PCB Design: Adding Batteries (CR2032) to Schematic

[_Bd_]

Hi, I am working on a PCB design and the mock-up prototype is now done, thanks for all your help by the way!
now its time to get it mobile. I am trying to add a battery (CR2032) to the design of the PCB,

the thing is that I have my voltage regulator set up as so:
http://img607.imageshack.us/img607/2811/ao8x.png

but I don't know how to hook up the batteries,

the voltage regulator is supposed to take a voltage in of 7.2V (from 2 CR2032 rechargeable batteries at 3.6V each)

and one stream is supposed to feed a RF transmitter at 3.3V (the bottom regulator)
the top one is a 7805 that outputs 5V to drive the other circuit.I was confused as the battery symbol (that comes in adafruit library) has 3 leads, from what I remember soldering, 2 of them are positive? and 1 is negative, but should I link the negative to the ground? or how should I wire it up
thank you!

 

[Okefenokee]

Your schematic symbol looks like it has two pins on the cathode. It's hard for me to tell from the image but I think the symbol has two bold vertical lines, one long one for the positive side and one shorter one for the negative side.

 

[_Bd_]

yeah the 2 pins are positive, and the single (middle) pin is the negative, my question is, should the negative go to ground? or where should I connect it to?

 

[davenn]

what is the bottom regulator ?
looks like a negative regulator ?
what's it outout voltage supposed to be ?

Dave

 

[_Bd_]

The top regulator is a 7805, it receives up to 13V input and outputs 5V

the bottom regulator is a UA78M00, receives up to 13V input and outputs 3.3V, or that's what I want it to do.

Thing is most of my board works on arduino (5V) but the RF transmitter works on 3.3V

I found these http://www.ebay.com/itm/Charger-4x-3-6V-Lithium-Li-ion-CR2032-2032-LIR2032-Recharge-Button-Battery-/360795733989, they are 3.6V and I am using 2 that I want to connect in series to give an input of 7.2V

what is the bottom regulator ?
looks like a negative regulator ?
what's it outout voltage supposed to be ?

Dave

should I set them up in series instead of parallel then?

VCC = 5V
VEE = 3.3V

 

[davenn]

OK
thanks for your clarifications

in this case you have indicated the lower 2 capacitors with the wrong polarity

here's a circuit I drew using the battery supply you wanted...

Now the major problem with your thought of using only 2 batteries is that you are not really going to sustain the 2 to 2.5V headroom difference between the input and output of the 5V regulator

I would suggest that you will find you need 3 of those 3.6V batteries in series, rather than the 2 you wanted and that I have shown

OHHH and you can see I have chosen the BA033T regulator for the 3.3V rail

cheers
Dave

 

[_Bd_]

So, given that I am limited in size, would you suggest I move to 3x CR1220 or CR1225 ? (3volts each) and get a 9V total voltage?

Also, is ground the same as 0V ? (I know they are similar in AC or DC, but I can't remember where? or maybe I am completely wrong)

 

[davenn]

So, given that I am limited in size, would you suggest I move to 3x CR1220 or CR1225 ? (3volts each) and get a 9V total voltage?

Im not too familiar with the 1220 or the 1225, you would need to consider if they have enough current supply capabilities for what your circuit requires
NOTE tho that none of those small button batteries have much current capability
the CR1220 for example is only 20mAH
you are not going to power much off that for any length of time
These styles of batteries are used in digital watches and for memory backup where the required current for the circuit is extremely low

I suspect you are going to have to reconsider your project size and a different type of battery

Also, is ground the same as 0V ? (I know they are similar in AC or DC, but I can't remember where? or maybe I am completely wrong)

in this case, the same thing

cheers
Dave

 

[Jony130]

OK

You put capacitor backward for 3.3V

 

[davenn]

OK
thanks for your clarifications
in this case you have indicated the lower 2 capacitors with the wrong polarity
here's a circuit I drew using the battery supply you wanted...

https://www.physicsforums.com/attachment.php?attachmentid=64262&stc=1&d=1385332430

Now the major problem with your thought of using only 2 batteries is that you are not really going to sustain the 2 to 2.5V headroom difference between the input and output of the 5V regulator

I would suggest that you will find you need 3 of those 3.6V batteries in series, rather than the 2 you wanted and that I have shown

OHHH and you can see I have chosen the BA033T regulator for the 3.3V rail

cheers
Dave

 

[meBigGuy]

The lm7805 is a voltage hog, needing 2V headroom. Use an LDO regulator like the http://www.ti.com/lit/ds/symlink/lm2940-n.pdf, which is more like 0.5V @ 1A.

Or, any LDO regulator that fits your needs. Just be sure to use modern LDO regulators, not another battery.