# Simple LED flasher with 2n3904

Discussion in 'The Projects Forum' started by schizo18, Oct 21, 2009.

1. ### schizo18 Thread Starter New Member

Oct 21, 2009
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I'm trying to build a simple alternating LED flasher:

I'm using LED pods that have 24 LED's each on them. I can connect them straight to battery power and have full brightness.

My LED's aren't nearly as bright as they should be with the above circuit. I'm hooking this to automotive power and I even took out the 470Ω current limiting resistors with little to no improvement.

Anyone know what I can do to make my led's brighter?

2. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Could you show a schematic of how your LEDs are connected?

In the diagram you posted above, you see the 470Ω resistors? Those determine the current going to each LED. So Take the voltage going from that point and subtract it from the forward voltage of the LED (you can find that on the LEDs specs). Then, according to the current with that forward voltage drop (typically 20mA), you divide it by that current. But if you have 24 LEDs then that circuit won't be able to power all of them.

3. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,671
898
First off, the base of the 2N3904 is not being driven very hard by the 100K resistors. A single LED may light well, but not 24, even at only 10 mA each.

The 2N3904 is limited to 200 mA collector current. You might consider using 2N2222's (about 800 mA, depending on version) and driving them harder.

If you reduce the resistors, say to 10K, then you will have to increase the capacitors 10-fold.

John

4. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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is there a way to make it power all of them?

5. ### ELECTRONERD Senior Member

May 26, 2009
1,146
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jpanhalt made a good suggestion. You're going to have to get transistors that can drive 24 LEDs. So the idea is to saturate the transistor so it can alternate between each of the LEDs (I'm assuming you'll have twelve LEDs on each side alternating?), and also be able to drive all the LEDs. So if you have 12 LEDs on each side in parallel (we're not doing series), then you'll need two transistors that can drive 240mA each (480mA total). That's quite a bit of current for one transistor.

6. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,671
898
I think our posts may have crossed. My suggestion might power all of them, if they don't draw too much current. As you observed, the 470 ohm resistor will probably have to be changed, depending on how you have the LEDs wired.

John

7. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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I updated with the 2n2222 transistor and it seems to be the same problem. I'm trying to drive a pod of 24 LED's on each side.

The manufacturer rates the led pods at:

Minimum Working Voltage / Working Current / Power Dissipation​
DC-12V / 330mA / 3.96W​
Typical Working Voltage / Working Current / Power Dissipation​
DC-13V / 360mA / 4.68W​
Maximum Working Voltage / Working Current / Power Dissipation​
DC-13.8V / 384mA / 5.31W​

8. ### ELECTRONERD Senior Member

May 26, 2009
1,146
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It helps to actually use the acronym "LED" instead of "led". So just for verification, those ratings are for the LEDs and not the transistors? If they are, they're high power LEDs. You would only be able to drive one LED or if any at all.

9. ### CDRIVE Senior Member

Jul 1, 2008
2,223
99
Have you actually measured the battery voltage while the FF is running? The 9V battery is considered the weakling on the block. You would be better off testing the FF with a 12V power supply or a 12V lantern battery.

10. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,671
898
OK, let me summarize what I am assuming:
1) The LEDs you have are designed to work directly from 12 to 13.8 V. That is, they do not need any additional current-limiting resistor.
2) You are using 12 V, not a 9V battery. With those current drains, a little 9V battery will not last long.
3) With those currents, your 2N2222 will get quite warm. It is within spec, but you will likely need a heatsink on it.

Suggestions:
1) Forget about the 470 ohm resistor
2) Reduce the 100K resistor to 1K; increase the capacitors to 1000 uF. If you don' t have that size, you want the product of the resistor and capacitor value to be constant; that is, 100K X 10uF = 1.0 seconds, 10K X100 uF = 1.0, etc. The unit of the product is in seconds. So, if you have a 470 uF or 330 uF, then a 500 ohm resistor might work. I would still try a 1K resistor with the largest capacitor you have that is 1000 uF or less.

John

11. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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Ok, I have the circuit hooked up with the 2n2222, removed the 470Ω resistor, increased the capacitor to 1000μf, put the 100k resistor at 4.7k.

The LED pods now have full power, but they don't go all the way off and the flash rate is super slow. Added a zener diode pretty much in place of the 470Ω resistor to try and drop feedback and it helped but the LED's still stay on a little.

Thanks for all your help guys.

12. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,671
898
Decrease the capacitors. Toss the zener. In all likelihood, it can't supply anywhere near the 300 mA your spec. showed.

John

13. ### CDRIVE Senior Member

Jul 1, 2008
2,223
99
IMO, you would be far better off just adding FETs or Darlington LED drivers.

14. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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I'm pretty well open to anything that would work. I know very little about creating electronic devices.

15. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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OK it seems I've gotten almost nowhere. Maybe someone can help me start over here. All I'm really trying to do is create a fast alternating flasher. I need it to drive the above mentioned pods though. I've tried an astable multivibrator using a 555 and I couldn't seem to get that to work quite right (see attached schematic.)

I even tried to go simpler with the above mentioned schematic and I finally got the LED's to light full power but now I'm stuck with either super slow alternating flash or they seem to go so fast that they don't even turn off fully. I moved anywhere from 1Ω up to 200kΩ on the resistor and from 0.22μf up to 1000μf. I guess I just can't find that magical number/ratio.

I'm a volunteer firefighter and I'm trying to cheapen the deal a little on emergency warning lights. I found a great deal on those LED pods and when I hook them to a commercially available flasher they work great, but that flasher brings the cost up to nearly the same as if I were to buy ready made lights. I've seen enough circuits out there that I have faith it can be done, but I'm at a loss as to how.

ANY and ALL help is greatly appreciated.

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16. ### Wendy Moderator

Mar 24, 2008
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There is no attached schematic.

17. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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fixed above

and of course keep in mind all of these schematics are actually hooked up to 12V auto supply

18. ### CDRIVE Senior Member

Jul 1, 2008
2,223
99
Even if you use a 555, as shown in your last schematic, you still need driver transistors because you're asking way to much from the output pin of the 555. National Semiconductor specs the output pin of the LM555 as Source and Sink 200mA Max!
Here's an alternative using IRF510 Power FETs, available at Radio Shack. Using FET drivers like this will enable you to use your original circuit values of your original discrete FF circuit, thus maintaining the frequency and making those values nearly independent of your load.

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19. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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I have some TIP42 from radio shack, would that do the job as well?

20. ### schizo18 Thread Starter New Member

Oct 21, 2009
25
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just really looked at the specs online and I guess not as it has:

• V(CE): 1.5
• V(BE): 2.0
. the one I bought however claims 40V apiece