555 Fading LED: Calculation

Discussion in 'The Projects Forum' started by vanglock, Oct 25, 2013.

  1. vanglock

    Thread Starter New Member

    Oct 23, 2013

    first thing, i really like the forum, gave me some good advice so far.
    Still, I'm doing my first project, it would be great, if you could help me.
    I do have some theoretical background, but would like to verify my thoughts.

    I need to fade in and fade out different LEDs with a fading time of something like 5 - 10 seconds.

    I found this circuit:



    And I'm trying to do the math:

    The C gets charged over the 20k R.
    Vcc is 9V, the 555 chip charges from 1/3 to 2/3 of Vcc, so:

    The voltage at R will be between 6 and 3 V
    The voltage at C will be between 3 and 6 V

    Looking and the right side of the circuit, the base of the transistor is connected to the C, which means the voltage at C will be at the LED and its resistor with a voltage drop at the transistor:

    U_{BE} = 0.7 V

    My LED parameters:

     I_{max} = 20 mA

    U_{f} = 2.1 V

    The maximum voltage at the resistor will be

    U_{R,led,max} = 6V - U_{f} - U_{BE} = 3.2 V

    To limit the current, the value of the resistor has to be at least:

    R_{led,min} = \frac{U_{R,led,max}}{I_{max}} = \frac{3.2 V}{20 mA}=160 \Omega

    Same is true for the PNP side.

    Now, dimensioning the RC Network:

    Since I'm using the the CMOS Version of the 555, the maximal sourcing current it given with:

    i_{R,max} = 10 mA

    Now with slow fading times, the current in the transistor is not negligible, and with the gain of the transistor:

    B = 100 - 500

    the maximal current:

    i_B = \frac{i_C}{B} = \frac{20 mA}{100} = 0.2 mA

    Since I'm using 2 transistors, there need to be a current trough the resistor of at least:

    i_{R,min} = 2 i_B = 0.4 mA

    with this parameters, I can give a range of the resistor:

    R_{min} = \frac{U_{R,max}}{i_{R,max}} = \frac{6 V}{10 mA} = 600 \Omega

    R_{max} = \frac{U_{R,min}}{i_{R,min}} = \frac{3 V}{0.4 mA} = 7500 \Omega

    so the values for R in the RC Network:

    R = 600 \Omega ... 7500 \Omega

    The charging + discharging time from the C between 1/3 to 2/3 is given with:

    T = 2 \cdot 0.7 R C

    Remark: current i_B is neglected in calculation, since it has opposing effects in charging and discharging.

    So e.g. a time of 10s and with R = 3300 Ω

    C = \frac{T}{1.4 R} \approx 2200 uF

    So, thanks so far for reading.
    I'd like tu discuss my thoughts and want to make to ensure, that I put it together in the right way.

    Thanks for any help!
  2. elec_mech

    Senior Member

    Nov 12, 2008
    I have not built a fading circuit to date, but Bill offers a wealth of knowledge on his blog.
  3. vanglock

    Thread Starter New Member

    Oct 23, 2013
    Hi. Thanks for the reply

    I know about the blog, and especially about chapter 12.
    Still want to verify my thoughts, it helps me to extend and modify my circuit.
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    Your calculations look fine. The circuit works in simulation.
  5. Bernard

    AAC Fanatic!

    Aug 7, 2008
    Looks about 3 s up & 3 s down. Could use darlington transistors to remove them from calculations. If the LEDs are blue & red, resistor values will be different.
  6. vanglock

    Thread Starter New Member

    Oct 23, 2013

    Thanks for the help!

    I built it so far and it works quite well.

    Still got a question:

    At the transistor is a voltage of 3-6V with a current of 0.02A. Isn't this a waste?
    Can I put another LED at the collector side?
  7. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    Should be ok, providing the Vf of the LED is < 3V (i.e. use a LED other than blue or white).
  8. Bernard

    AAC Fanatic!

    Aug 7, 2008
    On the Red side, with red Vf of 2.3V, the resistor now would be about 36 Ω. Not enough spare V on blue side? Sorry , see that you are using all red.
  9. Wendy


    Mar 24, 2008
    The classic formula is:

    F = .7 / RC

    The transistors will throw this a little off, but not by much. If it does not work you may need to go to a Darlington pair or a Sziklai pair, both of which offer much more isolation. I would Google them to look them up.