Wire Kit for Battery

Can anyone tell me how to wire this kit for battery use? (Please don't say read the PDF data sheet. Trust me, I have.) I would like to use 3 AAA batteries.

 

Right, I did that. (I just soldered all three of the 5V Regulator Pins together). What I'm curious about is how to omit the 2.5mm DC power input and use battery power instead.

 

Take the regulator and the 1n4004 diode out and connect the + of the batteries to the place marked X2 then put a jumper wire from X2 to Vcc.

 

or rather from anode of D2 pad to Vout pad of the regulator

 

Where's the negative lead go? And what is the Vout of the regulator?

 

after removing the regulator.
just give power across C12. Look at the diagram in the datasheet
Vcc is positive and the other is negetive

 

Alright, I'm really sorry to be redundant here, but I'm just not understanding.
1. Do I remove the diode D2?
2. I have a jumper connecting the two outside pins where the regulator used to be, and the middle pin is open. Is this correct?
3. Where does the positive lead of my battery connect?
4. Where does the negative lead of my battery connect?

 

I'll edit a pic for u in while...

 

Thank you!

 

check the pic.
and remember not to apply more than 4.5V, that is 3 1.5V cells.
If the battery drain is fast, do not increase voltage, instead find a single Li-on cell, I'll tell what to do if it comes to that

 

How can I distinguish which hole from the regulator is Vcc when looking at the PCB?

 

post a pic of the board and I will show u

 

The output pin of the regulator IC2 is also connected to one side of R7, which is right next to it.

 

Well I wanted to keep the diode, but if it comes down to it, I'll remove it. In that case, Do I just jump the two outer pins of the regulator? Also, I didn't know the polarity of my LED, so I just removed it and jumped the gap. Is that okay?

 

If you keep the diode, then you will reduce the available voltage by roughly 0.7v, and waste power. Is that really what you want to do?

Yes. However, if you want to save on batteries, I suggest that you remove the regulator from the circuit entirely, as it will be a constant 5mA load even if it is not doing anything. This will only serve to generate heat and kill your batteries much faster than you'd like.

NO!
Remove the jumper that you substituted for the LED.
The jumper will only serve to drain your battery a lot more quickly. It may also cause the port in the IC to be burned out due to excessive current.


If you want to do this the easy way, you can simply remove the input diode (closest to the power jack) and the regulator (IC2) and place a jumper from the non-banded hole of the diode (it's anode) to the left side of R7, which is Vcc.

You can use rechargeable NiCD or NiMH batteries; up to 4 of them in series. Let them settle for an hour or two after removing them from a charger. Four NiCD or NiMH batteries in series will measure 4.8v, which is within the specifications of your device. If you use primary batteries (alkaline, other non-rechargeable batteries etc.) then you can only use three in series.

 

And leave the two open holes where the LED was untouched? I broke the LED getting it out...

 

Yep, leave the LED pads (holes) open. Don't put anything there.

 

WOW this is frustrating. Still not working. Here's some photos. Maybe you can find my error. And yes, I know I have a black wire for the positive lead and a red one for negative on my power input.