Will this simple charger work for me?

[BobRCnut]

I found this circuit and writeup online. Can one of you electron wizards give me your opinion as to whether it will handle my particular application? First, the circuit writeup, then my application, then the circuit drawing:

"The circuit below illustrates using a LM317 variable voltage regulator as a constant current source. The voltage between the adjustment terminal and the output terminal is always 1.25 volts, so by connecting the adjustment terminal to the load and placing a resistor (R) between the load and the output terminal, a constant current of 1.25/R is established. Thus we need a 12 ohm resistor (R) to get 100mA of charge current and a 1.2 ohm, 2 watt resistor for 1 amp of current. A diode is used in series with the input to prevent the batteries from applying a reverse voltage to the regulator if the power is turned off while the batteries are still connected. It's probably a good idea to remove the batteries before turning off the power."

Now, here's my application: I have a modified PC power supply that puts out a solid 12.2 volts, with 10 Amp capability. I'd like to use this simple charger circuit (driven by the PC power supply) to occasionally balance the cells in my 7- and 8-cell 2600 NiMh packs by slow-charging them at 260 mAh. My questions: (1) will the 12.2 volts be high enough to slow-charge an 8-cell NiMh pack, since the fully charged voltage of the pack will be close to that of the supply? ..and (2) will this circuit (once properly adjusted via the value of 'R') really maintain a slow-charge current of 260mAh throughout the 14-16 hour conditioning charge cycle? (I normally charge these packs at .5C to 1C with a GWS MC-2002 delta peak charger off the same PC power supply, but its trickle capability is too limited for slow-charging).

[zagisrule!]

First off, remove the diode. You are probably a "big boy" now and can handle polarity duties...plus that diode will have to dissipate some heat if we just left it there.

Second of all, you will loose at least almost a volt with the circuit from the supply, so you should get at least a 15V supply, which might put out 14V.

Third of all, the circuit will indeed keep the current at the same level it is set at so long as you can maintain the proper gap in voltage from supply (output of charger) to the pack. Remember that you must maintain a certain headway of output voltage to keep current up, use ohms law to figure out what you need as far as that voltage goes. If you choose to use your 12V supply, current will drop to nothing once the battery reaches the ouput of the regulator's maximum of maybe 11V.

Also be sure that your resistor is suitibly rated in terms of W. Figure this out like this:

I^2 x R

So, if you have .5A flowing through 4, you would need a 1W resistor at the minimum. Get it? You might also check the thermal coefficient which is rated in PPM (parts per million) to see how much the temperature will effect the resistance...the lower the PPM the better, because that means the resistance will stay closer to it's spec'ed value as temp goes up.





-Matt

[BobRCnut]

Matt, thanks for the very helpful reply. As I suspected, it sounds a bit marginal using my existing 12V supply. I'd thought about eliminating the diode, but wasn't sure, because chances of (for example) a power failure during a 14 hour charge are much greater than the usual 1-hour fast charge. I think I'll scrap this idea and look for a more capable AC/DC type charger. I've heard that there's a workaround for the GWS to cripple the 90-minute safety shutoff, but I don't know if the GWS is reliable at that low a current rate. Letting the GWS go into trickle mode does no good because it simply switches on a 1-Amp current for short periods.

[zagisrule!]

You can leave the diode on there if you want, but you should parallel one or two decent sized ones. I just would not use it because of the 0.6V-0.7V drop they inflict because of the PN semiconductor junction...and you have a low overhead supply so the first goal is to reduce voltage drop to an absolute minimum if you can.

Radio Shack sells 15V, 1A adapters for under $20...which would do the trick nicely, even with the diode.




-Matt

[BobRCnut]

Matt, can you expand on that last statement? You're talking to someone who's kinda prehistoric when it comes to electronics these days. What kind of 'adapter' is it, and how would it apply here? I did some quick searches thru the Radio Shack web site but found nothing that appears to be related...

[blue_bull]

BobRCnut,

LM 317 needs a voltage difference from input to output of 3V at least, voltage drop across resistor must be 1.25V, which is the specific reference voltage level of LM 317. Resistor`s value is 1.25V devided by desired current 0.5A (e.g.) which gives 2.5 Ohms. Wattage is I*I*R = 0.5*0.5*2.5= 0.625 Watt.

Input voltage (without diode) must be at least 4.25V higher than highest battery voltage for proper constant current function.

Depending on voltage across LM 317 (input-output) and current value you may need a heatsink.

Herbert

[Bob Baker]

A simpler solution is to take out the lm317 and the diode and just use a resistor. With the correct resistor value, this will give you a tapered charge, where the current at the end of the charge is c/10, which is what is needed. It usually takes a bit of experimenting with the value, because different types of cells have slightly different volatage at the end of a c/10 charge.

This will work fine with the 7 cell packs; resistor value should be about: R=(12.2-7*1.45)/.27=8.5 ohms. Use 10 ohms which is readily available; 2 watt power rating or higher.

For 8 cells, you're running out of voltage, but it should work: R=12.2-8*1.45/.27=2.2 ohms. The problem is that with slight cell voltage variations, the current will vary a lot; so measure the current. In this case, the starting current will be pretty high, over 2 Amps, depending on the state of the pack. Use a 5 to 10 watt resistor; power=I*I*R or (V*V)/R.

To check the current, just measure the voltage across the resistor; I=V/R.

If you could get a bit more voltage it would help; is your 5V line on the pc supply loaded with a resistor? this will help bring the voltage up a bit.

[zagisrule!]

The adaptor would plug into the wall and output 15VDC. I did not know the 317's drop was so bad, so I would look at an 18V supply. These adapters are also known as "wall-warts" and have plugs on the end to power consumer devices. You can add a plug to your 317 circuit and have it come apart for easy use, or just chop the connector and solder it on.

Be sure you have a heatsink on the 317 too...





-Matt

[zagisrule!]

These sinks would work!


Hehehehe! :D

[BobRCnut]

Once again, thanks, guys, for all the good info...

Herbert, your update pretty much proves that it won't work with my PC power supply. It's an ATX type, so I have a resistor/pilot light parallel'ed across the +5v. output to provide the necessary load and keep the +12 steady.

Bob, I'd thought of that, too, just using a series resistor, but the tapered current starting off at a high level kinda defeats the purpose.. and I never know the exact state of charge of the battery to start with, it all depends on the low voltage cutoff of my ESC and how much I pushed it during the last flight.

Matt, now I know what you're referring to.. in fact, I have an 18v. DC output wallwart right here, that I use for my electric starter NiCd pack, but it only puts out about 200mAh, as does the one for my cordless drill charger. Since this is just a 'fill up some spare time' project, I think I'll hit the Goodwill store today (they have a whole shelf full of assorted wall-warts), and if no luck, Radio Shack is also nearby. I LOVE that heatsink, by the way... did that come from a Warp drive, or one of Ted Nugent's guitar amps? <grin>

[Mr.RC-CAM]

I'd thought of that, too, just using a series resistor, but the tapered current starting off at a high level kinda defeats the purpose..
Constant current sources are ideal, especially if you need to compute mAH intake. However, if you set up a compromise current (perhaps based on 1.35V/cell), then the effective average will probably exceed the necessary accuracy for slow charge and cell conditioning. From what I can see, this is all that your app needs.

And you call THAT a heatsink? Now this is a heatsink!

[BobRCnut]

OK.. managed to find a wallwart DC adapter rated at 16V., 900 mA, so I think I'll try it without the diode. Since the goal here is simply to safely balance the individual cells, I'm guessing that it's better to be on the low side of C/10 (260mA) than the high side, it'll just take longer. The 7-cell pack should be no problem.. the 8-cell I'm not so sure about, because at full charge it's gonna be about 11.5V... leaving little margin to maintain the 260mA desired current. We shall find out, I guess. Given this LM317T circuit, and no diode, how can I measure the actual current into the pack? My cheapie DVM is only ranged up to 200mA...

[zagisrule!]

Current could be measured with a shunt resistor...and why add another one when you can already use the resistor that is acting as feedback for the regulator?

Measure voltage with the red probe on the output of the regulator (before the resistor) and put the black probe on the feedback of the regulator (after the resistor). The voltage should be right around 1.25V. Depending on your resistance, you can calculate the current using ohms law. This is the same way any shunt resistor works, and will tell you quickly whether or not your unit is working.

Here is an interesting thing I am going to tell you...short the output of the circuit to ground to test it...nothing will smoke! This is because the output will be limited to your target current! Cool, huh? So, ground the output and do your voltage test to the resistor and you should have a 1.25V difference in the terminals of the resistor.

You are set at that point, sounds like you found a good adapter cantidate.

For your final design, I would recommend this or something similar:

The large caps are "filters" so to speak, and the smaller units are there to prevent the 317 from breaking into oscillation.



Hehehe...that is one HUGE sink Mr. RC-CAM. I bought this and a few others like it surplus for a shade under $10 each. I have been informed they are an old Wakefield unit... they are going into a 100W MOSFET audio amplifier that needs some beefy heatsinks because of the heavy bias.




-Matt

[zagisrule!]

RC-CAM,

That can has one huge shadow!

:D


-Matt

[Comatose]

Well of course it has such a huge (tapered) shadow, its only inches from the sun. WHy else would anyone need to heatsink a table?

[BobRCnut]

Thanks for all the help, guys.. it works! For R1, I used a pair of 10 ohm resistors in parallel, which calculated out to 260 mA of output current, and the actual measured current is a steady 240 mA, regardless of the load (I temporarily placed different low values of resistance across the output to test with). Close enough for balancing my 2600 mAh packs.

Interestingly enough, the sample circuit on the back of the LM317T package shows R1 in the 'ADJUST' leg of the circuit rather than in the 'OUTPUT' leg, with the 'OUTPUT' lead of the LM317T going directly to the output of the entire circuit.. and believe me, it does NOT work that way. I tried it, and a 1 ohm 10 watt resistor across the overall output got QUITE warm QUITE quickly because there were almost 2 AMPS flowing through it. Obviously, you guys are smarter than Radio Shack... <grin>

[blue_bull]

BobRCnut,

I think the application which RS has made on the package is the way to make a voltage regulation, you made a current regulation. Both things are possible with LM317.

Herbert

[BobRCnut]

Herbert, thanks for the clarification. One last question: before I started putting things together, I checked the DC adapter alone, which is rated at 16V, 900 mA. With no load on the output, it measured 13.7V. I was concerned that I might not have enough 'overhead' for the 8-cell pack. After completing and testing the constant current circuit, I found that the output of the adapter NOW measures 21V. Is this because of the additional input filtering provided by the electrolytic?

[blue_bull]

BobRCnut,

it is nearly impossible to me to give you a remote-diagnosis on that phenomenon without knowing the circuit of your DC adaptor. The thing I can say is, normally output voltage at no load is signifcant higher than rated value if it is a very simple (cheap), uncontrolled adaptor. Rated voltage is at rated current. Maybe you try it again to be sure to not having made a mistake at measuring- or simply forget about it. It works and that was your aim.

Herbert

[Mr.RC-CAM]

After completing and testing the constant current circuit, I found that the output of the adapter NOW measures 21V. Is this because of the additional input filtering provided by the electrolytic?
To save costs, nearly all consumer DC adapters have no capacitive filtering inside them. They rely on the bulk filtering input caps installed in the product that is powered by them.

So, the wall wart's output voltage will eagerly rise when you apply a bulk filtering cap on them. However, the voltage is not very meaningful unless it is measured with your device/load.

RC-CAM

[BobRCnut]

Thank you, RC.. I figured that was the case, but it's been a long time.. the last project I built was a Heathkit, which is a name that's probably totally unfamilar to many of you.

Anyway, the measured output voltage of the adapter, with the current regulator circuit in place and supplying 240mA to my battery pack, is 21 volts, so I'm no longer concerned about not having enough input voltage margin to balance an 8-cell pack. It works.. I'm happy with it.. I'm outta here.

[jperch]

BobRCNut,

I remember Heathkit. I was a kid and could never afford any kits they made but I remember the Heathkit store on 27th ave and Indian School.

Joe

[mt_100]

Will this circuit work for a six cell battery? I need a slow overnight charger and have six and eight cell betteries and will power from a power supply.

Also, when you show the ground in the circuit, is that the negative side of the power supply?

[jperch]

mt_100,

It should work for both 6 cell and 8 cell packs. You just need to make sure your input voltage is high enough to drive the current. Yes, ground is typically the negative side of things.

Joe

[mt_100]

Excellent! I have been looking for a good, simple design for an overnight charger. I have several packs that I would like to use this on.

Will 13.8 volts be enough for an eight cell pack?

Also, the resistors for a 200mah rate, will 1/4w work?

[Norman Adlam]

Herbert, thanks for the clarification. One last question: before I started putting things together, I checked the DC adapter alone, which is rated at 16V, 900 mA. With no load on the output, it measured 13.7V. I was concerned that I might not have enough 'overhead' for the 8-cell pack. After completing and testing the constant current circuit, I found that the output of the adapter NOW measures 21V. Is this because of the additional input filtering provided by the electrolytic?

Bob,

For your info, adding a decent high value capacitor to a full wave bridge rectified output will raise the voltage by 1.414 times, so your 13.7v will end up around 20v (depending on meter acurracy etc).

As Mr RC-CAM states, the meter is taking little load, so that may drop when a load is attached (depending on the ability of the transformer etc to supply the current demanded).

All good fun, ain't it! :D

Cheers,

[mt_100]

I was curious about my questions above before I build it, can anyone help?

[jperch]

mt_100,

From an earlier post in this thread, I see that the regulator has a drop out voltage of about 3V. This means that there needs to be at least 3 more volts at its input than it sees at its output. 8 cells is a nominal 9.6 volts but would have a peak voltage a bit higher. I am guessing around 12V. This would mean that you would need more like 15V input to keep the regulator regulating.

Now I don't mean to get picky on you, but the difference is somewhat important. Your question asks about the resistors for a 200mah rate. The units of mah are actually units of battery capacity. However, many folks use it interchangeably with units of ma, which is a current. So, are you asking for a 200ma current to charge your packs over night, or are you asking for a current to charge your 200mah packs over night? the difference is that you would use a 200ma current to charge 2000mah or larger packs over night. But you would use a 20ma current to charge your 200mah packs over night.

At any rate, let me tell you how to find out what resistor is big enough. The regulator maintains the voltage across the resistors at 1.25V. This is how is regulates the current through teh pack. The power dissipated by these resistors is going to be the product of the voltage across them and the current through them. So, the power would be 1.25V times the current (either 20ma or 200mah depending on your answer to my question above). It is okay to use resistors that are bigger (in wattage) than necessary. For example, if the answer to your question is 200ma, then 1.25V * .2 amps = .25 Watts. So, 1/4 watt resistors should work. They might get a bit hot, but they should work. I would recommend 1.2 watt resistors. If your answer above is 20ma, then 1/4 watt resistors would work very well.

I hope that all helps,
Joe

[Mr.RC-CAM]

A successful way to determine the minimum input voltage for this circuit is to use this formula:
Vin = (Cell Count x 1.6) + 3

To determine the minimum required resistor wattage:
R Watt = (1.25V x charge amps) x 1.25
Note: The 1.25 on the end allows for a 125% safety margin.

To determine the VReg's maximum expected power dissipation for a typical discharged pack:
VReg Watt = (Vin - (Cell Count x 0.9V)) x (charge amps)
Note: Heatsink as required.

RC-CAM

[jperch]

RC-CAM,
Are you sure that your equation for R Watt is correct? From that equation, it looks like the power dissipated in the resistor is equal to 1.25 times the voltage difference between the input and the battery pack times the charging current.

It seems to me that the voltage across this resistor is always 1.25V. That is how the regulator maintains the current. The current through this resistor is the charging current. This would make the power dissipated by this resistor equal to 1.25 times the charging current.

Joe

[Mr.RC-CAM]

Are you sure that your equation for R Watt is correct?
Oops, the formula was mangled. I fixed it, per the edits. Thanks for catching that.

RC-CAM

[mt_100]

Question on the capacitors.

I only have .1uf caps around here, can I substitute them for the 470uf or the 220uf?

[Mr.RC-CAM]

The .1uF would be fine on the input if it is a low ESR ceramic type. I would recommend at least a 1uF on the output for good stability (but, it would probably work fine with nearly anything there).

If this was a commercial design you would want to be choosy. But for a tossed together home brew job, you can usually get away with murder. :)

RC-CAM

[mt_100]

OK, made one of these today. First thing was I didn't have a four ohm resistor to get the 300mah charge rate I wanted so I serial connected four 1 ohm resistors.

I got capacitors from Radio Shack as all I had was .1uf and needed the 470uf and the 220uf. Both of those had two pins with one shorted than the other and black down arrows on one side.

I put it together carefully and it didn't do anything, no output at all. So I thought maybe the capacitors were backwards. I originally put them in with the down arrows on the ground side of the circuit. I reversed them and put them on the positive side. Hooked it into the power supply and a second later the 470 capacitor popped and smoked. Unhooked it and figured it was toast.

The LM317 pins are 1 - 2 - 3 right? That is what I assumed when I hooked it up. I think I had it right the first time with the capacitors but have no idea at this point.

Any ideas?

[jperch]

MT100,
Yeah, you had the caps right the first time. Typically, those black stripes have a "-" sign in them. They point to the negative side of the cap.

As for the pinout of the LM317, I don't know which is 1, 2 or 3. But as you look at the chip with the printing readable, the adjust pin is on the left, the output pin is in the middle and the input is on the right.

Here is the first page of the data sheet with a picture. I hope this helps.

Joe

[mt_100]

Thanks, I thought I had the caps in backwards, and I had the LM317 in wrong as well.

I have a new one working now putting out 300mah into a 6 cell 3000mah pack off my 13.8 volt supply.

I have an 18 volt adapter around here that I may use to do higher packs so it will work out great.

[arneansper]

Nice circuit and works well too. I built mine for forming small NiMH packs, so the charge current is only 38 mA. It is powered from PC PS (12V). Because of the small current I did not put the heatsink to the reg. It is warm, but you can keep it between your fingers, so I suppose it is not too hot.

One thing I noticed was the 3V dropout voltage of the LM317. I looked at the datasheet and there was a diagram that showed the relationship between current, temperature and voltage dropout. I have attached it for reference. For reasonable temperatures and current less than 1A the dropout stays below 2V.

Arne

[warren52nz]

I use two LM317's to do this properly. The first one is set up to regulate the current as shown above and the second one is wired as a voltage regulator with its input connected to the first one's output. I use this to charge Li-Po's. Works beautifully. You need 3 volts minimum for EACH LM317 and heatsinks. If you want to go above 1.5 amps, use LM350K which is a 3 amp version of the LM317, basically.

The current sits at the current set point until the battery is 95% charged (depends on the wire resistance downstream) and then quickly tapers off to zero at the voltage set point.

[jludwick]

Like this?

I use two LM317's to do this properly. The first one is set up to regulate the current as shown above and the second one is wired as a voltage regulator with its input connected to the first one's output. I use this to charge Li-Po's. Works beautifully. You need 3 volts minimum for EACH LM317 and heatsinks. If you want to go above 1.5 amps, use LM350K which is a 3 amp version of the LM317, basically.

The current sits at the current set point until the battery is 95% charged (depends on the wire resistance downstream) and then quickly tapers off to zero at the voltage set point.

[jperch]

Warren52mz,
The circuit you describe (and posted by jludwick) is good for doing lipoly cells. With those, you need a constant current and a constant voltage (at different times). So that circuit will work fine for that. There is a thread around here somewhere (been around a very long time) that covers that circuit in severe detail. I believe that this circuit was intended to be used for NiMH or NiCd cells. They don't need the constant voltage part of the circuit. You just need to make sure you either limit the current to C/10 or keep an eye on the charging so that you don't burn up the cells.

Joe