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Five LM217 current regulators in parallel is OK?

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Flyback

Well-Known Member
Hello,

Is the following 400mA NiCd battery charger, involving five LM217 current regulators in parallel OK?

Each LM217 is supposed to pass 80mA, but i wondered if parallel connection will make only one LM217 pass the 80mA, and the others will not be able to?
 

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  • NiCd Battery charger ..400mA (0.1C).pdf
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Why not use a single LM317? It will handle 400mA with the right heat sink.

this is just a charger for use in the lab, and we dont want to bother with a heatsinked 1.5A device and hassle with heatsink paste etc......its easier to just solder 5 TO220's in parallel.........the Vin is 15V, and the battery could be down as low as 0V, so five in parallel each carrying 80mA means theres definetely no heatsink needed for any of the five LM217's.

We need this charger for the lab because the sales and apps guys are constantly coming to us for fully charged batteries for the demo kits, and we cannot find any off the shelf NiCd chargers for five D cells in series.
We will get about 10 of these built , and we can leave the batterys on them overnight and they will be charged fully in the morning...(its a C/10 charger).

Min current is no problem as C/10 is 400mA for us.
 
Hello,

That should work because they are all current sources. Kind of an interesting idea because it spreads the heat out nicely.

It's too bad that the tab is Vout and not Vin. If the tab was Vin you could bolt all the tabs together, but it is Vout and those must be kept separate. But all the Vin pins can be tied together, and all the Adj pins can be tied together also because they all go to the load.

It doesnt hurt to build one up and test it either :)
 
this is just a charger for use in the lab, and we dont want to bother with a heatsinked 1.5A device and hassle with heatsink paste etc......its easier to just solder 5 TO220's in parallel.........the Vin is 15V, and the battery could be down as low as 0V, so five in parallel each carrying 80mA means theres definetely no heatsink needed for any of the five LM217's.

We need this charger for the lab because the sales and apps guys are constantly coming to us for fully charged batteries for the demo kits, and we cannot find any off the shelf NiCd chargers for five D cells in series.
We will get about 10 of these built , and we can leave the batterys on them overnight and they will be charged fully in the morning...(its a C/10 charger).

Min current is no problem as C/10 is 400mA for us.
With each one dissipating over a watt, they could sizzle spit if you don't have them bolted to something to remove some of the heat.
 
Yes but TO220s can disipate 1W without heatsink......and in any case, once the batt has charged up to 6V , (which is quite quick), then each TO220 only dissipates 400mw.
 
Personally I would go with one LM317 on a heat sink and avoid the hassle of wiring together all those parts. :rolleyes:
 
Personally I would go with one LM317 on a heat sink and avoid the hassle of wiring together all those parts

there on a pcb, so i dont have to wire them, just solder them in
 
Hello again,


This solution isnt as bad as it sounds at first because power losses are often split up like this and that means lower temperatures all around. But i see some other facts about this now that might help to simplify this a little.

First of all, it now sounds like you are charging maybe 4 cells in series for a combined total voltage of 6v. Correct this if wrong.

Well, charging 6v cells with a 15v source is not the best idea in the world. To minimize power lost the source voltage should be close to the total cell voltage. In other words, 14v would be better, 13v better than that, etc., etc., down to 6.01v which would be really great except that would be too low to be practical. So that's the basic theory of minimizing the power lost in the charging process.

To be practical we have to allow some headroom for the control. That means whatever the series pass device is it has to have some voltage differential to work properly. So say we lower the source to 10v instead of 15v. That leaves us with a max of 10 times 0.080 or just 0.8 watts in each device. That considerably lower than 1.2 watts with a 15v source. And since the voltage jumps up quite fast, we might only see half of that in the real life operation. So the key here is to lower the voltage supply that feeds the regulators.

But so far we've been talking about using linear regulators. But why force this on the application. A single chip regulator chip like one from the Simple Switcher line of products originally made by National Semiconductor (now part of Texas Instruments) can do the job with some small modifications. And it would work equally well at 10v and 15v or even higher, and not require a heat sink. These parts use a single ic chip, a couple electrolytic caps, perhaps a 100uH inductor, and a Schottkey diode. Add a transistor for current regulation and you've got a single ic chip current regulator.

So the linear regulator is simpler but should really have the input/output voltage differential matched a little tighter. The switching regulator is a little more complex but only requires that one IC chip and a few other parts.
 
Hello again,


This solution isnt as bad as it sounds at first because power losses are often split up like this and that means lower temperatures all around. But i see some other facts about this now that might help to simplify this a little.

First of all, it now sounds like you are charging maybe 4 cells in series for a combined total voltage of 6v. Correct this if wrong.

Well, charging 6v cells with a 15v source is not the best idea in the world. To minimize power lost the source voltage should be close to the total cell voltage. In other words, 14v would be better, 13v better than that, etc., etc., down to 6.01v which would be really great except that would be too low to be practical. So that's the basic theory of minimizing the power lost in the charging process.

To be practical we have to allow some headroom for the control. That means whatever the series pass device is it has to have some voltage differential to work properly. So say we lower the source to 10v instead of 15v. That leaves us with a max of 10 times 0.080 or just 0.8 watts in each device. That considerably lower than 1.2 watts with a 15v source. And since the voltage jumps up quite fast, we might only see half of that in the real life operation. So the key here is to lower the voltage supply that feeds the regulators.

But so far we've been talking about using linear regulators. But why force this on the application. A single chip regulator chip like one from the Simple Switcher line of products originally made by National Semiconductor (now part of Texas Instruments) can do the job with some small modifications. And it would work equally well at 10v and 15v or even higher, and not require a heat sink. These parts use a single ic chip, a couple electrolytic caps, perhaps a 100uH inductor, and a Schottkey diode. Add a transistor for current regulation and you've got a single ic chip current regulator.

So the linear regulator is simpler but should really have the input/output voltage differential matched a little tighter. The switching regulator is a little more complex but only requires that one IC chip and a few other parts.
I think you are oversimplifying the linear regulator requirements.
In his circuit, the batteries need to charge to about 8.25V (see graph).
NiCad Charge curve.PNG
This is from https://www.electrodynam.com/rc/totm/totm1100.shtml

The series diodes require about 1V total. The current set resistor requires another 1.25V. The dropout voltage of the regulator is about 1.7V typical at 80mA. This means that the supply voltage must be at least 12.2V.
 

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  • nicd-chg.gif
    nicd-chg.gif
    11.6 KB · Views: 184
I think you are oversimplifying the linear regulator requirements.
In his circuit, the batteries need to charge to about 8.25V (see graph).
View attachment 73872
This is from https://www.electrodynam.com/rc/totm/totm1100.shtml

The series diodes require about 1V total. The current set resistor requires another 1.25V. The dropout voltage of the regulator is about 1.7V typical at 80mA. This means that the supply voltage must be at least 12.2V.

Hello there,

I have no problem with that. If he really needs 8v instead of 6v then clearly that 10v i quoted must be raised by at least 2 volts. I didnt know how many cells he was working with so i stated some general information.

So he has five cells in series then?
 
yes its five D cell nicd's in series......no off the shelf charger solutions exist for battery packs at the required 400mA current.....offtheshelfers are all for single cells
 
Hello there,

I have no problem with that. If he really needs 8v instead of 6v then clearly that 10v i quoted must be raised by at least 2 volts. I didnt know how many cells he was working with so i stated some general information.

So he has five cells in series then?
It's on his schematic, but it's easy to miss.:)
 
Hi,

Wow, lots of stuff in series with the cells !
It's not just the cells, there are two diodes and a mosfet in series with the cells, plus a 1 ohm resistor.

The two diodes could require up to 0.8v each, and the 1 ohm resistor another 0.4v, so that's another 2 volts right there. So there is not much room for left.

The diode and resistor on the output consume power but they do take that power from the M317's so that's not too bad, and the input diode takes some too. So the dissipation of the LM317 is already limited to less than a watt.

Lets see, 15-0.8-0.8-0.4=13, so we are starting with about 13v or so not 15. 13 times 0.080 gives us 1.04 watts. So we end up with 1 watt plus a little more.

Now all we have to do is add a tree, a golf cart, and a pigeon coup in series with the 15v supply and we'll be all set. We've already got everything else so what the heck :)
 
Hi,

Wow, lots of stuff in series with the cells !
It's not just the cells, there are two diodes and a mosfet in series with the cells, plus a 1 ohm resistor.

The two diodes could require up to 0.8v each, and the 1 ohm resistor another 0.4v, so that's another 2 volts right there. So there is not much room for left.

The diode and resistor on the output consume power but they do take that power from the M317's so that's not too bad, and the input diode takes some too. So the dissipation of the LM317 is already limited to less than a watt.

Lets see, 15-0.8-0.8-0.4=13, so we are starting with about 13v or so not 15. 13 times 0.080 gives us 1.04 watts. So we end up with 1 watt plus a little more.

Now all we have to do is add a tree, a golf cart, and a pigeon coup in series with the 15v supply and we'll be all set. We've already got everything else so what the heck :)
The current set resistor has 1.25V across it. You forgot that.:)
 
Hello there Ron,


Well at least i didnt forget the golf cart :)

So we are down somewhere just over 11v. Some NiCd's could show more than 1.65v terminal voltage too so lets say 1.8 for some margin. That leaves us with about 2v. Another diode or two perhaps to drop some of that 15v ? :)

At least the power consumption in the LM317 is down to around 900mw, not too bad for a single package with no heatsink. I'd test it anyway first though. We dont always want smokin' hot parts either.
 
If one of those LM217s fails, it brings circuit safety into question. There is no thermal protection for this circuit.
 
If one of the LM317's fails open it will only mean that the charge current goes down, no other devices affected. If one fails short then there's a problem of course, but that's the same with any circuit like this. Maybe you are suggesting that a secondary form of circuit protection is to be added.
 
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