/hijack mode on/

The basic idea of the dual oscillator is definately going to make it the 555 projects. It is an excellent way to introduce someone to the concept of hetrodyne, as well as being interesting visually. It could be one of the better ideas I've had. You'll note the diode bridge strongly resembles a balanced modulator.

/hijack mode off/

I think the circuit is a fascinating exploration into a number of interesting concepts. All well worth the cost in parts and time for the newbie and the oldie. And you get a neat little fading LED module in the bargain. WIN, WIN, WIN!!!!

hgmjr

 

ok i just set everything up like in the single 555 that bill posted yesterday. Now i have nothing it not even coming on I have checked and double checked my connections. I used PnP for the transistors. and 10k pot for R1 and R2.

 

OK, this one is guarenteed to work. So now we're in troubleshooting mode. Assuming you are using a 555, what voltage do you have on pin 2 and 6? Whatever it is (it will be either +9V or gnd) the output, pin 3 should be opposite. If the 5555 is dead, or there is a wiring error killing the 555, pin 2 and 6 (which should be the same points) will be the same voltage as pin 3.

Things to look for, is pot R1 set at max value (the resistance without power between pin 3 and 2 is the greatest)? Is both of the capacitor's polarity as shown? Is the R2 pot also at max value (actually, this can be set midway)?

What transistor part numbers did you wind up using? Did your 555 get hot at any time?

 

I just seen your post after doing some troubleshooting myself
I used the 2n2907 PnP. What im getting is 8v and some change to the base side of Q1. The 1.8v at the emitter side if Q1. then 1.8v at base of Q2 then no voltage at emitter of Q2.

 

checked pins 2 and 6 they are reading .08. Pin 3 is reading 9v

 

Here is another idea:

R1 and R2 can be reduced to 1KΩ if the caps are upped to 2,200µF.

Bill,
I suggest it would be far better to increase R1 & R2 from 10k to 100k and reduce the caps from 220uF to 22uF. This will reduce power consumption. Besides, 2,200uF caps are a good bit larger than 22uF caps, while the physical size and cost of resistors & pots doesn't change much with their value.

 

once the rate is set for the fade could R1 and R2 be swapped with the appropriate resistor to keep the rate the same?

 

checked pins 2 and 6 they are reading .08. Pin 3 is reading 9v

If this is the case, then R1 is open, or C1 may be shorted. This (R1 open) could easily happen if you accidentally set it to too low of a value. The excess power dissipation in the resistor would cause it to overheat and burn out. If C1 is an electrolytic capacitor and you installed it incorrectly (polarity-wise), that would be the reason it's shorted.

Pots are generally rated for around 1/2 Watt, but that's across the whole pot. When you are using it as a rheostat (as in this application) it's much less when you have the resistance turned way down; the small area of resistance can't dissipate heat very fast.

I suggest the minimum value of resistance for R1 should be 300 Ohms; this would give a 20mA current when V(C1) was at 2/3 Vcc (6v) and pin 3 is low. Using a fixed 300 Ohm resistor in series with a pot would prevent the destruction of the pot if it were accidentally set too low.

 

R1 is at 8.5K R2 is at 9.8K. set to the max like bill suggested. Im thinking that the PnP might need to to be swapped for NpN. the bias of the emitter could be the cause.

 

You're on this schematic, right?

 

thats what we moved onto

 

OK.
If R1 is connected from pin 3 to pin 2 and 6, and C1 is also connected to pin 2 and 6, and the negative side to C1 is connected to ground...

Then except for the first few moments of being powered up, C3 should always have 3v to 6v across it.

Are you sure pin 7 isn't connected to pin 6? That would cause a problem.

Remove C1, and short it's leads together for several seconds. Use your voltmeter to check that it's discharged. Then see if you can measure continuity through C1. It may be defective (shorted).

 

Measure pin 5. If your Vcc is 9v, it should read about 6v.

Try replacing C1. As I mentioned before, you should see 3v to 6v on C1.

The transistors must be PNP. If you try to replace them with NPNs, it will not work.

 

I have 5.7v at pin 5, each cap is showing 3v.

 

Are you absolutely certain that you have your LED installed in the correct polarity?

hgmjr

 

OK, something's changed. Before you were showing 0.8v on C1, and now you're seeing 3v?

When C1 measures 3v, what are you reading on pin 3?

 

I have 9v at pin 3. but now i have .15 at c1 and .4 at c2

 

Do you have a spare 555 timer? You may have blown that one.

 

Just to verify - you ARE taking all of these measurements with the negative (black) meter lead connected to ground/battery negative terminal, right?

 

no i only have this one. i have 9v at pin 3 0v at pins 2&6, 6v at pin 5.