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I bought a cheap blinker that claims to be good for LED bulbs (so the frequency is independent of the load.)

It is true, but it has a flaw: the lamps go ON only after about a half second from when I turn the switch, which I don't like at all. I want my turn signal going immediately ON and then blink.

I opened the blinker, and this is what the schematic looks like:

Falstad circuit simulator

enter image description here

I used Falstad to try and understand the circuit and get a solution myself before asking - without result.

How could I modify the circuit so the lamp starts ON?

Please note that I have no other wiring other than what is represented: I have no a "pure" ground wire, just the + and the wire that goes to the handlebar switch.

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    \$\begingroup\$ 1. Returning C2 to Vss rather than V+ MAY help somewhat. 2. PLacing a resistor between the two inverters and a cap from inverter 2 input to V+ MAY help somewhat. This gets the first inverter driven on initially. || HOWEVER,, the issue is that the circuit starts off "dead" and MUST charge a "local" supply before it can turn the FET on. IF you can bring in "true ground" then it can be done "with ease". \$\endgroup\$ – Russell McMahon May 20 at 14:30
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The main issue is that the circuit starts off "dead" and MUST charge a "local" supply before it can turn the FET on. IF you can bring in "true ground" then it can be done "with ease".

Try reducing C1 to as low a value as will still allow proper operation. Startup time is in part related to C1 charge time. Also implementing suggestion 2. below will minimise the oscillator time to input low, which is needed to turn the FET on.

Suggestions:

I am not going to try to 'work these out' in detail without knowing if (literally) chassis ground can be provided or if an NC relay can be fitted in, but the following 'lines of attack' may be profitable:

  1. If you can add a relay with a NC (normally closed) contact it could be used to assist startup and then be disabled by a driver. The IC numbering suggests that you have an hex Schmitt inverter package available. If the other gates are available they could provide the required logic.

  2. Returning C2 to Vss rather than V+ MAY help somewhat.

3 Placing a resistor between the two inverters and a cap from inverter 2 input to V+ MAY help somewhat. This gets the first inverter driven on initially.

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  • \$\begingroup\$ Thanks. Yes, i can confirm that there's no ground available, neither by wiring or from chassis. I can fit some stuffs in the blinker case, there's already a relay (i don't know if NO or NC) which is useful for other SUZUKI bikes, but on mine is not wired - i mean, the blinker have 7pins, but on the bike socket there's just the two wires i mentioned. Tried to play on the emulator with your point 2 and 3 without seeing nothing useful, but i'm not an electronic so i should know the values of caps and resistors. \$\endgroup\$ – Parduz May 20 at 15:36
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Could you add a normally closed relay on the indicator wire? Then when the indicator line goes high the contacts will open. Add 12v to the switch and your done :)

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  • \$\begingroup\$ I'll do any modification inside the blinker case, but I'll avoid any modification of the bike wirings, as it will require a huge amount of work (even for just removing the tupperware - it's a Burgman 650) \$\endgroup\$ – Parduz May 20 at 13:50
  • \$\begingroup\$ You could build the relay into the blinker, only needs to toggle the output... - just have to make a larger over... @Parduz \$\endgroup\$ – Solar Mike May 20 at 13:57
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    \$\begingroup\$ Such a relay would require a ground connection, which the OP has already stated is not available. \$\endgroup\$ – Dave Tweed May 20 at 13:59
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    \$\begingroup\$ How could it be possible without changing the wirings? The relay will need a ground, which is not available. Pls explain (or draw the schematics) as i don't understand how it will be feasible. \$\endgroup\$ – Parduz May 20 at 14:01
  • \$\begingroup\$ For that « missing » ground you can use the indicator lamps - the coil current won’t even make the 21W bulbs glow dimly - a trick often used... \$\endgroup\$ – Solar Mike May 20 at 21:36
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My suggestion would be to take the existing circuit and reduce the time constant by a couple of orders of magnitude — so that it "blinks" at a rate that's too fast to actually see.

This reduces the startup problem to a negligible delay and gives you your virtual "Vss" supply. You can now add a second, slow timing circuit that actually establishes the visual blink rate. This probably involves changing the gate that drives the MOSFET to a NAND gate.

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  • \$\begingroup\$ this seems interesting and somewhat viable. What should i do? \$\endgroup\$ – Parduz May 20 at 15:37
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Simple fix: reposition C2. Connect the minus side to the minus of C1. Connect the plus to pin 13 of the IC ( where the minus used to be. ) This will pull the gate low when the circuit is first powered up. On further thought; I think the connection of R2 will also need to be changed. Cut the trace that connects to the FET and add a jumper so it ,R2, connects to the pins 12-11 on the IC. The FET will turn on just enough to stop charging C1 so first blink might be dim. This seems to be the most doable change.

Parduz reports that this didn't do the trick so new configuration. Move the plus side of C2 to the junction of R2 and R3. I can't read the value of R3; if it is 75K it can stay, but it is 7.5K it will need to be replaced with a resistor between 47K to 220K; whatever you have or can obtain. Then put a new capacitor in the spot that used to have C2; value will depend on the new R3, if 100K then .33uF to .47uF, .1 to .22 uF for 220K, etc. This will hold the FET off for 30 to 50 mSec, allowing C1 to charge up enough to fully turn on the FET. This won't address the complaint of too much variation in timing with voltage change, Unfortunately I am not in a position to completely re-engineer this circuit. This is the most I can think of for now.

Further thoughts: I have been assuming that C1 discharges when the blinker is off, but now realize that is not likely. The oscillator will always stop in the off state and to get it to reset, both C1 and C2 must go to zero volts. To do this a resistor should be connected across C1; 47K would would discharge it in about 15 seconds while causing about 10% ripple when on. This is likely to make the timing variation with voltage worse, so it will be a trade off.

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  • \$\begingroup\$ Thanks. I tried the C2 mod alone: the lamp still starts OFF; what it changed was that the frequency seems a lot more sensible to the voltage variation: while i'm running is faster and sometimes irregular (I guess it depends on what other loads are active on the plus wire, lowering the 14V to 13 or 12). The R2 mod i've tried just on the circuit simulator, alone or with the C2 mod also. The resulting "ring" (R2, R3, IC1-13 and12) starts oscillating on his own, so the lamp starts ON or OFF depending on the instant you activate the handlebar switch. At least, this is what the simulator show. \$\endgroup\$ – Parduz May 28 at 7:42
  • \$\begingroup\$ Thank you for trying my suggestion, I'm sorry it didn't work. I have added some new ideas to my answer. I believe they will make it blink on after a very short delay. Leave R2 be. I won't be able to respond for a few days, I apologise for the delay. \$\endgroup\$ – EinarA May 29 at 7:15
  • \$\begingroup\$ I found some time. I think I have discovered why the circuit doesn't start as intended and have added it to my answer. \$\endgroup\$ – EinarA May 30 at 6:42

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