You're not trying to block current flow with these diodes. You want the current to flow through them to make them light. They are definitely backwards. They should be "pointing" towards ground.

 

Yes, LEDs do need to be forward-biased to make them light up. The abbreviation "Vf" for the LED voltage drop means forward voltage. The arrowhead in a diode symbol points the way for conventional current (not electrons) in the forward direction.

Reading the forum article on special diodes, and also Bill Marsden's articles on the subject might help you to get an overview on how LEDs can be used.

http://www.allaboutcircuits.com/vol_3/chpt_3/12.html
http://forum.allaboutcircuits.com/showthread.php?t=19075

 

Some would say that blowing IC, transistors and capacitors is part of the learning process.

I have always done things correctly so I have never blown up an IC, transistor or capacitor and not even an LED.
So maybe I learned nuttin.

 

I have always done things correctly so I have never blown up an IC, transistor or capacitor and not even an LED.
So maybe I learned nuttin.

I have yet to blow LEDs on accident.

 

I have always done things correctly so I have never blown up an IC, transistor or capacitor and not even an LED.
So maybe I learned nuttin.

That's not what he said. He said that blowing components is part of the learning process. If you fry LEDs, it helps you learn. That doesn't mean that if you don't fry them, you don't learn. That's like saying "Only the good die young", and then looking at someone who died when they were old and saying "they must not have been good". That's not what the saying means.

 

I have always done things correctly so I have never blown up an IC, transistor or capacitor and not even an LED.
So maybe I learned nuttin.

I was just referring to this:
http://www.makershed.com/product_p/9780596153748.htm


Page x in the preface.

"Along the way, you will make some mistakes. This is good. Mistakes are the
best of all learning processes. I want you to burn things out and mess things
up, because this is how you learn the limits of components and materials.
Since we’ll be using low voltages, there’ll be no chance of electrocution, and
so long as you limit the flow of current in the ways I’ll describe, there will be no
risk of burning your fingers or starting fires."

But this is a little off topic...

 

I read the datasheets of components that say their limits and I do not exceed those limits. I avoid mistakes because that is learning the hard way.

 

Sheeeesh.....! That circuit would not light a LED...but it will blow the semi's

Put a 1K resistor between base and the 555, pin 3. This will limit the base current to a safe value.

Next Ground the emitter. Put the LED's in the collector.
LED negative connects to collector.
For a 12V Vcc to light a LED, R should be around 680Ω to 820Ω or else you'll blow the transistor if it's power is low and these values are for normal leds like red or green.

If you wanna use high bright LED's like white or Blue, you should reduce the Resistance value to like 220Ω. Any lower than this will increase LED current and limit it's usage life.

 

@ R!f@@: The OP is trying to use powerful LEDs which are rated for a maximum current of 140mA, so the resistors for a single LED at full power from a 12V supply would be more like 60 ohms. http://uk.mouser.com/Search/Product...irtualkey62510000virtualkey720-LDCN5M1R1S351Z

The LEDs would of course require proper heat sinking if they are going to be run continuously at this level without burning out.

Clearly the emitter follower set-up is no good, and the grounded-emitter arrangement will be better, as the OP has already been advised. However, since he wants three 140mA LEDs (or more) in parallel, a 1k base resistor will not give enough current to use an ordinary transistor in this way.

Of course, he could put two LEDs in series to cut down on the current a bit, and I have already suggested this. (The maximum forward drop is given as 3.8V, so three in series would not be advisable. )

 

hmmm. so it seems... I agree with you Adjuster

Did we scare OP away?

 

hmmm. so it seems... I agree with you Adjuster

Did we scare OP away?

Excuse me? I'm still here. Some how I've managed to blow a second 555. I looked over my circuit three times before I used any power.. I really have a bad grip on this all.

 

Excuse me? I'm still here. Some how I've managed to blow a second 555. I looked over my circuit three times before I used any power.. I really have a bad grip on this all.

Post a clear, close photo or two of your breadboard; someone may be able to spot the problem.

ETA: If your breadboard is built like the schematic in post 5, the 10k resistor you have labeled as R1 is in the wrong place. It should go from the switch to +12, not from the switch to pin 2.

 

Post a clear, close photo or two of your breadboard; someone may be able to spot the problem.

ETA: If your breadboard is built like the schematic in post 5, the 10k resistor you have labeled as R1 is in the wrong place. It should go from the switch to +12, not from the switch to pin 2.

There is already a 10k Ohm resistor going from the switch to the positive.. That 10k resistor that is label as R1 is to discharge the capacitor to prevent the circuit from firing for longer than the determined pulse.

I know what I did wrong with my circuit now, I just need to go get another chip.

 

I know what I did wrong with my circuit now, I just need to go get another chip.

Care to share?

 

Care to share?

I put my positive as the ground and my negative as the source.

 

Oh... That explains a bit. Would be nice to see the complete circuit/pcb when it is finished.

Good luck.

 

Oh... That explains a bit. Would be nice to see the complete circuit/pcb when it is finished.

Good luck.

I've still got a ways to go yet. I may make a PCB for the base circuit when I get that part finished. However, I still need to work on making the trigger work for the timer circuit. Then I have to solve my LED problem.

 

There is already a 10k Ohm resistor going from the switch to the positive.. That 10k resistor that is label as R1 is to discharge the capacitor to prevent the circuit from firing for longer than the determined pulse.

I know what I did wrong with my circuit now, I just need to go get another chip.

The 10k resistor you have labeled as R6 is a pull-up resistor to hold the trigger input high until it receives a negative going pulse from the switch. The combination of the resistor you have labeled as R1 and C1 is a conditioning circuit, and R1 is supposed to go to +12. Here is Bill Marsden's post that contains a schematic to that effect. http://forum.allaboutcircuits.com/showthread.php?t=14018

My post number 30 in this thread also shows it connected correctly.
http://forum.allaboutcircuits.com/showthread.php?t=63629&page=3

The way you have it connected effectively doubles the value of the resistor because R1 is connected to R6 which is connected to +12. Thus, it may work, but it should be connected directly to +12.

 

I put my positive as the ground and my negative as the source.

Is this U'r new Year Resolution... .

Advice Ur Neighbors to start packing
A rocket Engineer is living in the Premises.

 

Is this U'r new Year Resolution... .

Advice Uri Neighbors to start packing
A rocket Engineer is living in the Premises.

It's called people need to use one system. One of the schematics I looked at for my 555 used electron flow and didn't specify that while the other used conventional.