Reduce lipo charger Ma outout

[mrfliboy]

I'm currently running a 2500 Apache Charger in the gym where we fly.

My 2 available settings are 250 ma or 700ma. Is there a way to reduce the 700ma to charge a 400ma lipo??


Thanks.

[Miami Mike]

There's no need to do that. Just charge it with the 250 ma setting. The charging rate of a charger and the MaH capacity of a battery are two separate quantities that don't have to match.

[mrfliboy]

There's no need to do that. Just charge it with the 250 ma setting. The charging rate of a charger and the MaH capacity of a battery are two separate quantities that don't have to match.


at a rate of 250 we are looking at close to 1-1/2 hrs or so

at 450 or so we can be under a hour.

It's all about time LOL

[rich smith]

Use the 700ma setting. Small batteries like that can go way over 1C w/o getting warm. Aero Aces charge at 4C and, contrary to popular opinion, this does not shorten life. That's why they only take 15-20min. Heat shortens life. High discharge current and low voltage does this.

Some of my 450mah were charged at 400mah and others at 1.2a as a test. Both went couple hundred cycles, no difference.


I'm currently running a 2500 Apache Charger in the gym where we fly.

My 2 available settings are 250 ma or 700ma. Is there a way to reduce the 700ma to charge a 400ma lipo??


Thanks.

[mrfliboy]

Use the 700ma setting. Small batteries like that can go way over 1C w/o getting warm. Aero Aces charge at 4C and, contrary to popular opinion, this does not shorten life. That's why they only take 15-20min. Heat shortens life. High discharge current and low voltage does this.

Some of my 450mah were charged at 400mah and others at 1.2a as a test. Both went couple hundred cycles, no difference.


Rich,

I tend to agree with that statement. But I also like to charge at 1c whenever possible.

[Dan Baldwin]

I'm currently running a 2500 Apache Charger in the gym where we fly.

My 2 available settings are 250 ma or 700ma. Is there a way to reduce the 700ma to charge a 400ma lipo??


Thanks.

It's not a good idea to try to reduce the charging current of a charger like the Apache. You could add a resistor in series, and the charger would immediately go into constant voltage mode, but if the initial charging current is 400 ma, the current would begin to taper, and it would end up taking longer to charge than if you had just used 250 ma.

You could build a dedicated Henion charger, or re-work one of the chargers you built a while back, or you could build a charger from one of my old threads. How many cells are the packs you will be charging?

Dan

[rich smith]

I can verify from personal experience that adding current limiting resistor to chargers don't work. Best not to worry about higher current, what's a few milliamps between friends? :)

It's not a good idea to try to reduce the charging current of a charger like the Apache. You could add a resistor in series, and the charger would immediately go into constant voltage mode, but if the initial charging current is 400 ma, the current would begin to taper, and it would end up taking longer to charge than if you had just used 250 ma.

You could build a dedicated Henion charger, or re-work one of the chargers you built a while back, or you could build a charger from one of my old threads. How many cells are the packs you will be charging?

Dan

[gigelus2k3]

I can verify from personal experience that adding current limiting resistor to chargers don't work.

Depends on where you add the resistor. If you open up any charger, you will notice a large body resistor connected to the output, having a value in the range of tens of milliOhm. In CC mode, the controller feedback loop keeps a constant voltage across that resistor. If you double the resistor value, the current is reduced to 50%.

In the case of a multi-value CC current level, the most likely way to control it is by changing the regulated voltage across the resistor. If the current selection is done via switches, this involves resistor dividers of some sort; changing these dividers will change the regulated voltage, thus the current. This has no effect on the CV operation.

Back to the OP, can you get the charger's schematics?

[mrfliboy]

It's not a good idea to try to reduce the charging current of a charger like the Apache. You could add a resistor in series, and the charger would immediately go into constant voltage mode, but if the initial charging current is 400 ma, the current would begin to taper, and it would end up taking longer to charge than if you had just used 250 ma.

You could build a dedicated Henion charger, or re-work one of the chargers you built a while back, or you could build a charger from one of my old threads. How many cells are the packs you will be charging?

Dan


Single cells Dan. I've got way too many chargers floating around as it is.
I currently am running 2- 2500's, the eflite balance charger, 1- SH charger and your dual set up charger.

I kinda figured the resistor idea wasn't the way to go. Figured I ask.

BTW my last SH charger is still going strong along with the one of yours I built.

Thanks for the info guys.

[rich smith]

I've reverse enginneered a dozen or so chargers and designed a couple from scratch for my business. Usually current is controlled by very low R resistor (part of divider) deep in the charge circuit, not in series with the output. Putting one there at best has very unsatisfactory result like Dan mentioned. Usually don't work at all.


Depends on where you add the resistor. If you open up any charger, you will notice a large body resistor connected to the output, having a value in the range of tens of milliOhm. In CC mode, the controller feedback loop keeps a constant voltage across that resistor. If you double the resistor value, the current is reduced to 50%.

In the case of a multi-value CC current level, the most likely way to control it is by changing the regulated voltage across the resistor. If the current selection is done via switches, this involves resistor dividers of some sort; changing these dividers will change the regulated voltage, thus the current. This has no effect on the CV operation.

Back to the OP, can you get the charger's schematics?

[Bruce Abbott]

The SM-2500 uses a 0.1 Ohm power resistor to sense the charging current. If you change this to 0.175 Ohms, it will reduce the charging current from 700mA to 400mA (and 250mA would reduce to 143mA, etc.). 0.175 Ohms is a non-standard value, but you could create it using 2 x 0.15 Ohms in parallel to make 0.075 Ohms, then put them in series with the original resistor.

[rich smith]

It might make more sense to change the other resistor in that divider because it's a common value (910ohm). Low ohm resistors are hard to find and expensive. Or better still a pot like in some commercial chargers.

The SM-2500 uses a 0.1 Ohm power resistor to sense the charging current. If you change this to 0.175 Ohms, it will reduce the charging current from 700mA to 400mA (and 250mA would reduce to 143mA, etc.). 0.175 Ohms is a non-standard value, but you could create it using 2 x 0.15 Ohms in parallel to make 0.075 Ohms, then put them in series with the original resistor.

[Dan Baldwin]

It might make more sense to change the other resistor in that divider because it's a common value (910ohm). Low ohm resistors are hard to find and expensive. Or better still a pot like in some commercial chargers.

The voltage divider formed by R25 (910 ohms) and R20 (10K) only reduces the voltage from the current sense resistor by a factor of .916. If the 910 ohm resistor was shorted or the 10K resistor was removed, it would only reduce the charge current from 700 ma to about 645 ma.

The gain setting resistor ladder formed by R30 (10K) and R31 (910 ohms) could be modified, however. Adding another resistor across R31 would increase the gain of the current amp, and result in a lower current. The value of that resistor would probably be about 1.2K. Probably the easiest way to make sure you could hit the right current would be to put a 1K reistor in series with a 1K pot, and put that across R31. You could add a switch or another jumper to turn your hack on and off.

Dan

[rich smith]

Yes, adding a resistor across r31 is best fix. Personally I'd just use the charger the way it is for that battery.

The voltage divider formed by R25 (910 ohms) and R20 (10K) only reduces the voltage from the current sense resistor by a factor of .916. If the 910 ohm resistor was shorted or the 10K resistor was removed, it would only reduce the charge current from 700 ma to about 645 ma.

The gain setting resistor ladder formed by R30 (10K) and R31 (910 ohms) could be modified, however. Adding another resistor across R31 would increase the gain of the current amp, and result in a lower current. The value of that resistor would probably be about 1.2K. Probably the easiest way to make sure you could hit the right current would be to put a 1K reistor in series with a 1K pot, and put that across R31. You could add a switch or another jumper to turn your hack on and off.

Dan

[gigelus2k3]

I've reverse enginneered a dozen or so chargers and designed a couple from scratch for my business. Usually current is controlled by very low R resistor (part of divider) deep in the charge circuit, not in series with the output. Putting one there at best has very unsatisfactory result like Dan mentioned. Usually don't work at all.

Sorry, but Bruce's posted schematics shows that the sense resistor is in series with the battery and that's how all chargers work, assuming they use a resistor to sense the current. The voltage developed across the resistor is then compared to a reference voltage (given here by the PIC's PWM output on pin #12 and filtered by the LPF R32/C19); the comparison result is used to drive the DC/DC so that the two values become equal.

As Dan mentioned already, the easiest fix is to add a second resistor in parallel with R31.

B.t.w., how on earth is the power stage working? I've never seen such a topology before. Are the two inductors coupled?

Cheers,

Serban

[vintage1]

B.t.w., how on earth is the power stage working? I've never seen such a topology before. Are the two inductors coupled?

Cheers,

Serban
Something clearly wrong there..I'd say the second inductor would be in series with the line..and the back to back Zeners don't make sense either..should be Shottkies. And at least one diode should be clamping the flyback to ground..

But it might work as some kind of tuned circuit..still it looks dubious to me.

Normally you short the coil to deck. then let it fly..where upon output voltage is way above rail till it rings and comes below rail. Then it needs clamping, or you have to get the switcher on fast to stop reverse voltage on it.

[Bruce Abbott]

Adding another resistor across R31 would increase the gain of the current amp, and result in a lower current.However, that would unbalance the opamp's inputs, which could cause inaccurate current measurement. Also, soldering onto the tiny SMD resistor would be a fiddly job. I chose R14 because it is a large, easily identified component, with wire leads. Simply disconnect the ground leg of R14 and add the 0.075 Ohm resistor in series with it. If you want to be able to select the original current steps then wire a switch across the new resistor.

how on earth is the power stage working? I've never seen such a topology before. Are the two inductors coupled?SEPIC (http://en.wikipedia.org/wiki/SEPIC_converter)

BTW, please don't try to build a charger from that schematic - it may not be 100% accurate :o

back to back Zeners don't make senseThey're not zeners, and they're not back to back ;)

[mrfliboy]

Wow, lots of ideas a rolling now.

Thanks to all that have replied here.

The R14 idea seems the easiest. I'll open one of my chargers up and take alook see.

[rich smith]

Resistor is not current limiting resistor between battery and charger which was original subject. That fix just don't work.


Sorry, but Bruce's posted schematics shows that the sense resistor is in series with the battery and that's how all chargers work, assuming they use a resistor to sense the current. The voltage developed across the resistor is then compared to a reference voltage (given here by the PIC's PWM output on pin #12 and filtered by the LPF R32/C19); the comparison result is used to drive the DC/DC so that the two values become equal.

As Dan mentioned already, the easiest fix is to add a second resistor in parallel with R31.

B.t.w., how on earth is the power stage working? I've never seen such a topology before. Are the two inductors coupled?

Cheers,

Serban

[rich smith]

Actually back-to-back zeners make huge sense in some apps. The forward biased one exactly cancels the reverse one for perfect temperature compensation. In fact precision (1%) voltage references like LM series are little more than selected back-to-back zeners. This was one of my products as semi engineer at Sprague back in the 70s.

The cheap Apache is a great little charger and one of my favorite commercial units. My $5 Turnigys can't easily do 1 cell.

e..and the back to back Zeners don't make sense either..should be Shottkies. And at least one diode should be clamping the flyback to ground..

But it might work as some kind of tuned circuit..still it looks dubious to me.

[Dan Baldwin]

However, that would unbalance the opamp's inputs, which could cause inaccurate current measurement. Also, soldering onto the tiny SMD resistor would be a fiddly job. I chose R14 because it is a large, easily identified component, with wire leads. Simply disconnect the ground leg of R14 and add the 0.075 Ohm resistor in series with it. If you want to be able to select the original current steps then wire a switch across the new resistor.

SEPIC (http://en.wikipedia.org/wiki/SEPIC_converter)

BTW, please don't try to build a charger from that schematic - it may not be 100% accurate :o

They're not zeners, and they're not back to back ;)

While the battery voltage op amp is a differential to single ended converter because the negative lead of the battery is not necessarily at ground, the current amp has ground referenced input, and ground referenced output, so it's single ended on both sides. It has no balance to become unbalanced.

Dan

[gigelus2k3]

Bruce may be thinking about LM358's input bias currents (this is a bipolar opamp). However, that's mostly academic, because of the low-valued resistors used.

Anyway, thank you Bruce for the SEPIC link!

Serban

[Dan Baldwin]

The current gain stage is probably set up as a differential to single ended converter to eliminate any possible error caused by voltage drop across the traces, but that error would probably be very small (depending on circuit board layout), and it should be constant for a given current.

Typical input offset current for the LM358 is 5 na, so unbalancing the input impedances shouldn't cause any measurable differences in this circuit.

Although D1 and D2 seem to be drawn as zeners, they would in fact be schottkys. They are in parallel to increase the current handling capabilities, and decrease voltage drop. I have no idea why D2 is placed between the capacitor and the output. It only seems to be there to keep the capacitor from being charged by the battery when the charger is not active. Perhaps it's to eliminate the big current spike when the battery is first connected.


Dan

[gigelus2k3]

The second diode is probably an "anti-dummy" protection, in case the user connects the battery backwards.

If the impedance is different on the two inputs, the bias current starts playing a role (Ib is usually at least 10 times larger than Ios for bipolar based input stages). Anyway, for an impedance mismatch of 0.5kOhm, the LM358's Ib causes ~ 25uV of offset, which means nothing.

Serban

[mrfliboy]

Quote: The second diode is probably an "anti-dummy" protection, in case the user connects the battery backwards.


The charger is great in the fact that you can hook up a lipo wrong and all you get is flashing error. I use it for testing of weird lipos that have weird connectors and such for polarity.


Dan, whats your thoughts on the R14 idea??

[everydayflyer]

Seemns like you have a good supply of chargers and I would guess LiPlys also so why not just wait until you have two packs discharged and charge them in parallel at 750 mA.?

Charles

[gigelus2k3]

ok, ok!

[rich smith]

This is not true. It is obvious you have never actually measured the current between a fully charged and fully discharged in parallel.

Please don't put two LiPos in parallel! Even discharged, if they have different residual voltages, the higher voltage one will dump a very high and uncontrolled current in the lower one, probably damaging it.

[Dan Baldwin]

The second diode is probably an "anti-dummy" protection, in case the user connects the battery backwards.

If the impedance is different on the two inputs, the bias current starts playing a role (Ib is usually at least 10 times larger than Ios for bipolar based input stages). Anyway, for an impedance mismatch of 0.5kOhm, the LM358's Ib causes ~ 25uV of offset, which means nothing.

Serban

FET U4 is the "anti-dummy" reverse connection protection for the output. D2 only isolates C5 from the output.

Quote: The second diode is probably an "anti-dummy" protection, in case the user connects the battery backwards.


The charger is great in the fact that you can hook up a lipo wrong and all you get is flashing error. I use it for testing of weird lipos that have weird connectors and such for polarity.


Dan, whats your thoughts on the R14 idea??

Adding resistance to R14 should work just fine if you want to make the change relatively permanent. Because of the extremely low resistance involved, it would be difficult to switch the added resistance in and out. The added resistance of the switch and wire would affect the current.




Please don't put two LiPos in parallel! Even discharged, if they have different residual voltages, the higher voltage one will dump a very high and uncontrolled current in the lower one, probably damaging it.

Oh boy. Here we go.

Dan

[everydayflyer]

Dan I figured someone would challange parallel charging but also figured that since single cell 400 mAh packs are the subject and are being used by an experienced flyer that most would realize that with dischaged single cells of such low capacity there is just not a great deal of energy available to be transfered.

Parallel charging / discharging is a subject best addressed in the Batteries and Charger Forum. I was merely offering a simple and safe solution.


Charles

[Dan Baldwin]

Dan I figured someone would challange parallel charging but also figured that since single cell 400 mAh packs are the subject and are being used by an experienced flyer that most would realize that with dischaged single cells of such low capacity there is just not a great deal of energy available to be transfered.

Parallel charging / discharging is a subject best addressed in the Batteries and Charger Forum. I was merely offering a simple and safe solution.


Charles

We have certainly seen it argued a dozen times or so in the batteries and chargers forum. I haven't done any experimentation on parallel charging, so I generally don't weigh in on this battle.

I generally don't charge packs in parallel that weren't flown in parallel, but I have a few times when they were very close to the same voltage before they were put in parallel.


Dan

[mrfliboy]

Seemns like you have a good supply of chargers and I would guess LiPlys also so why not just wait until you have two packs discharged and charge them in parallel at 750 mA.?

Charles


Charles excellent idea.

My objective was to be able to charge my 400ma sgl cells using my 2500 charger. That'll do it.


And the fact that I am talking single 250 to 400 cells makes it perfect.

Thanks Charles

BTW I would prefer any discussions of parallel charging to stay in the Bat/Chg forum.