Modifications to 555 LED Fade Circuit

Question

I was looking for a circuit to fade an LED on and off using no microcontrollers, and found this nice circuit:

http://www.555-timer-circuits.com/up-down-fading-led.html

it seems like exactly what I was looking for, so I went ahead and used it with a few changes:

• I used a 2n2222 as the npn transistor instead of the BC547
• I used a 12V source instead of 9 volts because I'm using LEDs with from a 12V led strip (led + resistors all built into the strip).
• I used a 556 instead of a 555 because I had them on hand (and

I built the circuit as above (with the pinout changed accordingly for the 556), and it didn't work. I was a bit surprised, and double-checked all of the connections -- but they were all correct.

Would any of the changes altered how the circuit functions?

• I'm a bit concerned that changing the source to 12V might have effects, since that changes the threshold and trigger voltages -- could that be why the circuit isn't working?
• Should the leds and resistors be attached to the collector instead of the emitter of the 2n2222? I remember reading that LED driver circuits should be designed this way (something about V_be, IIRC).
• Any other ways that changing the circuit may have prevented it from functioning?

thanks!

Answer 1

A simple analysis of the original circuit and the substitutions you made goes something like this.

The 555 circuit shown is a squarewave (1:1) oscillator. The frequency being determined by R1 and C1. The voltage at the junction between these two components changes between 2/3 rds and 1/3rd of the supply (set by input pins 6 and 2). For a 9V supply this will be 6V & 3V. Changing the supply to 12V gives 8V and 4V. The voltage varies in an approximately triangular waveform.

This voltage is connected to the the base of an NPN transistor, Q1 (in this case a BC478.) Just about any small signal NPN would do so a change to a 2N2222 type would have very little effect on the operation. The transistor is connected as an emitter follower so the voltage at the emitter would be 0.6V lower than its base voltage. For a 9V supply this voltage would be between 5.4 and 2.4V. For a 12V supply this would be between 7.4V and 3.4V.

The 556 is just a double 555 in one package so using one half of the chip should still work (provided you have the correct connections).

Your last substitution is where it all goes wrong - using a 12V LED (with built in resistors.)

A normal (visible) LED needs between 2 - 4 Volts to turn on. This voltage depends upon colour (i.e. the energy gap). R3 is chosen to limit the maximum current through the LED (and transistor). The 9V circuit matches the 5.4V - 2.4V range very nicely for something like a RED LED (about 2V). For the 12V supply you would be better off using a COOL WHITE or BLUE LED (about 3.8V). The R3 value of 470R is fine. For a 12V circuit and WHITE LED (3.8V turn on) this gives a maximum current of (7.4 - 3.8)/470 A or about 7.7mA. Compared with (5.4 - 2)/470 A or about 7.2mA (RED LED, 2V turn on).

Answer 2

With 12 volts on the 555, the junction of R1 and C1 (U1 pins 2 and 6) will be oscillating between 8 volts and 4 volts.

Q1 is being used as an emitter follower, so the voltage on its emitter will be about 7.3 volts when U1-2 is at 8 volts, and about 3.3 volts when U1-2 is at 4 volts. there's also the voltage drop across R3 to contend with, but since we don't know the value of the ballast in your string - or the current through it - it's impossible to determine the effect of R3 on your string other than to note that it'll cause the string to be dimmer than if it (R3) wasn't there.

But, just for grins, if we assume that - when U1-2 goes high and Q1 emitter goes to 7.3 volts - your string draws 10mA when it's connected between R3 and ground, that means that R3 will drop:

\$ E = IR = 0.01A\times 470 \Omega = 4.7V \$

and the voltage available to your string will only be: \$ \ 7.3V - 4.7V = 2.7V\$

If your string expects to see 12 volts on its input for full brightness, but it sees 2.7 volts instead, that should be a pretty good indication of why it's not working.

If you could measure the voltage on the diode end of R3, loaded and unloaded, that would be a great help in solving the mystery. :-)

UPDATE

According to the link you provided and to the results of a search for the somewhat elusive 3528's specs, your array looks like this:

Note that the array comprises ten paralleled strings of a 120 ohm resistor and 3 LEDs wired in series.

That means that each series string will draw 20 milliamperes from the 12 volt supply, but all ten in parallel will draw 200 milliamperes.

So, the next step is to design a triangle-wave generator which can push 200 milliamperes through what looks like a 60 ohm load and then slowly(?) reduce that current until the LED dimness looks OK, then raise it again, slowly, to 200 milliamperes, and then repeat the cycle endlessly.

Answer 3

As he notes in the original description - Vmax for the LEDs is 2/3 x Vin - Vbe_Q1.
In his case this is ~= 9v x 2/3 - 0.5V = 5.5V

In your case with 0-12 V in you'd get a maximum LED voltage of about
2/3 x 12V - 0.5V = 7.5V.
12V LED strips have (almost always) 3 x LEDs in series.
That allows 2.5 V / LED at zero current

You do not say but, if these are white LEDs the 2.5V will be inadequate to make them more than glimmer faintly, at best.

IF you have access to the strip components you can test this by shorting out one of the three series LEDs in one module only (each strip is made up of many modules).

You could overcome this problem by using 12V for the Q1 supply and a higher variable dimming control voltage - about 19V is required to get full LED output as
2/3 a 19 - 0.5 ~= 12V

An LED strip will dim if you supply it with about 9V - 12V variable directly.

One method that may work in this case is

• Place LED strip in Q1 collector.
• Drive Q1 with a resistor divider from the 555
• Change R3 (still in Q1 emitter) to about R = V_555_max / I_strip_max

Starting with R3 is say 10 Ohms will show you if this works for you.

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There are better circuits available.
If you explained what you had available and what you wanted to do in good detail, rather than what you think may do something semi-unknown that you want, then we can probably help you.