Need help with an LED flasher circuit

[The Guru]

Sorry, I had compression on.

Here's a closeup of only the LEDs.
View attachment 61750

 

[Reloadron]

Nice little animation.

That is normal. The circuit you are using is a pretty common design. Here is an example using MOSFETS for the transistors. We are not turning the transistors On and Off as in instantly. If you were to look at the base leads using a scope you would see a sawtooth type waveform. So unlike banging the bases with a square wave we see a different result. Nice effect and actually normal and what would be expected of the circuit.

Ron

 

[The Guru]

Nice little animation.

That is normal. The circuit you are using is a pretty common design. Here is an example using MOSFETS for the transistors. We are not turning the transistors On and Off as in instantly. If you were to look at the base leads using a scope you would see a sawtooth type waveform. So unlike banging the bases with a square wave we see a different result. Nice effect and actually normal and what would be expected of the circuit.

Ron

Thank you.

That honestly looks a lot better! I like it.
I'll just order a few MOSFETs and make this circuit.

I'm going to experiment with it for a bit now.

Thank you so much for you all your help! (I keep saying this, haha)
I appreciate it extremely much.

 

[Reloadron]

No problem, anyone here in this forum is here to help. Incidentally the link I left with the video has a web page also out there where the author shows the actual circuit. Have a good day and let us know how things go .....

Ron

 

[The Guru]

Dang, it works absolutely perfectly! Even with a chain consisting of three LEDs in parallel replacing the single LED in the middle!
Brilliant!

I made the circuit and included an animation.

Thank you sir!

Yes I noticed the link in the description. I already copied the circuit.

Have a good day.

View attachment 61757

 

[The Guru]

Okay, I finally received my parts yesterday. So I decided to fiddle together the circuit.
However, it doesn't work. I've checked the connections 5 times, if not more, and haven't been able to detect a single error.

The circuit sticks with the third LED array and hasn't made a single other chain light up. It seems to go from bright to less bright but no other chain is triggered.

I tried to make it function by applying a voltage to the other transistors' gates. This made the connected chain light up, but eventually, the last remaining LED chain lit was the blue chain.

I've tried draining all the capacitors by shorting them, no result.

I figured that, since I've designed it to work with 4.5V, making it again on a breadboard designed for 9V (as used in the example) might be worth a shot. But nope, nothing. This time all the LEDs lit up dimly and didn't seen to be willing to oscillate.

Out of pure frustration and helplessness I've even driven up the voltage on the first chain to above 10V. (The only low power parts were the LEDs, but they all had resistor connected in series)
This resulted in the third chain's resistor smoking and turning brown. It appears that the resistance has skyrocketed. Not a big surprise but hey, I was desperate.

Does anyone perhaps have an idea of what might be going on? I'm totally lost.
I can't say I'm any bit of an expert, but wiring mistakes are out of the question, sorry to be blunt.
As mentioned, I traced the wiring back more than five times. Also, I checked for continuity on everything that is and isn't supposed to be connected: no results.

I've tested the LED arrays and they are functioning just fine. The problem must be in the oscillator circuitry.

I'll post the schematic here to eliminate any possible confusion as to which schematic I'm referring to.
View attachment 62500

Possibly worth noting is that I used electrolytic capacitors. It's running fine in my simulation program (which you can see in the animated image I've posted before), but as I mentioned in a previous post, my simulation program could be off.
It's quite old. It tells me that it is copyrighted from 1993-2000.

My apologies for the long read, but I want to provide as many details as possibly necessary.
Any help is highly appreciated!

 

[colin55]

The first circuit you posted is nothing like the final circuit you used.

You should read how the first circuit works to see why a re-arrangment will not work.

The LEDs in the first circuit produce a chasing pattern similar the running LEDs display in video shops.
In fact the effect is called: "Running Hole." All transistors will try to come on at the same time when the power is applied, but some will be faster due to their internal characteristics and some will get a different turn-on current due to the exact value of the 22u electrolytics. The last 22u will delay the voltage-rise to the base of the first transistor and make the circuit start reliably.
The 22u's between each of the transistors have a secondary purpose. They keep the next transistor OFF when the previous is ON. They do this by providing a negative voltage on the base.
This feature is not possible in the second circuit you used. That is why it will not work.

 

[The Guru]

The first circuit you posted is nothing like the final circuit you used.

You should read how the first circuit works to see why a re-arrangment will not work.

The LEDs in the first circuit produce a chasing pattern similar the running LEDs display in video shops.
In fact the effect is called: "Running Hole." All transistors will try to come on at the same time when the power is applied, but some will be faster due to their internal characteristics and some will get a different turn-on current due to the exact value of the 22u electrolytics. The last 22u will delay the voltage-rise to the base of the first transistor and make the circuit start reliably.
The 22u's between each of the transistors have a secondary purpose. They keep the next transistor OFF when the previous is ON. They do this by providing a negative voltage on the base.
This feature is not possible in the second circuit you used. That is why it will not work.

I know. Reloadron showed me the second circuit. The first circuit did not work since I'm using very large LED chains with the LEDs wired in parallel. For this reason I changed to the second circuit.

I didn't make a rearrangement: I used another schematic. You can see it functioning in this video Reloadron kindly linked me to.

Are you sure this circuit won't work? It is shown in action in the video and it works in my simulation program.

Do you have any ideas on how I can fix my circuit?

Thank you for your feedback!

 

[colin55]

Ok The second circuit works in a different way.

The circuit relies on discharging the 2u2 via the 1M resistor and the voltage characteristic of the FET gate.
You will have to see how long it takes to charge the 2u2 by taking the 1M to positive (via a LED) then take it directly to 0v to see how long it takes to turn the LED off. Do this as a separate, individual, section.

 

[Reloadron]

To avoid any confusion the attached is a simulation of the circuit I linked to. When Vout is Low the LED is ON. The last LEDs are a pair of two in parallel. The circuit is a 3 transistor ring oscillator. This circuit gives the effect seen in the linked video.

Ron

<EDIT> Hang on let me research this. </EDIT>

 

[colin55]

What FET's are you using?
If you are using 2N7000 the 1M resistors will need to be 47k and the 2u2 can be 10u or 22u

 

[janely]

nice! thanks for your share!

 

[The Guru]

What FET's are you using?
If you are using 2N7000 the 1M resistors will need to be 47k and the 2u2 can be 10u or 22u

I'm using IRFS630B MOSFETs. The datasheet can be found here.
I needed the MOSFETS to be able to safely handle the relatively high power demand, which is roughly 240mA for the most demanding LED chain.

The MOSFETs with TO-92 casings I found had current limits in the 200mA.
I had decided on using a heatsink at first, but later settled on these transistors to eliminate any current problems.
Besides this, I prefer the look of the TO-220 casing over that of the TO-90 casing.
This choice tripled the price, but that was fine as this is a one-time project.

 

[colin55]

Have you done this:

The circuit relies on discharging the 2u2 via the 1M resistor and the voltage characteristic of the FET gate.
You will have to see how long it takes to charge the 2u2 by taking the 1M to positive (via a LED) then take it directly to 0v to see how long it takes to turn the LED off. Do this as a separate, individual, section. Reduce the 1M to 47k if the circuit does not work.

 

[The Guru]

Have you done this:

The circuit relies on discharging the 2u2 via the 1M resistor and the voltage characteristic of the FET gate.
You will have to see how long it takes to charge the 2u2 by taking the 1M to positive (via a LED) then take it directly to 0v to see how long it takes to turn the LED off. Do this as a separate, individual, section. Reduce the 1M to 47k if the circuit does not work.

I'm sorry , I'm a bit lost. Could you please explain to me what I should do?

I'll try to replace the 1M resistors with 47K resistors first. Thank you for the input.

 

[colin55]

The circuit works with all sorts of combination of voltages, resistances and capacitances.
View attachment 62568

 

[Diver300]

You should put a resistor in parallel with the LEDs. I suggest 10 kΩ or so.

At really low currents, the voltage across an LED will be nearly its forward voltage (Vf). So an LED rated at 2 V might be 1.6 V at less than 1 μA. Just for fun, at no current an LED with a Vf or 2 V will have a voltage across it of around 1.6 V in bright light.

If you are running at 5 V, and loosing 1.6 V across your LEDs, you do not have the 4 V needed to turn on the IRF630B.

The 10 kΩ resistor will only have 10 mV across it at 1 μA, and it will absorb any current generated by the LED in bright light. That means you will get the full 5 V to turn on the IRF630B. When the LED is powered up, there will only be about 0.2 mA though the 10 kΩ resistor, so you won't loose much power.

 

[The Guru]

You should put a resistor in parallel with the LEDs. I suggest 10 kΩ or so.

At really low currents, the voltage across an LED will be nearly its forward voltage (Vf). So an LED rated at 2 V might be 1.6 V at less than 1 μA. Just for fun, at no current an LED with a Vf or 2 V will have a voltage across it of around 1.6 V in bright light.

If you are running at 5 V, and loosing 1.6 V across your LEDs, you do not have the 4 V needed to turn on the IRF630B.

The 10 kΩ resistor will only have 10 mV across it at 1 μA, and it will absorb any current generated by the LED in bright light. That means you will get the full 5 V to turn on the IRF630B. When the LED is powered up, there will only be about 0.2 mA though the 10 kΩ resistor, so you won't loose much power.

Hmm. I've tried this and it does seem to have a positive effect! I also replaced the fried resistor.
I have been trying out a few things to watch the response.

I have emptied the capacitors by shorting their leads. After this I connected the circuit to my bench power supply I'm running it at roughly 5V. It is designed for use with 4 AA batteries, at 4.5V effectively.

After I connected the circuit to my power supply, nothing happened for a while. After a second or two, the third, blue, LED array lit up rather quickly. It stayed lighting up the blue array alone for minutes.

Deciding minutes was long enough to conclude the circuit wasn't working properly, I connected an extra lead to my 5V in and connected it to the second array's transistor's gate.
This made the middle LED array light up. After roughly a second, the third array grew dim very quickly. Then, the first LED array lit up. After a few seconds, the middle array started to grow dim.
Another few seconds later, the first array started to fade rather quickly and the third array started to grow in light intensity reasonably slowly. It stayed stable in this position for minutes yet again. It seems that the fault must be here. (Cheapskate capacitors?)

I decided to now connect my extra 5V lead to the gate of the first array's transistor's gate. This caused the first LED array to grow in light intensity quickly, and after reaching its peak it started growing dimmer, until it was off again. This happened VERY rapidly. A seconds at most.
The state of the third array did not change at all.

To clarify, with first, second and third array, I mean the order of the LED arrays from left to right.
The first chain consists of 6 blue LEDs wired in parallel, rated for 3.2V at 20mA which gives us a total current draw of 160mA.
The second array consists of 12 red LEDs, also wired in parallel, rated at 1.8V at 20mA, giving us a total current draw of 240mA.
The third array consists of 8 blue LEDs, same ratings as the first array, and draws 200mA in total.

Does anyone have a clue what is going on? I will run over the wiring again.

EDIT: I connected the second array's transistor's gate directly to 5V again. The situation that occured wasn't any different from my previous experience. But this time I noticed that the first array did not grow dim until the second array was turned off completely.

On an added note, the second array seems to never stay at its maximum light intensity, but rather starts to grow dim very slowly.

 

[The Guru]

I now soldered the 10KΩ resistors BEHIND the LED chain resistors. I repeated my previous experiment and this time the second array was still lit up (albeit at a barely visible level) when the first array grew dimmer and the third array started to light up.

I did this right after my previous post. Results: the third array is still lit, turning my walls blue.
It seems to keep staying stable in this position and besides this, no apparant "interaction" with the second array seems to be present.

I could never have imagined this simple-seeming project to turn out to be such a nightmare...

 

[KJ6EAD]

As mentioned by Colin in post #34, since the timing is controlled by the charge/discharge rate of the capacitors and the gate characteristics of the FETs (all the same) but is affected by the load (unique) via the interstage resistors (all the same) I think you need to balance the load currents on all stages by adding a resistor from drain to Vcc on the two lower current stages so all stages draw 240mA. Alternatively, you may be able to achieve even switching times by fine tuning the interstage resistances individually.

I'm also concerned that you may not have enough gate voltage to turn the FETs fully on. This is where I'd want an oscilloscope to observe the gate voltage versus time. Can you run the circuit on 6V?