Need help with an LED flasher circuit

I'm trying to make a simple LED flasher but I'm stuck.
It's basically three chains: one chain with 12 red LEDs in parallel (1.8V @ 240mA) and two chains with 6 blue LEDs in parallel (About 3.2V @ 120mA)

I want the chains to light up in a simple flashing pattern.

I started of with the idea of a regular flasher like **broken link removed**.

However, since the draw from the red chain is different from the draw from the blue chains, this probably won't look all too great.

Besides that, I'm a complete n00b. Started out with the basics a few weeks ago.

I've had a few ideas, but honestly no idea how I can make them work...

1: Using a transistor in some way so that the flasher circuit will only need to provide a
base current
2: Using a relay. Maybe a reed switch with a nice inductor next to it.
3: Using some kind of IC to get this done.
4: Increasing the value of the capacitor for the red chain. (I've tested my NPN transistors
at double their rating and they didn't get very warm. They only started to get real
warm when I set my variable supply to roughly 5V @ 1A. I thought I might be able to
fix this heat problem with a simple heatsink (I have plenty of small heatsinks).
When I tried to add larger value capacitors I had some issues with the circuit itself
in my simulation program.)

I don't really mind spending a bit of money: this is supposed to be a one-time project.
The relay method seems somewhat expensive though... Maybe a bit over the top...

I don't really like idea #3 though... I would prefer a solution without ICs if possible.

Can anyone help me out here please?
Or maybe give me a good IC to work with if there's no other way? I've been looking for a few decade counters but I got lost in the options and functions.

Any help is highly appreciated!

Hi and welcome to the ETO forums.

Where did you get the currents from for the LEDs? Do you have a link to the data sheets for the LEDs?

Based on what I think you want to do, using the circuit you posted it should be doable depending on the LED current. If you just want to sequentially turn the LED strings On / Off. There are several ways to go about it. However, the actual LED forward voltage and current must be known on an individual LED basis.

Ron

Thank you.

Well... I got them all on ebay... The red ones where in a package of 5 sets of different colored LEDs, containing 20 each.
I can't really find any info on them on the seller's page, so I tried them out at the voltage and current I found elsewhere: they worked well enough. I was trying them at roughly 1.8V

The blue are also from ebay, but the seller included a table:
Product Number
LE-DS152
Product Name
100 pcs 5mm Blue LED 6000mcd & Free Resistor
Emitted Color
Blue
Size (mm)
5mm
Lens Color
Water Clear
Peak Wave Length (nm)
460 -470
Static Sense
Yes
Forward Voltage (V)
3.2 - 3.4
View Angle
5 - 25°
Luminous Intensity (mcd)
5000 - 6000
Maximum Current
20mA Continuous, 50mA peak for 10% Pulse Width
Item Net Weight
75g / 2.7oz

I found that most LEDs are said to be rated at 20mA and it seemed fine so I just multiplied 20mA by the amount of LEDs in the chain.

Do you mean that I should measure the forward voltage and forward current of every single LED?
It's not a very precise project though. Nothing special, just a little gift for someone.

But yes, I would like the LED chains to flash in a sequential pattern if that's not too complicated.

Thanks again.

Try one LED in each section. Get this to work first.

Add a 22u between base and emitter of the third transistor to make the circuit reliable.

It works with one LED in each section like you suggested.

Thanks for the capacitor suggestion!

Now add one more LED at a time in series or parallel

I put two LEDs in parallel in the middle position to double the current required, which is basically (and roughly) what's happening in the circuit I'm trying to make.

Changing the middle cap's value only increases the length of the ON-interval for the middle LEDs.
And making the middle resistor 235Ω makes the middle LEDs go weird. Its intensity increases twice and decreases once (Increase, decrease, increase) in one interval.

However, changing the middle 39K resistor to 19K makes the middle LEDs light up just as intensely as the other LEDs.

Is this is a good solution?

It looks like this now.
**broken link removed**

The Guru said:

I put two LEDs in parallel in the middle position to double the current required, which is basically (and roughly) what's happening in the circuit I'm trying to make.

Changing the middle cap's value only increases the length of the ON-interval for the middle LEDs.
And making the middle resistor 235Ω makes the middle LEDs go weird. Its intensity increases twice and decreases once (Increase, decrease, increase) in one interval.

However, changing the middle 39K resistor to 19K makes the middle LEDs light up just as intensely as the other LEDs.

Is this is a good solution?

Click to expand...

Wait... the interval for the the middle chain is shorter now (obviously).
Sorry my bad.

Any other ideas?

Providing the circuit starts-up, any changes can be made.

colin55 said:

Providing the circuit starts-up, any changes can be made.

Click to expand...

But I don't know what to change.
Leaving it the way it is but adding an extra LED in parallel makes the middle chain draw double its normal current and the middle LEDs grow dim.
Lowering the 39K resistor makes the middle LEDs light up briefer.
Lowering the middle 470Ω resistor to 235Ω makes the middle LEDs oscillate in their interval.

OK, I see you have been busy.

I would not have done what you did in your last drawing with the parallel configuration. The Blue LEDs have a forward voltage of 3.2 to 3.4 volts so lets call it at about 3.3 volts. The forward current is a maximum of 20 mA so lets try them at about 15 mA. Your supply voltage is 6 volts so we get Vsupply 6 volts - Vled 3.3 volts / Iled 15 mA = 180 so you want a 180Ω series resistor per single Blue LED. You don't want the LEDs in parallel as you have them, it really is not a good practice. You want each LED to have its own series resistor. So try placing a pair of blue LEDs each with about a 180Ω series resistor in parallel.

From your original posted currents I assumed you had added the currents and the currents were not per LED but wanted to be sure.

Ron

Reloadron said:

OK, I see you have been busy.

I would not have done what you did in your last drawing with the parallel configuration. The Blue LEDs have a forward voltage of 3.2 to 3.4 volts so lets call it at about 3.3 volts. The forward current is a maximum of 20 mA so lets try them at about 15 mA. Your supply voltage is 6 volts so we get Vsupply 6 volts - Vled 3.3 volts / Iled 15 mA = 180 so you want a 180Ω series resistor per single Blue LED. You don't want the LEDs in parallel as you have them, it really is not a good practice. You want each LED to have its own series resistor. So try placing a pair of blue LEDs each with about a 180Ω series resistor in parallel.

From your original posted currents I assumed you had added the currents and the currents were not per LED but wanted to be sure.

Ron

Click to expand...

I try.

Yes I understand that each LED should have its own series resistor.

My only real problem, though, is that I'm not sure how to let the three LED flasher circuit handle the difference in draw from the chains.

And yes, I added the currents.

Thank you very much for your help though!

Attached is an image of a circuit based on pretty much what you have. For the last stage I added a LED. I used an LED with a forward voltage of about 3.6 volts and limited the current to about 20 mA. The transistors I used are very common 2N2222 NPN transistors. The collector current on Q1 and Q2 is about 20 mA and the collector current of Q3 is about 40 mA (two LEDs @ 20 mA each). The graphs show Q1, Q2 and Q3 collectors. When the collector voltages are low the LEDs are on.

The only trick to all this would be the transistors need to be able to handle the current for their loads. The 2N2222 can handle about 800 mA max and we would be well below that even with 10 LEDs running 20 mA each. I would not try 10 each 20mA LEDs using the BC147. Also I used a 12 volt power source so my series resistors are based on 12 volts.

My only real problem, though, is that I'm not sure how to let the three LED flasher circuit handle the difference in draw from the chains.

Click to expand...

The only difference will be the current for each transistor. The actual LED current being limited by each LED series resistor.

Actually using 12 volts with 3.6 volt (Vfwd) LEDs I could have made up two LEDs in series with a single current limiting resistor for each series of two. The resistors would be 12 volts - 7.2 volts (two LEDs in series) = 4.8 volts / 20 mA = 240Ω. I could repeat that five times in parallel for ten LEDs.

Ron

Thank you very much!

I've been rather dumb (or lazy, it depends on what you want to call it): I've already assembled 2 of my chains, without any series resistors.
I assume that for theoretical functionality, I could put a replacement resistor in series with a chain of n LEDs with the formula 1/Rr=1/R1+1/R2+...+1/Rn

I understand that if one or more LEDs break, the current on each LED will increase and damage might occur.
I don't mind taking the risk though. Even though it's not smart and correct.

I hope I'm not rude here. Thank you very much for your honestly amazing help.

It seems, however, that when I put three LEDs with series resistors in parallel, they light up slower. (Either that or my simulation program is off)
Is there any way to fix this problem?

I really hope I'm not asking too much.

Thanks again.

The Guru said:

Thank you very much!

I've been rather dumb (or lazy, it depends on what you want to call it): I've already assembled 2 of my chains, without any series resistors.
I assume that for theoretical functionality, I could put a replacement resistor in series with a chain of n LEDs with the formula 1/Rr=1/R1+1/R2+...+1/Rn

I understand that if one or more LEDs break, the current on each LED will increase and damage might occur.
I don't mind taking the risk though. Even though it's not smart and correct.

I hope I'm not rude here. Thank you very much for your honestly amazing help.

It seems, however, that when I put three LEDs with series resistors in parallel, they light up slower. (Either that or my simulation program is off)
Is there any way to fix this problem?

I really hope I'm not asking too much.

Thanks again.

Click to expand...

Screwing things up a little is part of the learning curve. Sometimes we tend to get a little ahead of the curve.

Attached are a few LED examples. Placing LED directly in parallel is a bad idea because while they should be the same in the real world they aren't. My example uses LED with a Vf of 3.6 volts and a If of 20 mA. The problem is that any two LEDs of the same part number and lot of manufacture will not quite be the same. One LED due to manufacturing imperfections will draw slightly more current than the other. When in direct parallel it will starve the opposite LED(s). This is just the nature of the beast. Thus in a parallel LED configuration each LED has a series resistor. There is no way to know exactly how the LEDs will behave and what the current split will be through each LED. While it will look fine in a simulation it will not work in a real world design. The LED drawing more current will continue to do so and eventually become toast. The single series resistor was selected based on an even current split. Eventually the transistor will likely become unhappy.

In your circuit the 39K resistors combined with the 22 uF capacitors form the RC timing network. Those components alone determine the On / Off time for each stage of the ring oscillator. The load on each transistor collector should not have much to do with the LED timing. That assumes the transistors current ratings can handle the loads and the Vsupply is adequate.

It seems, however, that when I put three LEDs with series resistors in parallel, they light up slower. (Either that or my simulation program is off)
Is there any way to fix this problem?

Click to expand...

My best guess here is that since the load has changed the transistors may be turning on slower based upon the base drive of the transistors. If you look at the base signals they are not instantaneous change from Off to On. So the transistors are not driven into saturation like an On / Off switch.


Ron

That's true. Thank you so much for your patience and guidance.

Ah yes, of course! I should have thought of that... It seems that I was getting ahead of myself in my enthusiasm...
There's one thing I don't really understand though. Why will the transistor become unhappy? Because the voltage and thus the current will increase?
I've heard of the V,I curve of LEDs. Will that be the cause of trouble?
Sorry, I'm a bit lost on this one.

Hmm, okay. I guess the extra load steepened the ON/OFF graph. Somewhat similar to the sketch I included.
Is this correct?
I'll give changing the timing resistors a shot.

View attachment 61742

You made a good point that I failed to mention earlier. The LED is a current device, the LED is also a non linear device. For example a typical generic red LED can have a forward voltage of about 2.0 volts and a forward current of about 20 mA. At 1.7 volts the LED would likely glow faintly. The current draw about a few mA. At 2.0 volts the current would be about 20 mA, the normal Vf and If. However, if that voltage increases just a little as we pass 2.1 or 2.2 volts that current will be above 50 mA. A sharp non linear curve as to voltage and current plot. Knowing the Vf and If we can use Ohms Law to calculate the series resistance to limit the LED current but Ohms Law does not work for a LED beyond that. It's not like the LED has a fixed resistance. Since the transistor is handling the LED current it would not take much before we exceed the current ratings of the transistor. Thus I say an unhappy transistor.

Ron

Reloadron said:

You made a good point that I failed to mention earlier. The LED is a current device, the LED is also a non linear device. For example a typical generic red LED can have a forward voltage of about 2.0 volts and a forward current of about 20 mA. At 1.7 volts the LED would likely glow faintly. The current draw about a few mA. At 2.0 volts the current would be about 20 mA, the normal Vf and If. However, if that voltage increases just a little as we pass 2.1 or 2.2 volts that current will be above 50 mA. A sharp non linear curve as to voltage and current plot. Knowing the Vf and If we can use Ohms Law to calculate the series resistance to limit the LED current but Ohms Law does not work for a LED beyond that. It's not like the LED has a fixed resistance. Since the transistor is handling the LED current it would not take much before we exceed the current ratings of the transistor. Thus I say an unhappy transistor.

Ron

Click to expand...

Ah yes, I see. That will give some trouble.

Decreasing the timer resistor values makes the intervals shorter (as you explained). It makes the fact that the LEDs aren't fully turned on right away much less noticeable. Maybe I can get away with this at the right values.

I've tried replacing the 39K resistor with a 30K resistor and now the circuit seems to be working reasonably well.
I've added a GIF file.
View attachment 61744

It's very hard to tell from the animation I just added, but the LEDs increase slightly in brightness before the end of their interval.
I added a closeup. Let's hope the change in brightness is more visible now.
View attachment 61748