General question about powering an arduino

What voltage is the capacitors providing to the micro just after being fully charged?

This is what I have currently:

Code:

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$ 1 4.9999999999999996E-5 105.18789638808724 50 5.0 50
z 384 272 384 176 1 1.2 5.1
r 320 272 320 176 0 15.0
r 544 192 544 384 0 100.0
c 352 272 352 320 2 10.0 2.4298641778297494
c 352 320 352 384 2 10.0 2.4298641778297494
w 352 384 448 384 0
w 448 384 544 384 0
w 544 192 544 176 0
g 448 384 448 432 0
w 384 272 352 272 0
w 352 272 320 272 0
w 320 176 352 176 0
w 352 176 384 176 0
R 432 80 432 32 0 0 40.0 5.0 0.0 0.0 0.5
w 384 176 544 176 0
s 352 112 432 112 0 0 false
w 544 144 544 176 0
178 432 80 544 80 0 1 0.2 0.011428571428571427 0.05 1000000.0 0.0114 1050.0
w 432 128 432 304 0
w 432 304 448 304 0
w 448 304 448 384 0
w 544 96 544 144 0
R 352 112 304 112 0 0 40.0 12.0 0.0 0.0 0.5
o 3 64 0 35 2.5 0.05 0 -1
o 2 64 0 35 5.0 0.05 1 -1

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The switch represents the computer being turned on and off. But this circuit currently prevents the capacitors trying to power everything in the computer via the relay. I've lowered the charging resistor from 100 ohms to 15 ohms to charge it a little faster. The other resistor represents the load on the system. I have another idea on how to get things working better... but I'm unsure on how to do it.

EDIT: Simulator is showing me that each capacitor is charging to 2.27v in 3 minutes. There could be another way of cutting the power to the arduino and servo after it finishes it's power down sequence using another relay if you're interested.

I see what you're saying. this is what I sketched up and breadboarded:

Code:

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$ 1 5.0E-6 382.76258214399064 99 5.0 50
r 192 112 240 112 0 100.0
w 192 112 192 160 0
d 192 160 240 160 1 0.805904783
w 240 112 240 160 0
w 192 160 192 240 0
w 240 112 304 112 0
c 192 240 192 288 0 10.0 0.23950676780649208
c 192 288 192 336 0 10.0 0.23950676780649208
w 192 336 304 336 0
g 304 336 368 336 0
s 304 112 368 112 0 0 false
R 384 112 464 112 0 0 40.0 5.0 0.0 0.0 0.5
w 384 112 368 112 0
181 304 256 304 208 0 2779.341976072886 100.0 5.0 0.4 0.4
w 304 256 304 336 0
w 304 208 304 112 0
o 6 64 0 35 0.3125 0.05 0 -1

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Quote:

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Originally Posted by crenn

What voltage is the capacitors providing to the micro just after being fully charged?

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I'll find out tonight :up:

Quote:

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EDIT: Simulator is showing me that each capacitor is charging to 2.27v in 3 minutes. There could be another way of cutting the power to the arduino and servo after it finishes it's power down sequence using another relay if you're interested.

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It doesn't matter to me if the caps drain fully afterwards, my computer is always on for at least 3 mins anways lol. As long as it's got enough juice to move the servo closed it can sit like that until they're drained. Caps don't have a memory like batteries do they? or a certain number of charge/discharge cycles?

I'm not 100% sure about super capacitors. But normal capacitors don't have an issue with memory, and it's not really a defined number of cycles, just how old they are it seems.

They put out 5.02v total when fully charged. One was 4.999v and the other was 5.024v. I unhooked it and cycled the servo with its sensor pin (7) about 5 or 6 times before it didn't have enough juice to move it, so that will be more than enough.

The only problem I find is that it still feeds power back through the PSU. I had the sensor hooked to the 5v on the floppy connector and the arduino hooked to the 5v on a molex, and even when I unplugged the PSU the servo didn't move until I unplugged the sensor wire from the PSU. This may be the need to set it up with a relay like you suggested

Yeah, I was afraid about that happening so that's why I tried a transistor and eventually a relay. I didn't power the relay with 5v because the power would still feed back into the system, so that's why it's powered with 12v. If you want, we can still discuss giving the design 2 relays so that it can disconnect itself (and hopefully save charging time next cycle).

btw, took a look at your design in the simulator. You placed a diode where the zener diode should have been.

whats the difference between a regular diode and a zener? I see the ones that I bought were in fact zener diodes.

I'm going to try hooking the signal wire to 12v, that may work just as well

Don't hook it upto 12v... you'll damage the pin on the micro.

The different is a zener is designed for a specific reverse break-down voltage (in your case 5.1) but can work as a normal diode as well.

ah ok. so it looks like a relay is the way to go then. have the 12v trigger the relay, so when the 12v shuts off the relay will open and shut off any 5v to the sensor pin

Quote:

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Originally Posted by SXRguyinMA

ah ok. so it looks like a relay is the way to go then. have the 12v trigger the relay, so when the 12v shuts off the relay will open and shut off any 5v to the sensor pin

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Actually it won't shut off 5v to the sensor pin directly. It isolates the system so it's not feeding back into the PSU (and then the sensor pin).