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

1. ### vanglock Thread Starter New Member

Oct 23, 2013
3
0
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:

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
1,513
193
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
3
0

Still want to verify my thoughts, it helps me to extend and modify my circuit.

4. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,971
1,135
Your calculations look fine. The circuit works in simulation.

5. ### Bernard AAC Fanatic!

Aug 7, 2008
4,240
414
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
3
0
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?

7. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,971
1,135
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
4,240
414
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 Moderator

Mar 24, 2008
20,772
2,540
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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