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120v to 3.7v circuit

joshzstuff

Jul 4, 2010
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My project is to build a power supply that will allow devices powered by battery to run on continuous power (mains).

Objectives:
-120AC to 3.7v DC
- smallest package possible
- small power requirements (<150ma output)


One Idea I'm researching now is to use a USB wall adapter.
[I'll post a picture in my next post]

Should I tweak this device, or add a few components to drop the voltage from 5 to 3.7? (i.e. diodes, regulator)

I'm just getting started in learning electronics, so any recommendations on research that will help me understand the principles will be appreciated.

Thanks in advance.
 
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joshzstuff

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Here is a candidate for the job:
Wall%20USB%20charger.JPG


Any suggestions?
 

Resqueline

Jul 31, 2009
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That there seems to be a USB charger and so may have a specific current limit, but may very well be suited for the job you have in mind.
In the upper right hand corner you find a zener diode between C4 & R6. The markings on it tells the Voltage rating (could be 3.9V). Compare that to the actual output.
Choose a new one being a Volt or so less (such as 2.7V) and try it out. Zeners have a wide tolerance but I guess your application will also tolerate running on 3 to 4.2V.
 

joshzstuff

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Hello Resqueline,

Yes, it is a USB charger.
That is just the kind of tip I was looking for!

It's been a while, so I'm going to go back and review my zener diodes & "Zener knee voltage"

? How do I interpret the label?
I'll include a close up of the diode.
It looks like it says "C3" and then on the side "4"

Zdiode.JPG


"C3"
Zdiode%20C3.JPG


"4"
Zdiode%204.JPG


I guess your application will also tolerate running on 3 to 4.2V.
I'm powering a camera and replacing the Lithium Polymer battery.
I think I need to get as close to 3.7 as I can because the camera will not function while it is charging (4.1-4.2 volts)

Also, the battery has a "protection circuit" across it, and I'm not sure how it will react when I remove it along with the battery.

@ zener value
Is there a good reference that you know of to research how to find the values?
Does "C3 4" mean anything, or must I remove the diode to identify it?

Thanks for your help Resqueline!
 

(*steve*)

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1) measure the output voltage (let's assume it's 5V)

2) Remove the diode from the circuit (noting which way around it went)

3) connect it in series with a 9V battery and a 1k resistor, and measure the voltage across it. If it's under a volt, change the polarity of the battery. You should get a reading like 3.7V, 5.6V, etc. Here is a page that lists the common zener voltages.

4) get another zener with the same relationship to the output voltage you require. e.g. if the zener is 4.3 volts for a 5.0 volt output, then for 3.7 volts, you need a 3v zener.

5) fit the new zener and measure the output voltage. Don't be surprised if it's not exactly what you expect (but it should be close)
 

Resqueline

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What steve says.
The labeling is to be read as 4C3, meaning 4.3V. If you add to this the 1.2V forward voltage drop of the IR-diode in the optocoupler you get 5.5V out which is a common value.
Yes, zeners are quite "loose" regulators. They have an absolute tolerance of 5 or 10%, a temperature coefficient, and a knee plus a slope on their voltage regulation.
The camera not working when charging is most likely due to some other reasons than voltage.
Removal of the protection/sensor circuit may lead to some objections by the camera. I guess you'll just have to try & see.
 
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joshzstuff

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Ok, first off, let's get one thing strait . . .
You guys rock!

This was exactly what I was looking for, and a great explanation for the aspiring novice.

I've been doing my research so that I can make a reasonably intelligent reply to your most informative posts.

@ steve
Thanks for the link, it was most informative.

4) get another zener with the same relationship to the output voltage you require. e.g. if the zener is 4.3 volts for a 5.0 volt output, then for 3.7 volts, you need a 3v zener.

5) fit the new zener and measure the output voltage.

Just to make sure I'm interpreting this correctly steve, you mean that once I verify the relationship between the zener voltage and final output voltage, I may then expect that my final output voltage will vary in proportion to the difference between the old & new zener diodes?

(e.g. replacing a 4.3v with a 3.9v should result in the output dropping from 5.0v to 4.3v ??)
[note, this is taking into consideration your comments about zener tolerances:]

Don't be surprised if it's not exactly what you expect (but it should be close)

zeners are quite "loose" regulators. They have an absolute tolerance of 5 or 10%, a temperature coefficient, and a knee plus a slope on their voltage regulation.

Odd voltage readings with USB cables
~5.12 directly from the charger

5.31 from a 4' USB-A extension
5.37 from the extension + a 4' USB-B cable
5.38 from the USB-B cable only

Might this be because this simple power supply is not "regulated" (with exception to the zener) and I will only get a steady voltage rating when it is connected to the load?
- if so, then should I make an effort to test the voltage under load? or
-will the fluctuations scale proportionally?

(a side note, plunging in cables and testing of the charger resulted in HAVOC on my AM radio [1020 Khz] )


Battery or Capacitor ??
The camera not working when charging is most likely due to some other reasons than voltage.
Removal of the protection/sensor circuit may lead to some objections by the camera

@ Resqueline
this detail may be more helpful to me than you know.
In 'phase 2' of this project I was planning on replacing the battery with a capacitor capable of powering a 150ma load for ~30 sec.

It now occurs to me that I might instead build a charging circuit, and allow the native battery to preform this function.
(although I would think that there would need to be overcharge protection considerations, possibly a factor for the capacitor idea)

Project's main objective​
This project centers around a motion triggered camera I will be using for surveillance outside that is connected to a light. When the light turns on, I want it to power on my camera. (the camera starts and stops recording based on motion)
The problem, however is that if the power fails before the camera saves the video, it will not be saved to the memory.
Thus I will require a buffer long enough to allow the camera to time out and save the video before powering down.
[advantages I perceive for the capacitor over the battery are frequent and brief usages, resulting in what would be a tough cycle for the Lithium Polymer battery.]

I am content at this point to focus on the constant power aspect of this project. (first stand, then fly ;-)
It has been educational as well thanks to your pithy replies.

I stumbled on to another candidate device in my research, if you can give your opinion on it's feasibility:

LED drivers . . . an alternative?​
This unit promises 3.5~3.8v @ 2amps.
http://www.dinodirect.com/5W-3-5V-3-8V-2A-Constant-Current-LED-Driver-110V-240V-AC.html


Since I'm looking for the smallest package possible, this solution seems very desirable.
Since this circuit was not designed for the purpose I want it for, do you think this is a good idea?
Can you point out any drawbacks?

Next research item, optocouplers
Although I am eager to complete this project, I also intend it to be a learning process.
So in an effort to better understand this charging circuit, I will next be doing the necessary research to fully understand how the remaining components affect the output voltage as summarized by Resqueline:

If you add to this the 1.2V forward voltage drop of the IR-diode in the optocoupler you get 5.5V out which is a common value.

I'm off to do more optocoupler research to determine why it adds voltage instead of lowering it. [I take the 817B to be between the usual CTR of 10-50%]
 

Resqueline

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With the charger being so simple it's almost amazing that it works & regulates at all. Like you noticed it has some shortcomings: voltage instability, and radio interference.
Be glad if you manage to make it keep an output between 3.6 and 4.2V. Try a zener of 3.0V or less. The optocoupler IR-LED is in series with the zener, that's why it adds.

LED driver: don't even think about it. It's a constant-current supply and doesn't regulate the voltage at all (even if a nominal voltage is stated). It'll kill the camera.

Will the camera wake up & record on power-up just like that btw.?

Super-capacitors might very well be able to replace the duty of the Li-Po cell.
Notice though that the voltage ramp-up on application of power may be painfully slow.

I'd try to connect everything directly to the battery terminals and see how it works out.
 

(*steve*)

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ust to make sure I'm interpreting this correctly steve, you mean that once I verify the relationship between the zener voltage and final output voltage, I may then expect that my final output voltage will vary in proportion to the difference between the old & new zener diodes?

(e.g. replacing a 4.3v with a 3.9v should result in the output dropping from 5.0v to 4.3v ??)
[note, this is taking into consideration your comments about zener tolerances:]

It's the absolute difference. So if a 4.3V zenner gives a 5V output, then the zener voltage needs to be 0.7V less than the output voltage you desire.

Note that low voltage zeners have very poor knee characteristics, so the regulation will get worse as you use a lower voltage zener (all other things being equal)
 

joshzstuff

Jul 4, 2010
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Sorry for the delay in replying, I've been working on your suggestions on every spare scrap of time I could find.
I'll tell you where I'm at in the project, but first I'll reply to your comments:


some shortcomings: voltage instability, and radio interference.
Be glad if you manage to make it keep an output between 3.6 and 4.2V.
I see your point. I was hoping that the zener option would get me close enough to satisfy the camera, it appears it is not. (see below)


Will the camera wake up & record on power-up just like that btw.?
YES! it has 2 switches, (double throw) that can be set to "on" and "motion detection" so when power resumes, so does the recording, provided there is motion.
That is why I was interested in super capacitors (I've purchased some)


Super-capacitors might very well be able to replace the duty of the Li-Po cell.
This sounds promising! so I purchased a few:
4x 1Farad 5.5volt
4x .33 Farad 5.5 volt
I have no idea how to determine how may mAh this will store, but I can try it out.

Notice though that the voltage ramp-up on application of power may be painfully slow.
? can you explain this in more detail?
I really only need to supply the camera with power for a minute tops so it can close the video file.
Would this cause a problem here?

I'd try to connect everything directly to the battery terminals and see how it works out.

Well, unfortunately the 'protection circuit' turns out to be the Achilles-heel in this project . . .

Here is what it looks like:
Cam%20&%20Battery.JPG


Here is a close-up of the components comprising the circuit.
There are 3 SMT components and 2 IC's
- The first is labeled: "G3JR" the only thing I could find on it was this link: http://www.515ic.com/shortcut/G3JR.aspx
- the second is labled: "8205A" it appears to be a "Dual N-Channel Enhancement Mode Field Effect Transistor" http://www.alldatasheet.com/datasheet-pdf/pdf/260211/KEXIN/KI8205A.html
Protection%20Circuit.JPG


Here is the rear of the circuit, it it helps identify it's function.
Protection%20Circuit%20rear.JPG





It's the absolute difference. So if a 4.3V zenner gives a 5V output, then the zener voltage needs to be 0.7V less than the output voltage you desire.

I think we're on the same page on this.
I replaced the zener on the usb charger and got the voltage close to 3.7v , however the "battery circuit issues I'm having seem to negate this victory. (see below)


Issues with replacement power:​
Here is the thing,
the motion cams will NOT function properly without the battery "circuit"
When I attached the larger battery the cam started malfunctioning!

At first it was very subtle, the motion part would start, but would not close the file.
What I've notice lately is that it won't even record now (prolonged usage of the larger battery may have damaged the cam?)
-it's a lithium ion (3.7 volts)( but actually around 4.12 when charged)

My though is, first I'm going to try to get the cam to run on a larger battery, then I'll try to get it to work off of transformed AC.

-Do you recognize the components in the "circuit"?
-(is the one indeed a "Dual N-Channel Enhancement Mode Field Effect Transistor"?)
If so:
Is this consistent with the thought of this circuit being a protection for the batt. OR regulation for the camera??

Mini Regulation circuit??​
My other though is that this may be a regulator circuit for a camera with extremely picky input requirements.
Could this be a possibility?
OR are you saying that I was right, and it was only a "protection circuit" . . . for the battery only?


if so then I'm back to square 1 again, because I can't get the cam to work from any other input.
If that is the case, I may have to can the idea of the motion cams :-(
and build a sequence circuit with another camera that doesn't seem to have this circuit built in.
(although this is a Lot more involved, and I'd rather work my way up to this level)

possible plan of action:​
unless I am totaly off base about my "mini regulator" idea . . .
My plan is to build Variable power regulator based on a LM337t
I would then be able to experiment with the voltages to see if I can make the camera behave properly. (without the battery "circuit")
 

joshzstuff

Jul 4, 2010
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Oh, one thing I forgot to mention about the battery "circuit"
I attempted to salvage the circuit and put it inline with the new battery, however I've had a few attempts fail because I overheated the tiny components in an attempt to remove the old battery and solder a new one on.:eek:
I may have one or 2 more chances left before I run out of cameras with working "battery circuits"
I thought about snipping the leads off instead (one lead is crimped, so it's easy)
Perhaps then I can hold the circuit together long enough to determine if a transplanted "battery circuit" will work with a new battery . . .

If it tests successful, I'll then need an alternative method to soldering. (perhaps with a much smaller tip?)

IF this is a mini regulator circuit, I'm thinking of a few options:

- I could try to obtain larger, more solder friendly components to replace them. OR
- I could take care of the fine regulation with my new regulator circuit.
 

Resqueline

Jul 31, 2009
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There seems to be just two connections between the protection circuit and the camera so I don't think there is any communication going on there.
The circuit should only be responsible for disconnecting the battery on charging voltages over 4.2V, and perhaps also on disharging it below 3V.
So I can't see why it should interfere with the operation of the camera, but simply working on the camera could induce problems like static discharges, cable breaks etc..

A capacitor of 1F being discharged with 1A will drop with 1V per second. You have only 4.2V-3V=1.2V to work with so 1F will amount to only 0.33mAh in your case.
 

joshzstuff

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Selecting (or modding) viable 'unregulated' power supply

I really appreciate any help in selecting a candidate to use for my upcoming Variable Voltage Regulator circuit.

I've thought about starting from scratch, but I'm trying to keep the package as small as possible, so I am considering using or modding these AC to DC power adapters for the "unregulated Power supply" section of my circuit based on the LM337t

Reg%20Power%20Supply%20Schematic.JPG



Here is what I'm looking for from the supply:
-Small size
(I've rejected other alternatives such as wall wart plugs for their bulk)
-Only low current required
(I only require ~120mA or so to power small devices)
-Clean/ consistant power
(I will be using sensitive equipment that may react adversely to noisy or unstable voltages)


Here are the candidates I'm considering:
Unregulated%20Candidate_a.jpg


I think it is interesting the different paths the manufacturers took to reach the same goal.
I'm sure there are drawbacks to the choices that were made because of cost.

Any insight into advantages of any of these designs would be helpful.

Issues:​
-Supply "D" is a very noisy circuit as it is. Using a telephone toner receiver, It seems the noise is focused in the area of the 222 1Kv capacitor.

-Some of these designs are so simple that I'm sure I'll have issues with them.

I've posted some of the components labels in the picture, but I can post any others if needed.

Thank You.
 

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joshzstuff

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There seems to be just two connections between the protection circuit and the camera so I don't think there is any communication going on there.
I meant that the circuit might be a self contained power regulator (requiring no other connections than input and output)
My thoughts were that possibly the fluctuations of the battery from fully charged, to dead may be too much for the sensitive camera circuitry.

(i.e. if this regulation circuit was contained on a chip it would only need:
-2 wires from the source (battery/ power supply)
-and 2 wires to the load (camera)
Reg%20Power%20Supply%20Schematic.JPG


Of course, I reserve the right to be wrong, :)
but I was hoping that by Identifying the components in the circuit it might confirm one theory or the other.

The circuit should only be responsible for disconnecting the battery on charging voltages over 4.2V, and perhaps also on disharging it below 3V.

Your probably right, I'm just trying to get to the bottom of why my cameras won't seem to accept alternate batteries

So I can't see why it should interfere with the operation of the camera, but simply working on the camera could induce problems like static discharges, cable breaks etc..

This is true.
I'm sure this is a good possibility despite my care taken.
The frustrating thing is that this has happened more than once, so I'm second guessing the function of the "battery circuit"

I will try to remove Only the circuit with the next camera I get and see if it will operate with the only change being the removal of this circuit.

A capacitor of 1F being discharged with 1A will drop with 1V per second. You have only 4.2V-3V=1.2V to work with so 1F will amount to only 0.33mAh in your case.
Thank You
I'll come back to this once I get alternative power working on my camera.
 

(*steve*)

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What range of output voltages are you after. It looks like you want 1.25 to 24 volts.

At 120mA, the LM317 will be dissipating in excess of 2.3W so will require a heatsink.

The input voltage needs to be about 28 volts. Which of those power supplies can do that?

A 0.1uF capacitor on the input of the LM317 is probably way too low. 100uF or 1000uF would seem more appropriate. However the 0.1uF may be useful if the source id DC with high frequency noise.
 
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(*steve*)

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Oh, great. The same problem in 2 different threads. :-(

Let's close this one then.
 
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joshzstuff

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Oh, great. The same problem in 2 different threads. :-(
I'm sorry Steve.
I was looking to focus on the differences in design and their advantages/disadvantages.
I thought that including it here would side-tract the thread.
I didn't mean to multi-post.:eek:

What range of output voltages are you after.
I am still focusing on replacing a lithium ion battery, so I don't really need the upper range. (only ~ 3-6volts)

The purpose of this variable voltage circuit is to "tune" the voltage optimal to the device I'm replacing the battery in.
After which my intentions are to measure the Pot ohm setting, and replace it in the circuit with permanent resistors.

However the 0.1uF may be useful if the source id DC with high frequency noise.

This reminds me of something Resqueline said:
The camera not working when charging is most likely due to some other reasons than voltage.
Removal of the protection/sensor circuit may lead to some objections by the camera. I guess you'll just have to try & see.

Tell me what you think . . .
what are the chances that devices like this will not operate when plugged in is because they lack the filters necessary to handle 'dirty' power that will cause them to malfunction, or damage?

I don't have access to an oscilloscope, so I can't really tell what the output of this circuit is going to look like, or what kind of ripples there will be.
(although the LM317 paired with a cap is supposed to reduce ripple)
A friend told me to check the noise with a telephone toner receiver. (see above)
My thoughts were that someone with more knowledge could infer by the circuit design the status of it's output.

Clean power is a priority.
I'd like to get as reasonably close to pure dc as the camera is used to getting.
I know I'm in the dark about tolerances like this, but only testing will tell me what the long term effects will be using ac to replace the battery.

(this doesn't mean I'm negating the possibility that I messed up my cameras by my handling them:()

@ Steve
Do your recommendations about input voltage and capacitor rating still apply under this criteria?
 

(*steve*)

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OK, lets go back a few steps.

I believe that you modified a power supply for approx 3.7 volts.

Did you connect that to your device INSTEAD of the battery. i.e. did you remove the battery (and it's little supervisory circuit) and connect the power to the red and black wires?

I had assumed that the battery pack would have been removable. Now I see that it's wired to the camera directly. That must mean there is a charger socket on the camera itself. Why can't you power it via this?

I did re-read the thread again for an answer, but I didn't find one. Please excuse me if you've actually explained this earlier in the thread and I missed it.

If you're only looking to fine-tune the voltage then I would look at a low dropout three terminal regulator powered directly from one of those (unmodified) 5V regulators.

The datasheets for them will inform you of suitable amounts of filtering and the resulting voltage will be much more stable.

You should choose resistors so that the desired voltage is about mid-range for the trimpot and so that the extreme travel of the trimpot allows for a guaranteed range of voltages (taking into account component tolerance). That can all be a pain to calculate :) Figure out the resistance required for 5 volts and get a trimmer as close as possible to that. That will allow you to change the voltage between 1.25 and 5 volts which covers your desired range nicely without crowding it all up at one end of the trimpot's range.

The recommendation about input voltage was based on the values you put into the circuit diagram. Most of them do not apply since (after merging the threads) it is now obvious what the output voltage range should be).

Note that the lm317 requires an input voltage 2.5V higher than the output voltage to maintain regulation. At low currents this falls a little (but not dramatically). At 200mA you need about 1.8 volts. That means with a 5V input, the best you're going to get at the output (and have it stay regulated) is 3.2 volts.

However there are a plethora of low-dropout regulators which can regulate to voltages much closer to the input voltage. You might consider using one of them. I will mention a couple (based on nothing more than they are some I happen to have) MC33269DT-ADJ, and LM1086CSX-ADJ. Note that these are not a panacea -- they typically have much more stringent constraints on input and output capacitors.

Another word on that board attached to the battery. It is most likely to be the monitoring/protection circuit that prevents (by disconnecting the battery) gross overcharge or discharge. In some cases they also count the number of charge/discharge cycles and prevent recharge after a fixed number of complete cycles (not likely in this case).
 

joshzstuff

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I believe that you modified a power supply for approx 3.7 volts

I did, thanks to the help here.
I assumed that my larger replacement battery would work without issue,
however I ran into a snag and I have not successfully got the cameras to work after disconnecting the original batteries.(and protection circuit)

Another word on that board attached to the battery. It is most likely to be the monitoring/protection circuit that prevents (by disconnecting the battery) gross overcharge or discharge.

I believe that you and Resqueline were right about this all along.
You can imagine my caution after It seemed that the removal of this circuit caused damage to my camera :eek:
(I'm not 100% sure that the 'damage' is permanent, but I am acquiring more test subjects in the mean time)

Also, I'm fairly certain that the replacement battery that I'm using also has such a protection circuit built in.

That must mean there is a charger socket on the camera itself. Why can't you power it via this?

It's a long thread Steve, I'll summarize the thought:
The following 3 re-quotes the relevant items for easy reference:

Originally Posted by Josh
I think I need to get as close to 3.7 as I can because the camera will not function while it is charging (4.1-4.2 volts)

Originally Posted by Resqueline
The camera not working when charging is most likely due to some other reasons than voltage.
Removal of the protection/sensor circuit may lead to some objections by the camera. I guess you'll just have to try & see.

Originally Posted by Josh
Tell me what you think . . .
what are the chances that devices like this will not operate when plugged in is because they lack the filters necessary to handle 'dirty' power that will cause them to malfunction, or damage?
That was the reason for my question above, I think the camera disconnects all operations when charging to protect if from, say the possibility of being connected to a cheap $5 noisy charger.




If you're only looking to fine-tune the voltage then I would look at a low dropout three terminal regulator powered directly from one of those (unmodified) 5V regulators.

I'm listening.
Does any of the designs (unmodified) A-C have a particular advantage? (i.e. cleaner power)

You should choose resistors so that the desired voltage is about mid-range for the trimpot and so that the extreme travel of the trimpot allows for a guaranteed range of voltages (taking into account component tolerance). That can all be a pain to calculate :) Figure out the resistance required for 5 volts and get a trimmer as close as possible to that. That will allow you to change the voltage between 1.25 and 5 volts which covers your desired range nicely without crowding it all up at one end of the trimpot's range.

I'm new, so I am just learning how to size these components.
Thanks for your help & advice when it comes to this.


The recommendation about input voltage was based on the values you put into the circuit diagram. Most of them do not apply since (after merging the threads) it is now obvious what the output voltage range should be).

That means with a 5V input, the best you're going to get at the output (and have it stay regulated) is 3.2 volts.
A thought I had concerning this issue:
Change out the zener to a HIGHER value so as to give the regulator more head room.
One possible issue, though, I don't know what my upper limit is.
I could experement with it though, since I have the components on hand.

As you state below, there are indeed a plethora of components to choose from.
I have bought the ones recommended to me by people that can help me with the supporting circuitry.

The LM337 was suggested to me, however without a diagram.
So I found a diagram online:
[See web page & video below]
http://www.instructables.com/id/Variable-DC-Power-Supply-for-15/

He appears to use 4X AA batteries with a control range of 7.2V to 0.06V
It seems like he is really turning his potentiometer, but I guess you really can't tell from the video.

However there are a plethora of low-dropout regulators which can regulate to voltages much closer to the input voltage. You might consider using one of them. I will mention a couple (based on nothing more than they are some I happen to have) MC33269DT-ADJ, and LM1086CSX-ADJ. Note that these are not a panacea -- they typically have much more stringent constraints on input and output capacitors.

It's almost overwhelming the variety of components available to a newcomer.
Ideally I'd like to have the best one for the job, but especially in the early circuit building stage, I will settle for one that is adequate.

Again, I'm willing to buy any component that I can get support for. So if you can give me any insight about the models you have on hand I'll follow you with purchases and circuit design.

I do lean towards smaller components (so long as they are manageable) The regulator I have was a little larger then I expected.

I've had some bumps in the road, but Thanks for all of your help.
I really feel I'm getting somewhere in this project (albeit slowly)

Now, If I can just get a camera working on a non-stock battery . . .
then I can test AC power!
 
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