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I'm currently building a project that involves a 2-stroke engine. I have tapped into the wires where you would normally place a kill switch in order to get power for some electronics. The voltage is 9VAC. The first half wave is used to supply my battery and logic circuitry with power. The second half wave, I've discovered, that if you draw more than a few mA it will mess with the ignition and stop the engine, so I didn't use it for power. But then I found the need to be able to use my MCU (an ATtiny84A) running on the first half wave's power to short the other half wave so it can stop the engine if something goes wrong.

  • I can't use a relay: it draws too much power.
  • I can't use a MOSFET: the two waves have the opposite potential.
  • I can't use a latching relay: not in stock where I live (can't buy online due to shipping restrictions).
  • I can't use a reed relay because the ones I find are NO but I need it to be NC.
  • I can't use an SSR: only available for higher voltage range.

What do I do? I came up with the idea of using an optocoupler with a MOSFET but I'm not sure if it's gonna work. Here's the schematic that I created but if you find a better solution let me know! My schematic

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    \$\begingroup\$ By "kill" I assume you mean "turn off"? I initially parsed this as saying you wanted to destroy the engine. \$\endgroup\$ – Hearth Apr 4 at 14:49
  • \$\begingroup\$ You haven't explained how the engine is meant to start. \$\endgroup\$ – Andy aka Apr 4 at 14:52
  • \$\begingroup\$ Just reading this schematic I can't understand what you're doing. Do you expect Q1 to ever cut off current flow through it? Because it won't. What's D2 there for? Why are you not just sending a signal to the ECU? \$\endgroup\$ – Hearth Apr 4 at 14:52
  • \$\begingroup\$ Wait, is the "engine magneto" a power source? I took it to represent the starter motor, but the circuit makes a little more sense if you assume it's a generator. Though those might be the same thing in a combustion engine; I'm not too familiar with engines. \$\endgroup\$ – Hearth Apr 4 at 14:55
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    \$\begingroup\$ What about a latching relay? That only requires power briefly to change state. \$\endgroup\$ – Unimportant Apr 4 at 15:29
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I think the magneto peak voltage is definitely too high for D1/D2, probably the optocoupler and maybe the MOSFET (no part numbers being provided).

Try replacing D1 with a 1N4007 and and replace the optocoupler/MOSFET with an opto-MOS SSR.

enter image description here

Edit (for general info): The objective is to replace this kind of "kill switch"

enter image description here

A magneto is a permanent magnet generator that forms part of an internal combustion ignition system that requires no battery or electronics of any kind, just the magneto itself, contact points that open at the appropriate time to generate a spark, and an ignition coil. They are used on small engines and on many light aircraft (in a dual configuration) where it is very undesirable to have a failure that causes the engine to cease rotating.

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    \$\begingroup\$ Yeah and the irony is that I tested with 1N4007 though I planned to finally use the ones on the schematic. Still didn't work. Now about the opto-MOS SSR. Seems exactly what I'm looking for but I don't know where to find it in my country. \$\endgroup\$ – Lefteris Garyfalakis Apr 4 at 15:12
  • \$\begingroup\$ If the goal is to short out the magneto, what about an opto-triac across the terminals? Should be easier to find than an optoisolated MOSFET-type SSR. \$\endgroup\$ – Synchrondyne Apr 4 at 15:31
  • \$\begingroup\$ @Peter It might work, maybe just an MOCxxx optotriac. Worth a shot. \$\endgroup\$ – Spehro Pefhany Apr 4 at 15:33
  • \$\begingroup\$ Thanks a lot guys! I'm gonna buy one TLP3042 tomorrow and see if it's doing what I want. Thanks! \$\endgroup\$ – Lefteris Garyfalakis Apr 4 at 17:21
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schematic

simulate this circuit – Schematic created using CircuitLab

Use the same rectifier LDO combo that worked for you. I just showed it as D1 C1, U1, C2, while R1, Q1 shorts the DC side of the AC bridge which is normally no-load to suppress the magneto voltage with 1.5V + 2 Ohms (Rce for PN2222A) which should be more than enough. Estimated values depend on Magneto inductance.

This circuit uses stored energy from the +ve wave to shunt the magneto on both half waves for a duration of R1 C1 = T approximately regulated and longer with ripple.

Edit

On 2nd thought a PNP Darlington switch will work better like a Pch FET between +5V and bridge V- only (V+=open). In this mode , the uC then drives Vkill=0V to gate or R1 to PNP-b with PNP-e to +5. Then Drain or PNP collector with LDO source, pulls up AC bridge towards 5Vdc so effectively only the -ve AC half wave gets loaded thus killing ignition while continuing +ve AC charge cycles. Then R1 may be chosen like 10k with a Darlington PNP or FET.

Comments? @tomnexus

Sorry can’t edit schema on iOS...

Assumptions

Source is 9Vac with low current rating and AC- drives ignition is load sensitive while AC+ can supply a few mA easily . Cutoff =? <50%Vac?

Then a series-shunt e-switch (SPDT) will attenuate 9V better as engine voltage and ignition cutout better if this fails. Unless you use battery backup to drive shunt.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Apr 7 at 18:48
  • \$\begingroup\$ Comments anyone? \$\endgroup\$ – Tony Stewart EE75 Apr 7 at 19:17
  • \$\begingroup\$ @TonyStewartEE75 I just saw your message. Fantastic! Check the chat if you wish! \$\endgroup\$ – Lefteris Garyfalakis Apr 20 at 0:07

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