555 Fading LED: Calculation

Thread Starter

vanglock

Joined Oct 23, 2013
3
Hello,

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:



http://www.simple-electronics.com/2011/09/fading-led.html

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!
 

Thread Starter

vanglock

Joined Oct 23, 2013
3
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.
 

Bernard

Joined Aug 7, 2008
5,788
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.
 

Thread Starter

vanglock

Joined Oct 23, 2013
3
Hello.

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?
 

Bernard

Joined Aug 7, 2008
5,788
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.
 

Wendy

Joined Mar 24, 2008
22,646
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.

 

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