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

---

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).

well yea that's what I meant lol. I'll snap up a small relay @ radio shack after work and try it out tonight

something like this should work right?

http://www.radioshack.com/product/in...lterValue=SPST

I think something more like this: http://www.radioshack.com/product/in...ductId=2062481

You need it to handle the current of charging the capacitor and also for the arduino and servo to work.

it doesn't need to be a SPDT does it? They have this 1A SPST as well.

http://www.radioshack.com/product/in...lterValue=SPST

:EDIT: that SPST 1A is only a 5v relay, so I will need that one you listed

:EDIT 2: they're got it in stock at my local store, I'll scoop it up after work tonight :up:

crenn, here is the datasheet for that relay:

http://relays.tycoelectronics.com/datasheets/OUAZ.pdf

what would be the best way to wire it up? I was thinking the 12v would go to the switching pins (bottom left 2), 5v from PSU to bottom right pin, 5v sense wire to top right pin. This would close the 5v circuit when 12v is applied (computer on), and take away 5v when the computer is off. the 5v charge I was going to hook directly to the PSU as it is breadboarded now, or would this cause problems?

ok here's what I've got. Obviously the 5v and 12v will be switched together, but to do that in this program would make it look complicated and messy

Code:

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$ 1 5.0E-6 1.7725424121461644 99 5.0 50
c 112 240 112 288 0 10.0 0.002935472230715953
c 112 288 112 336 0 10.0 0.002935472230715953
178 384 224 400 224 0 1 0.2 0.03749999999999956 0.05 1000000.0 0.0375 320.0
x 445 258 554 261 0 10 5v sense pin to arduino
x 364 144 423 147 0 10 5v from PSU
x 249 283 314 286 0 10 12v from PSU
x 159 201 196 204 0 10 Arduino
s 320 256 352 256 0 0 false
g 256 336 304 336 0
w 256 304 256 336 0
w 256 336 224 336 0
s 272 112 304 112 0 0 false
w 112 336 224 336 0
w 384 256 352 256 0
w 288 256 320 256 0
w 112 160 112 240 0
w 112 112 112 160 0
w 160 112 224 112 0
w 160 112 160 160 0
w 224 112 224 208 0
w 224 256 224 336 0
w 336 112 304 112 0
w 384 304 256 304 0
w 384 272 384 304 0
w 400 240 480 240 0
w 272 112 272 224 0
w 272 224 384 224 0
w 224 112 272 112 0
181 224 208 224 256 0 1906.636257543845 100.0 5.0 0.4 0.4
z 112 160 160 160 1 0.805904783 5.6
r 112 112 160 112 0 100.0
R 336 112 400 112 0 0 40.0 5.0 0.0 0.0 0.5
R 288 256 272 256 0 0 40.0 12.0 0.0 0.0 0.5
o 0 64 0 35 0.009765625 0.05 0 -1
o 1 64 0 35 0.009765625 0.05 1 -1
o 31 64 0 35 5.0 25.6 2 -1
o 32 64 0 35 20.0 0.05 3 -1

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You can hook the 5v directly to the sense pin on the arduino. Then connect the closed position (ie when the 12v is applied) to the voltage supply of the arduino and charging circuit. So when the computer turns off, the arduino, servo and capacitor is isolated from the rest of the system and the arduino still can sense when the computer has turned off as the 5v supply will disappear from the sense pin.

EDIT: Goto Page 5 and my top post. That's more accurate to what you want.

I just edited the code again

:EDIT: I'll cehck what you've got on pg 5 :up:

ok I took your pg5 code and edited it so it wouldn't send voltage from the caps back through the 5v sense pin

Code:

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$ 1 4.9999999999999996E-5 1.6308177459886661 50 5.0 50
z 224 272 224 176 1 1.2 5.1
r 160 272 160 176 0 15.0
c 192 272 192 320 2 10.0 1.5976931634448988
c 192 320 192 384 2 10.0 1.5976931634448988
w 192 384 288 384 0
w 288 384 384 384 0
w 384 192 384 176 0
g 288 384 288 432 0
w 224 272 192 272 0
w 192 272 160 272 0
w 160 176 192 176 0
w 192 176 224 176 0
R 272 80 272 32 0 0 40.0 5.0 0.0 0.0 0.5
w 224 176 384 176 0
s 192 112 272 112 0 1 false
178 272 80 384 80 0 1 0.2 0.0 0.05 1000000.0 0.0375 320.0
w 272 128 272 304 0
w 272 304 288 304 0
w 288 304 288 384 0
R 192 112 144 112 0 0 40.0 12.0 0.0 0.0 0.5
x 421 292 458 295 0 10 Arduino
z 384 128 384 176 1 0.805904783 5.6
w 384 96 384 128 0
x 461 115 531 118 0 10 5v to sense pin
g 480 96 512 96 0
r 384 96 480 96 0 1000.0
181 384 256 384 272 0 1462.1431707282563 100.0 4.0 0.4 0.4
w 384 192 384 256 0
w 384 272 384 384 0
o 2 64 0 35 2.5 51.2 0 -1
o 3 64 0 35 5.0 51.2 1 -1
o 26 64 0 35 5.0 51.2 2 -1

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You can remove that zener diode as the sense pin should be connected before the relay (aka directly from the PSU itself). I'll draw up an accurate representation before I go to bed.

ahhh ok. Thanks again crenn, without you I don't know if this project would have even gotten off the ground!! +rep

like this then?

Code:

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$ 1 4.9999999999999996E-5 1.6308177459886661 50 5.0 50
z 224 272 224 176 1 1.2 5.1
r 160 272 160 176 0 15.0
c 192 272 192 320 2 10.0 0.0034025929640006517
c 192 320 192 384 2 10.0 0.0034025929640006517
w 192 384 288 384 0
w 288 384 384 384 0
w 384 192 384 176 0
g 288 384 288 432 0
w 224 272 192 272 0
w 192 272 160 272 0
w 160 176 192 176 0
w 192 176 224 176 0
R 272 80 272 32 0 0 40.0 5.0 0.0 0.0 0.5
w 224 176 384 176 0
s 192 112 272 112 0 0 false
178 272 80 384 80 0 1 0.2 0.037499999999999964 0.05 1000000.0 0.0375 320.0
w 272 128 272 304 0
w 272 304 288 304 0
w 288 304 288 384 0
R 192 112 144 112 0 0 40.0 12.0 0.0 0.0 0.5
x 421 292 458 295 0 10 Arduino
x 461 115 531 118 0 10 5v to sense pin
g 480 96 512 96 0
181 384 256 384 272 0 1045.742863484448 100.0 4.0 0.4 0.4
w 384 192 384 256 0
w 384 272 384 384 0
w 384 96 384 176 0
w 480 96 416 96 0
w 416 96 416 32 0
r 416 32 352 32 0 1000.0
w 352 32 272 80 0
o 2 64 0 35 0.009765625 1.6 0 -1
o 3 64 0 35 0.009765625 1.6 1 -1
o 23 64 0 35 5.0 102.4 2 -1

---

This is what I came up with, I'll look at yours in 2 ticks:

Code:

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$ 1 4.9999999999999996E-5 1.6308177459886661 50 5.0 50
z 336 320 336 224 1 1.2 5.1
r 272 320 272 224 0 15.0
r 496 240 496 432 0 100.0
c 304 320 304 368 2 10.0 2.410060209449615
c 304 368 304 432 2 10.0 2.410060209449615
w 304 432 400 432 0
w 400 432 496 432 0
w 496 240 496 224 0
g 400 432 400 480 0
w 336 320 304 320 0
w 304 320 272 320 0
w 272 224 304 224 0
w 304 224 336 224 0
w 336 224 496 224 0
178 384 128 496 128 0 1 0.2 0.011428027236798245 0.05 1000000.0 0.0114 1050.0
R 128 144 80 144 0 0 40.0 12.0 0.0 0.0 0.5
x 533 340 570 343 0 10 Arduino
x 549 74 619 77 0 10 5v to sense pin
w 496 144 496 224 0
178 272 64 320 64 0 1 0.2 0.30000000000000004 0.05 1000000.0 0.02 20.0
178 272 144 320 144 0 1 0.2 0.30000000000000004 0.05 1000000.0 0.02 20.0
R 128 64 80 64 0 0 40.0 5.0 0.0 0.0 0.5
w 128 64 176 64 0
w 176 64 272 64 0
w 320 80 384 80 0
w 320 160 352 160 0
w 352 160 384 160 0
w 128 144 272 144 0
w 128 144 128 176 0
s 128 176 208 176 0 0 false
w 208 96 208 176 0
w 208 96 272 96 0
w 272 112 240 112 0
w 240 112 240 176 0
w 240 176 272 176 0
w 384 48 432 48 0
r 432 48 512 48 0 100.0
g 512 48 560 48 0
x 237 36 353 39 0 10 Representation of a PSU
x 139 197 205 200 0 10 On/Off Switch
w 272 192 224 192 0
w 384 176 320 176 0
w 320 176 320 208 0
w 320 208 224 208 0
w 224 192 224 208 0
g 224 208 224 240 0
w 384 48 384 80 0
w 384 80 384 128 0

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Like that, I'm not sure why yours isn't working fully (voltage on the bulb wasn't 5V straight away) however that's pretty much the setup.

alright so i got it boarded up...and it doesn't work. it's losing too much voltage through the relay apparently. I've got 5.01v out of the PSU, and only ~3.6v out of the relay, and thats not enough to power the arduino. I'm going to keep messing with it and see what I can do. maybe put 12v into the relay instead of 5 and use a resistor after the relay to et it to ~4 or ~5v?

I'd say too dangerous. The voltage drop over the relay should be minimal :/

I think you were right...I think I fired my atmega328 chip. It won't run any loop and won't let me upload anything. the power led is on and everything though

and another strange problem arose...when I disconnected the PSU the relay didn't turn off until I unhooked either the 5v or 12v ground :?

ok well upon further testing, it would seem that with anything over 4.5v total in the caps the relay won't close. I hooked up my multimeter, and with the caps charged to 4.6v I shut off the psu. it dropped to 4.53 before the relay shut. I charged the caps to 4.75, and it went to 4.5 before the relay shut. :? it's shutting though, and 4.5v should still be more than adequate to move the servo to the closed position

If you power the arduino from USB, does it still work?

the power light comes on but it won't do anything. I tried USB, the 2.5mm jack and the 5v and gnd pins on the board. I'm hoping it's just the chip that's fried and not the whole board :think:

alright crenn I soldered up and tested it and it works great! I charged the caps up to full voltage (5.02) and shut off the PSU. The voltage trickled down from 5.02 to 3.37 before the relay finally clicked off. I'll have to wait for my new atmega328 to see if it'll be enough to move the servo. Maybe the draw of the servo itself will make the relay work right, who knows. I was thinking of working in a 5v relay that runs off the main relay and see if that will kill the 5v instantly. I'm not sure yet. oh well. I'll see what happens when I get my new chip in

Are you getting a raw ATMega328P or a preprogrammed ATMega328P with the Arduino bootloader?

preprogrammed. I could have gotten either though as you can burn the bootloader through the arduino and it's software

You need an ISP programmer to burn the bootloader, you can't with just the arduino and it's software (however there is ways around that).

Well I've got a programmer that will do ICs from 8-pins and up, I used it to write and program my lighting sequences to my PIC 12F683 on my RGB controller. But either way, I got one with the bootloader in it, so I'm all set. I just gotta try to figure out this funky relay thing. It's like it's leaking voltage backwards through the ground, through the PSU and then back through the 12v line and keeping the relay open. I may get a 5v relay and use the 12v to turn on the 5v and go that way, we'll see though.

The only solution I can think about is to use a 5v regulator to try to solve the problem, but I'm not sure how well that will work.

alright I got my new chip in today and all is good. now I'm going to do some more sketching with that circuit simulator and figure out how to work in that 5v relay

alright crenn what do you think about this instead? I'd need to get another diode but that's no biggie. could I use a transistor in this circuit instead maybe?

Code:

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$ 1 4.9999999999999996E-5 1.6308177459886661 50 5.0 50
z 224 288 224 192 1 1.2 5.1
r 160 288 160 192 0 15.0
c 192 288 192 336 2 10.0 2.0885785848445715
c 192 336 192 400 2 10.0 2.0885785848445715
w 192 400 288 400 0
w 288 400 384 400 0
w 384 208 384 192 0
g 288 400 288 448 0
w 224 288 192 288 0
w 192 288 160 288 0
w 160 192 192 192 0
w 192 192 224 192 0
w 224 192 384 192 0
178 272 96 384 96 0 1 0.02 0.015625 0.05 1000000.0 0.0010 320.0
x 414 344 451 347 0 10 Arduino
x 437 42 507 45 0 10 5v to sense pin
w 384 112 384 192 0
R 224 128 176 128 0 0 40.0 5.0 0.0 0.0 0.5
r 384 64 464 64 0 10000.0
g 464 64 512 64 0
s 224 128 272 128 0 0 false
w 272 144 272 384 0
w 272 384 288 384 0
w 288 384 288 400 0
181 384 208 384 400 0 1205.7972822685413 100.0 5.0 0.4 0.4
w 272 96 272 64 0
w 272 64 384 64 0
d 272 128 272 96 1 0.805904783
o 2 64 0 35 5.0 102.4 0 -1
o 3 64 0 35 5.0 102.4 1 -1
o 24 64 0 35 5.0 204.8 2 -1

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or how about this? why do I even need the resistor and diode on the caps?

Code:

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$ 1 4.9999999999999996E-5 23.47059216675035 54 5.0 50
c 192 272 192 320 2 10.0 2.440884490230046
c 192 320 192 384 2 10.0 2.440884490230046
w 192 384 288 384 0
w 288 384 384 384 0
w 384 192 384 176 0
g 288 384 288 432 0
w 192 176 224 176 0
w 224 176 384 176 0
178 272 80 384 80 0 1 0.02 0.015625 0.05 1000000.0 0.0010 320.0
x 414 328 451 331 0 10 Arduino
x 437 26 507 29 0 10 5v to sense pin
R 224 112 176 112 0 0 40.0 5.0 0.0 0.0 0.5
r 384 48 464 48 0 10000.0
g 464 48 512 48 0
s 224 112 272 112 0 0 false
w 272 128 272 368 0
w 272 368 288 368 0
w 288 368 288 384 0
w 272 80 272 48 0
w 272 48 384 48 0
w 384 96 384 176 0
162 384 192 384 384 1 5.0 1.0 0.0 0.0
w 192 176 192 272 0
z 272 112 272 80 1 0.1 4.9
o 0 64 0 35 2.5 0.8 0 -1
o 1 64 0 35 2.5 0.8 1 -1
o 21 64 0 35 5.0 0.8 2 -1

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1st circuit: Won't work due to all the current passing through a single diode. That diode is going to have a voltage drop of at least 0.7 depending on the construction of the diode, if you can find a diode which can handle the current (around 1-1.5A) has a voltage drop lower than 0.3v, that will work (remember, you have to give the arduino at least 4.5v to work and the servo may require a little more.

2nd circuit: If your power supply can support a surge of 60+ amps on the 5v line, then the only problem would be the diode as it would be very unlikely to support 60+ amps.

Quote:

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Originally Posted by crenn

1st circuit: Won't work due to all the current passing through a single diode. That diode is going to have a voltage drop of at least 0.7 depending on the construction of the diode, if you can find a diode which can handle the current (around 1-1.5A) has a voltage drop lower than 0.3v, that will work (remember, you have to give the arduino at least 4.5v to work and the servo may require a little more.

2nd circuit: If your power supply can support a surge of 60+ amps on the 5v line, then the only problem would be the diode as it would be very unlikely to support 60+ amps.

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ok so is that why I need the resistor and diode in before the caps? what if I put in just the 15ohm resistor and no diode? In the diagrams the diode doesn't seem to do much anyways, it all goes through the resistor :?