Best way to dim a LM8560 based LED alarm clock

Discussion in 'The Projects Forum' started by Johnnz, Mar 6, 2009.

  1. Johnnz

    Thread Starter Active Member

    Dec 31, 2008
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    Hi again,
    I have an alarm clock that is too bright at night. I want to use an LDR switching circuit to automatically dim it at night time. However, first I need to find the best way to do so. I have identified the driver chip for display/clock to be a LM8560 and have found a datasheet on it. It is designed for use with a common cathod Duplex LED display which has two connections for cycle-1 and cycle-2 connection to AC power.

    If I just used a resistor in each cathod connection to 0v, I suspect that the individual segments on the display will dim unevenly. So, I intend to make a dimming circuit like this one: http://www.reuk.co.uk/OtherImages/ne555-pwm-led-dimmer-circuit.gif

    I am unsure whether this will be suitable for PWM of the cathode connection to AC?

    Thanks for any ideas/advice.
     
  2. Johnnz

    Thread Starter Active Member

    Dec 31, 2008
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  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    It's a guess, but the cycle 1 & 2 resistors might have a big effect. Try increasing on and see what happens.
     
  4. Johnnz

    Thread Starter Active Member

    Dec 31, 2008
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    Sorry, I dont quite get what you mean by "increasing on". Do you just mean to try different values of resistors in this path of the circuit?
     
  5. Johnnz

    Thread Starter Active Member

    Dec 31, 2008
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    Ok, I have put 68ohm resistors in the 1 and 2 cycle path ways. This creates the desired effect of appearing to dim the led display by a factor of about 4 (seems to be 1/4 of the brightness to my eyes). Next, I will have to find a way to use a variation of the following circuit to automatically dim the display:

    http://www.circuitstoday.com/wp-content/uploads/2008/04/light-activated-switch-circuit.jpg

    I want to avoid using a mechanical relay (noisy and too large), I am wondering if I will be able to use a solid state relay such as this one:

    http://www.jaycar.co.nz/productView.asp?ID=SY4090

    However, it may be possible I dont need a relay to do this job? I measured the current through each common cathod and it is no more than about 90mA (depending on what the time is/segments are lit).

    When it is dark in the room I want the light dependant resistor circuit to imped the flow of current by bringing a 68ohm load in each pathway, but when it is bright in the room the current will flow unimpeded as is factory standard. Designing or modifying such a circuit is my next challenge. Any ideas or suggestions welcome :)
     
    Last edited: Mar 6, 2009
  6. Johnnz

    Thread Starter Active Member

    Dec 31, 2008
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    Basically, my only question now is this:
    Is the only way to short out two independant resistors in a circuit by using two seperate relays connected to another independent circuit? Besides two solid state relays, are there any other semiconductors that could do this job any cheaper?
     
  7. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
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    What are the two resistors connected to on each end?
     
  8. RobUrban

    New Member

    Nov 13, 2013
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  9. RobUrban

    New Member

    Nov 13, 2013
    2
    0
    Hello,

    I have a sure bet for you, but you will have to do the math for the components. email me personally rmu2@yahoo.com so I can send you a schematic. I do not have one prepared at this moment. But basically, you can use transistors like the 2n3904 NPN to drive the segments. Use a 5k or similar at the base of each transistor. Connect the collector to your positive power source. Connect your load resistor from the emitter to the anode of the segment.

    That was the first part.

    Part 2: the dimming....you need one more transistor, I chose a tip120 because it can handle 5 amps...maybe a little over kill but readily available. Build an emitter follower circuit with the tip120. Connect the collector to the positive power. Build a voltage divider between the collector, base and ground. Use a photo cell or Light Dependent Resistor as R1 from the collector to the base. Now the math can be tricky depending on supply voltage and resistors used. But the most common mistake that s made is that using this device to power your segment driver transistors is that they will either be on or off depending on how dark your room is. So to fix this you have to put a resistor in parallel with the photo cell to set the low light threshold for your display. you want it to dim, not to turn off.

    If you are interested, I will draw a schematic and send you a video of a working model if you'd like.

    Sncerely,

    Rob
     
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