vendors likely have additional choices, Digikey was just the first vendor in my bookmarks list.
     
  14. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    No. A shunt is a resistance in parallel used to manage current levels. It won't turn anything off or on.
     
  15. akirby

    Thread Starter New Member

    Oct 4, 2007
    7
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    Update - 1.8V with LED OFF, 2.0V with LED ON. I'll remove the LED and retest it.
     
  16. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    Since it has a resistor built in, it will light just like a LED with an external resistor.

    The LED Vf is 2.2V. So is it a green LED? Try putting a normal diode in series with the LED and the resistor. It should solve the problem.

    You don't need to retest. The LED is probably good and it is natural that it lits. Vf is normally given at 20mA. So if your LED has a Vf of 2.2V it doesn't mean that it will not be lit when crossed with a voltage of 1.8V. It only means that will have less current biasing it.

    See this datasheet from a typical green LED from Kingbright:
    http://www.kingbright.com/manager/upload/pdf/L-1503GD(Ver1189142457.8)

    If you analyse the V/I curve for that LED, you will see that at 1.6V will start to be biased, so it will start to emit light. At 1.8V you will have 2mA biasing the LED. You can notice the glow quite clearly with such current biasing it.

    http://www.kingbright.com/manager/upload/pdf/L-1503ID(Ver1189142329.9)
    A GaAsP/GaP red LED will behave even worse. It just needs 1.45V to start emmiting, and at 1.8V will have 3mA biasing it.

    You can always opt for an expensive (and ugly) blue LED, or a red one and an additional rectifier diode (to behave as a Zener, but forward biased, of course). Those last two will be a cheaper alternative, despite the blue LED's price late decrease.
     
  17. akirby

    Thread Starter New Member

    Oct 4, 2007
    7
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    Ok - I think I found the actual LED that was on the circuit board. It's a small right angle PCB LED:
    http://www.ledtronics.com/ds/rab362/

    I also measured the current without the LED attached and it looks like just over 10V when the LED should be lit and around 3V when it's not lit. Does this help?
     
  18. mrmeval

    Distinguished Member

    Jun 30, 2006
    833
    2
    Two 1N4148 signal diodes will give 1.2 - 1.4, should do it. What is the specification of the diode?

    Two resistors and one transistor can also work and will track supply/load changes and with the LED it is in effect a two wire device. The circuit follows the two transistor limiter.
    http://caladan.nanosoft.ca/c4/ccorner/4.php
     
  19. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    You only need one actually. 1.8V - 0.6V and the LED won't lit. And the LED brightness will be more contrled with just one rectifier diode, as will be less sensitivity to any voltage variation while it is in it's on state.
     
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