Help with fading LED night light project

Discussion in 'The Projects Forum' started by Gaetano, Apr 28, 2014.

  1. Gaetano

    Thread Starter New Member

    Aug 14, 2010
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    Hello

    I have a book by R. N. Soar, 50 Simple LED Circuits. I found one that I would like to try to make. I made it on the breadboard and it did not work as stated in the book. The LED lights up for a short time (<0.5 seconds) and then stays off. I made the circuit in Circuit Wizard and Crocodile Clips and it worked. The premise of the circuit is that the LED will light and slowly fade after about 30 minutes based upon the capacitor. I have attached a picture from the book as well as the schematic from Circuit Wizard and a picture of the breadboard. I would really like to make it and I would appreciate any guidance on this, thank you
     
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Probably stray leakage currents between the rows/columns of contacts in the breadboard are charging the cap more quickly than intended. Try increasing the cap value to, say, 100uF.

    Edit: low transistor gain would also shorten the 'on' time of the LED.
     
  3. wayneh

    Expert

    Sep 9, 2010
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    I don't see how that circuit could ever work. There is not enough resistance in the base current path to slow the charging of the capacitor. The 390Ω resistor gives a time constant under 1ms. I suppose the right transistors might have enough internal resistance to allow this to work.

    You could try adding a 10k resistor on the first base pin.
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    It works exactly the way a Darlington transistor circuit works - not a commercial Darlington part with the internal BE resistors, but the original Darlington circuit.

    This might be another leakage current problem. Because each transistor has the collector connected to supply the majority of the emitter current, the three BE junctions are not merely diodes. 9V batt, 2V LED, resistor.....17 mA. Assuming hfe = 100, that 17 nA to the capacitor. ec=it.... With zero leakage current it will take about 11 min for the cap to charge to 5V, the approximate off point for the LED. Close enough. Maybe the 30 minute number assumes leakage current to slow the charging rate.

    ak
     
  5. wayneh

    Expert

    Sep 9, 2010
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    If you replaced C1 with an ammeter, what would it read?
     
  6. ErnieHorning

    Member

    Apr 17, 2014
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    It’s probably an optical illusion but the cathode of the LED looks like it off by one row though it wouldn’t light at all then.

    The base of the last transistor (before the LED) also looks like it’s off by one row and not connected. You’d probably get a short flash from the LED with a floating base, when power is first applied.
     
  7. BobTPH

    Active Member

    Jun 5, 2013
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    The calculation looks like this:

    3 * Vbe + 390 * Ie + Vf = 9V

    390 * Ie = 9 - 3 * Vbe - Vf

    Guestimating 0.6V for Vbe and 2.0V for Vf we get:

    Ie = (9 - 1.8 - 2.0) / 390 = 13.3mA

    Now calculate the base current as 13.3mA / 100 / 100 / 100 = 13nA

    This is basically a very high gain voltage follower with near infinite input impedance.


    Bob
     
  8. wayneh

    Expert

    Sep 9, 2010
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    I guess my problem is that I don't believe the gain will be 100 in each stage at the moment the charging begins. With no base resistors, the path to ground is a short except for the 3 Vbe drops plus the LED and its resistor.
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    Think emitter follower. The base to ground resistance is β Q1 X β Q2 X β Q3 X Re. The Base to Ground resistance would be quite large.

    With a single transistor the base to ground resistance is β Q1 Re.

    [​IMG]

    If the gain (β) of this transistor were 50, LED Vf = 2.5V, and Vcc = 12VDC the LED current would be 9.5ma, the base to emitter current would be 190μA. The equivalent resistance Base to Ground would be 63KΩ.

    No base resistor needed.

    I'm thinking bad transistor somewhere.
     
    Last edited: Apr 28, 2014
  10. wayneh

    Expert

    Sep 9, 2010
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    Thanks to all for trying to get my head straight. I think I'm getting close to accepting it.

    My confusion can best be illustrated by imagining severing the top (in the schematic) wire that connects the first transistor's collector to Vcc. Wouldn't the LED flash briefly until the cap was charged (as the OP is observing), with all the current being Ibe?
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    We may need to split this off.

    http://www.allaboutcircuits.com/vol_3/chpt_4/6.html

    The thing is, the OP's circuit is going to have β X β X β = βt. A reasonable beta is 50, so 50 X 50 X 50 = 125K. The B-E current will be 1/125000 of the C-E current.

    The protoboard is also suspect, but they generally are more reliable than the components that go into them.
     
  12. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Yes. Disconnecting the collector of either Q1 or Q2 from V+ will have the effect of reducing the 'on' time by a factor beta. So it's possible one or other transistor isn't seated/connected properly in the breadboard.
     
  13. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    The BC108 has a typical gain of 200 to 500 at 2mA, depending on grade, and this drops by 50% at very low Ic. I'll stick with 100^3 for the total beta.

    ak
     
  14. wayneh

    Expert

    Sep 9, 2010
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    I figured out what was wrong with my thinking: I had incorrectly thought you could drive a large base current regardless of the collector current. Not true. Base-emitter resistance is an inverse function of collector-emitter current.

    So the circuit should work. Something's wrong. Transistor pin-out?
     
  15. BobTPH

    Active Member

    Jun 5, 2013
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    I tried this circuit out with 3 x 2N3904. The LED came on and I could see no discernible change in the voltage across it or the current through it over about a minute. I reduced it to two stages and it faded over about 5min. It faded over 15 seconds with a 0.1uF cap.

    These times are in reasonable agreement with LTSPICE, which shows the current falling from 12mA to 1mA in about 250 seconds (this is with two stages,) With three stages it fell from 10mA to 9mA in 10000 seconds!

    So, the circuit works, but is probably highly sensitive to beta and to leakage in the cap.

    Bob
     
  16. BobTPH

    Active Member

    Jun 5, 2013
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    It is the negative feedback that limits the current. If you put the resistor and LED between the collector and V+ it would not work at all, i.e. it would fade very quickly.

    Bob
     
  17. Gaetano

    Thread Starter New Member

    Aug 14, 2010
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    Hello All

    Thank you for your input. I tried out all of the suggestions and found it a good learning experience, especially because I am not trained in this area. After speaking to a colleague, I tried using ceramic capacitors and found that it worked. I have no idea why, but after 15 minutes the LED went off. I have taken before and after photos showing the circuit. If any one could tell me why the ceramic 220nf achieved the desired result I would be grateful. I can now use the circuit as a class project.
     
  18. wayneh

    Expert

    Sep 9, 2010
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    No idea? Capacitor leakage has already been mentioned as a suspect. I think you've showed it, and learned the hard way about leaky capacitors. Try interchanging the ceramic and the electrolytic, to prove it to yourself.
     
  19. BobTPH

    Active Member

    Jun 5, 2013
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    Yes, I am pretty sure leakage would be the key difference. With 3 amplification stages it takes only a very small current to keep the LED on forever.

    Bob
     
  20. Gaetano

    Thread Starter New Member

    Aug 14, 2010
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    When I used the electrolytic, the LED never turned off. The ceramic on the other hand worked, perhaps there is less leakage? I am not trained in electronics so I am working out the trial and error way, thank you for your input

    Gaetano
     
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