Alright so i have an electronics question and i'm not really sure what to do as i've never worked with this much power before. We have a 17 amp 12VDC battery connected to a power distrubution board, the board has 40 and 20 amp 12v outputs. We purchased some of the cold cathodes from parallax to put some lights on our robot however i have no clue how to down amp the power from the board.
When I hook up the lights the converters starts burning due to overampage (i assume). My electronics knowledge is really limited, can anyone offer a simple fix? We just want the lights to work.
But that doesn’t seem to make sense because it doesn’t factor in the original amperage unless i’m missing something. And i assume the resistor value should be different if you’re using 20 amps or 200 amps. Also a 40 ohm resistor seems really small. Does amperage or something change when DC is converted to AC?
Do you have specs or a link for your power distribution board? Typically the current consumed is dictated by the load, not the supply. If your distribution board can supply a 40 amp and a 20 amp output, that doesn’t mean anything connected to those ports will consume that much power. It means the board can supply UP to that much power on those ports.
Your DC to AC converters are designed for those CCFL lights, so they should in theory match just fine. The web links you provided indicate the power consumption:
12V / 300 mA power consumption per CCFL through inverter
The converter is going from 12 VDC to ~660 VAC, but the current consumption of these bulbs is low.
Without seeing your setup, may I suggest bypassing your power distribution board as a test and just connect the 12 V battery to the converter directly? I suspect either something wrong with your board or some sort of bad/missed/wrong connection.
Amperage = the number of electrons that pass a given point per second. Amperage is the same throughout the circuit.
The way to reduce amperage is in ohm’s law (voltage=current * resistance, or current = voltage / resistance). You can either decrease voltage or increase resistance.
That said, you shouldn’t need anything to run the CCFL’s with that inverter. The capacity of the battery may be 17Ah (it would provide 1 ampere’s worth of electrons at 12 volts for 17 hours), the inverter provides resistance to keep that many electrons from flowing all at once. What happens if you skip the power distribution board and connect the inverter directly to the battery?
If the 12V battery is If the 12V battery is genuinely supplying 12V to your inverter, and the connections aren’t backwards or anything, then you’ve got a faulty inverter. As mentioned above the inverter won’t draw any more current than it needs, unless it’s running at a higher voltage than it’s supposed to.
I’m wondering if it has I’m wondering if it has something to do with his undefined “power distribution board”. Thus the suggestion he bypass it and connect right to the battery.
i think you’ve got it right, i think you’ve got it right, just touching the leads to the battery has the inverter working fine. The FIRST competition requires a much lower level of knowledge then what most people learn from the start here project, the main aspect of the FIRST robot is the physical build itself. That being said teams are not supposed to add additional things however i thought some lights would make it look nice. The only question is - if i have two leads coming off the battery, one going to the power board (which i know nothing about by the way, there aren’t really specs given out for it), and the other is wired directly to the inverters which would allow them to run fine, however if the power were to draw power from the battery would the pulled current be dragged through the inverters as well? We have 4 cim motors running with only 1 bat for logic and power so there may be a lot of draw
If the power board and If the power board and inverter both have a positive and negative line each, connecting them to the battery in parallel, then they won’t affect each other (except for the fact that they’re both consuming power from the same source).
