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Newbe Q's regulators


By stylen, Section Homebrewed Electricity
Posted on Mon Apr 5th, 2004 at 08:21:51 PM MST
I need to make a battery pack for our camara.

I need to make a battery pack for our camara. I have 7.2 volt bat. and need to regulate down to 4.2v. what do I need and how to put together.
Newbe Q's regulators | 11 comments (11 topical, 0 editorial)

Re: Newbe Q's regulators (none / 0) (#1)
by kell on Mon Apr 5th, 2004 at 08:45:48 PM MST
(User Info)

In a case like yours the usual approach would be to use a "linear" regulator.  This is just a chip with three pins.  One is the ground pin (negative), one is for power in (from your 7.2 volt battery), and the last pin puts out a regulated voltage with respect to ground.  The standard chip with the closest voltage to what you want is a 5 volt regulator chip.  Every Radio Shack has them.  You can run the output through a diode to drop the voltage down about 0.7 volts, and you have about 4.3 volts, just about what you need.  
The only catch to this plan is that the regulator chip can only put out a limited amount of power.  What is the current draw of your camera?



Re: Newbe Q's regulators (none / 0) (#2)
by stylen on Mon Apr 5th, 2004 at 09:20:17 PM MST
(User Info)

The camara is 4.2v at 1.5 amp DC

[ Parent ]


Re: Newbe Q's regulators (none / 0) (#3)
by MP on Mon Apr 5th, 2004 at 09:50:39 PM MST
(User Info)

Hmm...4.2V seems like a strange voltage. I wonder how critical this is. If you need a fair amount of current and exactly 4.2V, you can tie the 7.2V battery positive lead to the collector of an NPN transistor. Also, on the collector, a 2 resistor voltage divider to ground on the 7.2V battery and also it will be the ground side of your 4.2V circuit. The voltage divider will be calculated to have 4.2V at the point where the two resistors are connected together. This point will be tied to the base of the transistor. (one of the resistors can be a pot for precision tuning). Now, the Emitter side of the transistor will give you the voltage you need.
What NPN transistor? Depends upon how much current you need. You can use anything ranging from a 2N3904 to a 2N3055. Since it is from a battery, and powering a camera, you are not going to need anything as hefty as a 2N3055. I am betting on the 2N3904 as being enough.



Re: Newbe Q's regulators (none / 0) (#4)
by dozer on Tue Apr 6th, 2004 at 01:50:22 AM MST
(User Info)


In terms of linear 3-terminal regulators, the one to use is the LM317 family.  Much simpler than NPN xsistors, etc..

All you need is the 3-pin IC and 2 resistors and a non-critical input capacitor.  Do some google-searching on LM317 circuits and you'll find all the info you need.

1.5 amps is barely within the capacity of a 317, but with such a low delta-V across it, the typical part will do it.  A minimum spec part might current-limit.

An LM338 is a beefed-up version of the 317, capable of 5 amps.

Both of these parts (and all the LM78xx series as well) require 1.5 to 2.5 volts of "headroom".  I.e., the input must be higher than the desired output by this amount, or they'll go out of regulation (the output will start to drop).

So the 78xx and 317/338 might be marginal as the battery voltage drops.  There are "LDO" 3-pin regulators as well (Low Dropout Voltage).  There are dozens of different part #'s, and they need more care with capacitor selection.  Most LDO's in this current range will still need around 1 volt of headroom.  

I can't remember the adjustable-version part# off the top of my head, but you might do some searching on LM2930, LM2931, and see what you find.  Better yet, hit the National Semiconductor website and search by category.  I.e. "voltage regulators", subcategory "low dropout", etc. etc...



how 4.2v? (none / 0) (#5)
by Norm (peppysue@suite224.net) on Tue Apr 6th, 2004 at 07:46:40 AM MST
(User Info)

Do not hook your camera up to 7.2 volts on the basis of what I am about to relate!  
I have a camera that says on the input adapter plug in (5v) the camera takes 4 alkaline batteries each when brand new are about 1.7v ....so 4 x 1.7 = 6.8v.

4 NiMHd rechargable batteries in series total are about 5.6v (about 1.4v each) and they'll keep on running the camera even when the voltage drops down to 4.5 (maybe even 4.2v when the camera is working)

 I can take 4 alkaline AA batteries the multimeter indicates like 4.8 volts...yet when I try to run the camera....Zilch!

 Of course you know the reason why...it's open voltage...not voltage under load.

 I thought of using a 6v SLA battery(if I could have regulated the voltage to 5-6volts) but since the 5v input doesn't work anymore I've found it's much easier to just keep 2 or 3 sets of NiMh AA batteries charged up.

 So are you sure you need exactly 4.2 volts?
 Mine works just fine on 4 NiMh AAs. 6.8v - 4.5v.(open voltage)  
   
( :>) Norm



Re: how 4.2v? (none / 0) (#6)
by brock (nevermab at uwgb dot edu) on Tue Apr 6th, 2004 at 07:54:01 AM MST
(User Info)

I would try 5 NiMH AA's in series, or maybe even 6.  If you can check the voltage open on the power supply for the camera, I would bet it is about 7v anyway.

[ Parent ]


Re: Newbe Q's regulators (none / 0) (#7)
by kell on Tue Apr 6th, 2004 at 07:25:11 PM MST
(User Info)

I believe the LM317 and LM338 are the adustable regulators.  You use a voltage divider consisting of two resistors to set the output voltage.  You can actually wire any three-pin regulator exactly the same way and get adjustable ouput, but in your case it's not necessary; as someone stated before, camera typically run on a fairly wide range of voltages (I know mine does) so all you have to do is get close.  A five volt regulator (7805) would do it if it could carry 1.5 amps, but they are rated for 1 amp, with a heatsink.  You can use a pass transistor to carry the current.  Use a NPN transistor like the 2N3904 or 2N3055, as mentioned.  Connect the positive of your 7.2 volt battery to the 5 volt regulator input AND to the collector of the transistor.  Connect the regulator output to the base of the transistor.  The ground terminal of the regulator (7805 chip) is connected to the negative of the 7.2 volt battery, and also serves as the negative for the final output of the circuit.  The transistor's emitter is your positive output voltage; it will give you somewhat less than 5 volts, because there is the equivalent of a diode drop across the transistor's junction.  As a result it will give you 4.3 volts nominally.

You may have to mount the transistor on a heat sink.  The heat sink would have to be isolated electrically from ground, unless you can get an insulator for the transistor.  It's a mica washer.  Radio Shack used to carry them for the 2N3055 and other transistors with the same shape as the 2N3055 (a flat oval called the TO-3), but they discontinued it.  Some stores that carry a lot of inventory may still have a few of them left.  It's called "transistor mounting hardware".  If you use a transistor that has the TO-220 shape which is rectangular with a tab and hole for a screw, Radio Shack still carries mounting insulators for them.  So that might be the way to go.  The TO-220 transistors are easier to deal with anyway, since you don't need to drill holes for the leads to go through the heat sink, like on a TO-3.

With a pass transistor your regulator chip won't even get warm, at the current level we're talking about.



Re: Newbe Q's regulators (none / 0) (#8)
by kell on Tue Apr 6th, 2004 at 08:20:18 PM MST
(User Info)

There's a quick and dirty way to drop voltage.  If you put four diodes in series you will drop your voltage down to about where you want it.  Just connect them end-to-end so the current from the 7.2 volt battery goes through all the diodes one after another.  Each diode has a striped end and an end without a stripe.  Take the first diode in the string and connect the end without a stripe to the positive from your 7.2 volt battery.  Then connect all the others so they're pointing in the same direction.  The striped end of the last diode will be at (approximately) 4.4 volts with respect to the negative post of the 7.2 volt battery.



Re: Newbe Q's regulators (none / 0) (#9)
by RobD on Wed Apr 7th, 2004 at 07:32:46 AM MST
(User Info) http://www.dsgnspec.com

When using pass transistors it is better to use a PNP configuration over an NPN.
Because you're not versed in electronics I would either use the diode dropping method, use silicon diodes, as not all diodes have a .7 volt drop (you can use 1N5401 diodes rated 3 amps) or a linear three terminal regulator. If you decide to use the linear regulator I advise the NS ( National Semiconductor) LM1084 or LM1085 with a heat sink.They are a TO-220 package and are easier to mount over the TO-3 package. The power dissipated by the device is (7.2-4.2)x 1.5 or 4.5 watts.
If you don't want to get the LM1084 you can go with a couple of standard Radio Shack LM317T regulators. Run them in parallel and use a .1 ohm 1/4 watt or better resistor from each output to adjust for any small voltage difference between the 317s. Each will drop about 2.25 watts and you should be able to get away without any heatsinks.
I did this circuit on the simulator but the values should work fine but check the final voltage before attaching it to your device.
RobD





Re: Newbe Q's regulators Rob D (none / 0) (#10)
by stylen on Wed Apr 7th, 2004 at 06:55:02 PM MST
(User Info)

In the pic. what are the  240 and 560  and the lm317 is positive  where dose the neg. come in could you draw a small pic of what it would look like finished and lable it.

[ Parent ]


Re: Newbe Q's regulators (none / 0) (#11)
by kell on Wed Apr 7th, 2004 at 09:50:02 PM MST
(User Info)

At the very bottom is the symbol for ground.  That is the negative, for both input and output.
The zigzag lines with the numbers next to them are resistors.  The numbers are the ohms of the resistors.
We've gone about as far as we can in explaining, the rest is up to you.  Look for a tutorial on elementary electronics on the web or a book.  You'll have lots of fun, take the first step of the journey now!



Newbe Q's regulators | 11 comments (11 topical, 0 editorial)
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