Hi,

I'm using this schematic

It's supposed to fade in and out on a timed interval, but in my case, it's simply fading on and then staying on permanently. Is this schematic incorrect?

the only difference between my components and the components in this schematic are, i'm using a pn 3565 transistor and the 555 timer I've got is a TLC555.

 

The output capabilities of the 555 & C555 are quite different, but a 33k load should be no problem for either, so maybe you have a leaky cap not allowing cap v to rise to 2/3 of 9V, so reset level is never reached.

 

Part of the problem may be the gain of the transistor. We've done something extremely similar here.

You must use two separate transistors, and not one darlington. This is because the darlingtons have a built in resistor that we can't use between base emitter.

A TLC should work, but the 1/3 and 2/3 points aren't quite as precision.

The gain of the transistor controls the loading in a key point of the oscillator. The difference between one transistor and two is a gain of 150 (one transistor) is about 18KΩ loading and a gain of 22500 (two transistors) is about 2.7MΩ loading.

 

The TLC555 just can't source enough current from pin 3. It can only source 10mA when Vcc=15v. However, it can sink quite a bit more current than it can source.

If you flipped the circuit around, using a PNP voltage follower to control the ground side of the LED, that would work.

 

Actually, the two transistors will address that problem nicely, I think. I was still adding to my original post while you were typing.

 

I agree that a discrete (2 transistor) Darlington is needed. Two PNPs or 2 NPNs should work equally well (with the LED installed in the right direction). You could also use a Sziklai pair, but the low-to-high illumination ratio might be less, due to having only one Vbe drop instead of two with the Darlington.

 

Actually the numbers work surprisingly well for a 9V battery. 2.5V for the LED, 1.2 V for the BE drop, and 3V swing for the LED.

At max voltage (which will turn off the LED on my schematic) the transistor is getting 6V, plus 1.2V (BE), which will leave 1.8V for the LED (which is off). One the other end the transistor base is 3V, add 1.2V BE, which leaves the resistor/LED dropping 5.8V on the LED circuit.

This creates a smooth transition going all the way dark and back again. I've built this one.

I think it is interesting the circuit is symmetrical, NPN and PNP work equally well either way.

 

Hey all,
thanks for all the info.

Here's how I worked it out:

I avoided the 555 all together and went with this similar but simpler circuit.

PS, someone on here showed me the gEDA site, where I could get all sorts of programs for drawing, simulating and PCB designing circuits. AWESOME.