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I am trying to recreate the circuit for a TWO PIN 12 Volt Electronic Flasher unit for LED turn signals. It is the inline flasher "relay" that I need. Where power comes in the B (Battery) pin and out the L (Load) pin.

I have searched the net to find a 2 pin circuit but all I can find is the 3 pin.

I found this guy's Youtube video, where he diagrams one but he does not give all the values and does not show some details. Like what MOSFET and Transistor.

https://youtu.be/mWQCSdkiYXg?t=83

Product he is referencing.

Here is his diagram. I am just guessing at the MOSFET and transistor values. I need help getting this to work.

schematic

simulate this circuit – Schematic created using CircuitLab

Thank you.

Need to add this. The load is not the issue here. Please stop bringing this up. The device obviously works and is demonstrated in the video. The need to add resistance is merely because he does not have it wired in a vehicle, where the wiring will add the resistance needed. That's how they are designed.

The issue, is the DIAGRAM correct or is it not? Will it work? Is it a viable circuit? The guy in the video drew up the diagram but he could have drawn it incorrect. That is what I am here to find out. That is what I am asking.

Pretend that the load is perfect. Assume that there is something wrong in the circuit. Where is it wrong. I have the circuit in my simulator and no matter how much resistance is added the flasher circuit itself does not work. I have used a LED tail light configuration with plenty of load, relays, solenoids, motors and it does not work. So Again, The load is not the issue here. Please stop bringing this up.

To the people who keep trying to edit my question. Stop. Do not remove this since it is to save time and effort. It needs to be here, even tho it shouldn't need to be.

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4 Answers 4

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Potential reason the first circuit isn't working is that the "transistor" may not be what it seems.
Here is a very similar YT video of a 2-pin flasher breakdown. The circuit this guy drew up is very similar to the first circuit above. The transistor in this case is a Programmable Unijunction Transistor (PUT). These work with a programmable input voltage a bit like a more modern comparator chip. The claim is that the PUT transistor is similar to a 2N6027.
Below is this guys rendering of the circuit:
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schematic image screenshot from the linked YouTube video

Schematic source from video: YouTube - bigclivedotcom - "Two wire flasher module teardown with unexpected find"

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    \$\begingroup\$ The difficulty OP is facing is making a solution that works with an LED lamp, which is both lower current and also has a stand-off forward voltage of perhaps 10V or more. \$\endgroup\$ Feb 13 at 19:46
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    \$\begingroup\$ @hacktastical Right on. I think a low resistance in parallel with the load will make it behave much more like a resistance, so there is hope for this design, even with an LED load. \$\endgroup\$ Feb 13 at 23:23
  • \$\begingroup\$ @SimonFitch Yeah, given the values I see (and assuming an appropriate NFET) perhaps 1k Ohm would be in the vicinity. PUJT's are notoriously variable (and I think harder to get these days), though. And it would prefer the 2N6027, less the 2N6028, here, I think. The 2N6027 has a higher minimum valley current. (Though perhaps a 2N6028 could mean a higher parallel resistance, I guess. Not sure right now.) \$\endgroup\$ Feb 13 at 23:33
  • \$\begingroup\$ The Big Clive video shows that this circuit does not work with an LED lamp. It has the same problem as all the other suggestions (555, etc.): not enough 'off' state current. -1 \$\endgroup\$ Feb 16 at 2:36
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In the video, the guy says that he can get the flasher to work with LEDs, but only if he places a resistor across the LED module. He states that it's because the LEDs draw too little current, and while that's a concern, it is not the main problem.

The reason it fails is that the LED modules are several LEDs in series, with a combined forward voltage close to perhaps 10V, and this prevents the capacitors in the circuit charging more than 2V or so. I can demonstrate this in a simulation:

schematic

simulate this circuit – Schematic created using CircuitLab

The first circuit shows the module with the kind of load it was designed to operate with, a standard incandescent lamp. The middle employs an LED-based lamp. On the right is a fix for the problems we are about to see:

enter image description here

These show voltages across the capacitors. The blue trace is with the incandescent lamp. It's perfect, because the load is resistive, and doesn't have the problematic diode behaviour you see for the orange trace, which is the unmodified LED load. As you can see, that capacitor's voltage tends to level out well below the 12V we would like. Eventually it would reach 12V, but it would take a long time.

It is the voltage developed by this capacitor which powers the oscillator circuitry, so if it is insufficient to bias the transistors, the oscillator cannot work. I suspect that the required capacitor voltage is around 3 or 4V, and if the LEDs' combined forward voltage exceeds 9V, that's not going to happen, at least not quickly enough.

The added resistor R3 causes the lamp module to behave more like the incandescent lamp, by permitting charging current to flow, even when the diodes are well below their forward voltage, passing next to no current. This is shown in brown.

This resistor will dissipate quite significant mean power \$P\$:

$$ P = \frac{V^2}{R_3} = \frac{(12V)^2}{220\Omega} = 660mW $$

Use a resistor rated for 1W.

All this is not to say that the circuit works, it's just a reason why it wouldn't. There may be other issues to resolve.

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  • \$\begingroup\$ I am not concerned with the resistance of the LEDs. I am trying to find out if the CIRCUIT itself is viable. The diagram, the guy in video provided, is it correct? Will it work? Is what I am here to find out. Yes the Flasher it self works, and it needs the resistor for those lights. However, the flasher is designed to be used in a vehicle with lengths of wire that add resistance. So it is not an issue. Is his diagram (and recreation of it) a viable circuit? Pretend the LED module has perfect resistance. Will the circuit Work? It Doesn't work in simulator. No matter how much resistance I add. \$\endgroup\$
    – Mouthpear
    Feb 12 at 16:47
  • \$\begingroup\$ 1. What signal / power is coming through the 1uF capacitor? 2. What other issues might there be? \$\endgroup\$ Feb 13 at 23:08
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    \$\begingroup\$ @MicroservicesOnDDD This is just one momentary path from +12V to ground which initially charges C1, until it has enough potential difference to operate the rest of the circuit. Presumably MOSFET M1 is supposed to "take over" by providing an alternative low impedance path following this, but M1's current wiring, as I see it, won't be able to do that correctly. \$\endgroup\$ Feb 13 at 23:13
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    \$\begingroup\$ @MicroservicesOnDDD I spent a couple of hours trying to make this work, but I suspect Q1 is not a BJT, and I gave up. Nedd's answer seems to confirm this. \$\endgroup\$ Feb 13 at 23:19
  • \$\begingroup\$ The circuit, as shown in Nedd's answer (from the Big Clive video), is not viable with an LED. It requires a pull-down load in parallel to make enough current + voltage for the oscillator to work. \$\endgroup\$ Feb 16 at 2:40
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For an incandescent lamp, the approach to a two-terminal flasher is simple: use a low-power flasher circuit that is charged up during the off-time through the lamp. A CMOS 555, a low-power micro, a Silego Greenpak (take a look this one), or a fistfull of discretes will do that.

Question for you: have you considered how your flasher will make the click noise? We'll come back to this.

I built such a thing for my 1959 VW, using a 7555, many years ago. It used a diode + capacitor approach like you show with your discrete solution to hold up the power during lamp 'on' time. It also used a relay which made an audible click as it worked. As I remember I really struggled with this thing because, like you, I was trying to make it work without a ground wire. But I did get it to work.

Flasher for LED lamps? Not so much. Commercial ones always seem to have a third ground terminal. This is because there isn't enough residual overhead voltage to operate the timer circuit, let alone the noisemaker + switch relay. So you need to run a ground wire. Is that such a big deal?

The Silego I mentioned above can work down to 1.65V, so this might be able to work with the residual overhead voltage available with 12V LED lights. But, it still needs something to make the audible click. Perhaps making a pulse to a speaker / transducer would do the trick.

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Given the answer @nedd provided that the transistors is, more to the point a Programmable Unijunction Transistor, was the key to why the circuit was not working. (Had nothing to do with the load).

So I was able to find an alternative/equivalent circuit to use, since my simulation software does not have a Programmable Unijunction Transistor in its library.

Here is what works. I am however giving the answer to @Nedd for understanding the question and giving the reason for it not working, and linking to the video.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Sounds good to me. \$\endgroup\$
    – Nedd
    Feb 14 at 4:39
  • \$\begingroup\$ In the above circuit is the shorted B-E connection on the NPN correct? \$\endgroup\$
    – Nedd
    Feb 14 at 4:43
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    \$\begingroup\$ The way you connect makes Q2 basically useless. To correct it: (1) the Emitter of Q2 should not be connected to its Base. (2) The Gate of M1 should not be connected to the Base of Q2, but it should be connected to the Emitter of Q2. \$\endgroup\$
    – kaosad
    Feb 14 at 17:23
  • \$\begingroup\$ @kaosad Whether they should or shouldn't be, is irrelevant. They circuit works the way it is. The two transistors in that configuration are apparently a common way, for some fields of electronics. Here is where I got it from. eevblog.com/forum/repair/… Take a look and see if you still have issue with it. Again it is working as i have diagramed it. \$\endgroup\$
    – Mouthpear
    Feb 17 at 6:14
  • \$\begingroup\$ @kaosad Ok so to test all theories out. I removed Q2 completely in simulation and the the circuit still works. However I remove Q2 from the breadboard and it completely stops working IRL. Maybe you can do the diagram and submit as answer? \$\endgroup\$
    – Mouthpear
    Feb 17 at 6:32

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