so im trying to figure out how to wire up my gun... fan and ignition. i have two spark strips, and i bought the electric spark circuit from launch potatoes.com. its powered by two individual battery packs, both 8 AA so 12 volts. i was wondering if anyone could draw me a diagram to make my electric spark circuit work with two lighted rocker switches one for safety and the other for ignition. and my fan switch is a regular rocker with an individual led to show when its on. a diagram for that would be great too... sorry im kinda a caveman with electronics. (for anyone that has time)
Thanks
electric circuit
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jimmy101
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Why 2 pack of 8 batteries each? Seems a bit overkill, 8 AAs should be more than enough oomph and should be able to run the fan for hours. The ignition circuit probably draws fairly little power and it isn't on that long anyway.
What is the sparker's input voltage rating? I would suspect it is either 1.5V, 3V or 9V. Feeding it 12V (at perhaps as much as 20 AMPs which your double power pack may well provide) will probably release the "magic smoke" from the sparker circuit.
Anyway, attached is a start. I assumed you will use the battery pack in parallel so the systems voltage is 12V. I didn't have a schematic symbol for the battery powered BBQ sparker so I used a generic one for a photoflash unit. Basically, any black box that has two input voltage wires and two output high voltage wires.
Lighted switches are cool but they may not work all that well in this application. The problem is visibility. In bright sun it takes a hell'a bright indicator light for it to be visible. Most LEDs aren't bright enough. A blinking LED is often a better idea, especially if it blinks fairly fast, like a couple times per second.
You need to find the voltage rating of the internal lights of the lighted switches and that has to match your power supply. (Or you can add a resistor in series if the power supply is a higher voltage). Shown in the diagram is a single LED (assumed to be a red one) and the proper size resistor for using it in a 12V circuit. The actual resistance doesn't have to be all that exact, +/- 25% is OK. The LED shown is on when the fan switch (S3) is on.
Switch S1 is the igniter main power and safety switch. Probably want a slide or toggle switch. S2 is a "momentary on" pushbutton switch and is the firing trigger. The sparker will spark as long as S1 is on and you keep pushing S2.
Let me know if you want anything else added, like the wiring for switches with built in indicator lights. (Include the part number for the switches and/or your source since there are different configurations.)
EDIT: OOPS, the resistor for the LED should be about 515 Ohms, not 600. The equation for a red LED assuming 20mA current is;
dropping resistor = (supply voltage - 1.7V)/0.02A
What is the sparker's input voltage rating? I would suspect it is either 1.5V, 3V or 9V. Feeding it 12V (at perhaps as much as 20 AMPs which your double power pack may well provide) will probably release the "magic smoke" from the sparker circuit.
Anyway, attached is a start. I assumed you will use the battery pack in parallel so the systems voltage is 12V. I didn't have a schematic symbol for the battery powered BBQ sparker so I used a generic one for a photoflash unit. Basically, any black box that has two input voltage wires and two output high voltage wires.
Lighted switches are cool but they may not work all that well in this application. The problem is visibility. In bright sun it takes a hell'a bright indicator light for it to be visible. Most LEDs aren't bright enough. A blinking LED is often a better idea, especially if it blinks fairly fast, like a couple times per second.
You need to find the voltage rating of the internal lights of the lighted switches and that has to match your power supply. (Or you can add a resistor in series if the power supply is a higher voltage). Shown in the diagram is a single LED (assumed to be a red one) and the proper size resistor for using it in a 12V circuit. The actual resistance doesn't have to be all that exact, +/- 25% is OK. The LED shown is on when the fan switch (S3) is on.
Switch S1 is the igniter main power and safety switch. Probably want a slide or toggle switch. S2 is a "momentary on" pushbutton switch and is the firing trigger. The sparker will spark as long as S1 is on and you keep pushing S2.
Let me know if you want anything else added, like the wiring for switches with built in indicator lights. (Include the part number for the switches and/or your source since there are different configurations.)
EDIT: OOPS, the resistor for the LED should be about 515 Ohms, not 600. The equation for a red LED assuming 20mA current is;
dropping resistor = (supply voltage - 1.7V)/0.02A
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Technician1002
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Disposable camera flash units run on 1.5 volts, not 12.
Think what will happen if you plug in a 120 volt appliance into a circuit 8 times higher in voltage. The flash circuit will do the same.
The dropping resistor for the LED will need to be a standard value. If you calculate a value and try to buy it, you will be disappointed.
Some standard values are 330 470 680 1,000. Going slightly higher in value to keep the current below the maximum is a good idea. A 680 or 1K resistor is suggested for a 20 ma LED on 12 volts.
Think what will happen if you plug in a 120 volt appliance into a circuit 8 times higher in voltage. The flash circuit will do the same.
The dropping resistor for the LED will need to be a standard value. If you calculate a value and try to buy it, you will be disappointed.
Some standard values are 330 470 680 1,000. Going slightly higher in value to keep the current below the maximum is a good idea. A 680 or 1K resistor is suggested for a 20 ma LED on 12 volts.