PDA

View Full Version : AC or battery power for charger


Pages : 1 2 3 [4] 5 6

kamtsa
10-29-2008, 12:05 PM
Any thoughts? Add a diode to the led feed?

I don't think that this will help since the LED is directional anyway.

Possibly a diode between the 5V feed and the charger, such that the resistor to the LED is connected on the feed side of the diode. This way the backfeed from the battery cannot reach the LED.

Caveats:
1. You may need to use higher voltage than 5 V due to the 0.7V or so drop on the new diode.

2. The LIPO will not backfeed the LED but may still backfeed the charger circuitry, draining the LIPO slowly.

I have the same problem and I just make sure not to disconnect the power. Worst case, the $10 LIPO will die.

Kam

shaggybirdman
10-29-2008, 12:14 PM
well as long as you don't leave the battery plugged in there's no problem. i hope my new charger shows up today.

bladelearnin
10-29-2008, 01:40 PM
this is what I got to do this mod

http://i465.photobucket.com/albums/rr16/blademan1963/RESISTOR.jpg?t=1225301916

bladelearnin
10-29-2008, 05:48 PM
i dd this mod and everything works but right before it gets to the end of the charge the led will start a slow flicker and the flicker will turn into a flash abut every 1-2 seconds but it won't go out.My flight time is over 7 minutes though so I guess no problem ttyl blade

shaggybirdman
10-29-2008, 08:52 PM
woot woot the charger is in, but couldn't get it. the battery was dead when i went to get it. by the time i got her started it was to late to get to the hobby shop before closing. oh well. i'll get it tomorrow.

i'll swap the tops, and be modded in no time flat.

gardingh
10-29-2008, 09:38 PM
my problem was just that. i didn't pay attention to the required input polarity. was there others that had a problem? i thought i was the only one who dropped the ball on input polarity.

i'll consider KE4NYV's board, but i already have the parts to mod mine like aristo's diagram. i mite add another switch for charging via my triton. a 12V option is nice, but charging off my car battery really seems like i'd be intending on flying outside where this heli isn't designed to fly. plus i can put the wall wort inside the original box, and carry only 1 thing. my 12V power supply is rather big, and bulky.

I was in the process of wiring my jack and was wondering why he was running things backwards from the way it seemed to me that they should have been. My charger works great though with the mod, thanks a ton Aristo.

So shaggy, did the reversal of polarity end up frying your charger entirely? Could a person put a diode on the supply to prevent reverse polarity issues? I imagine with the additional voltage drop you would have to jack up the supply voltage which is no big deal with my adjustable supply. That brings me to another issue, how would you protect the unit if someone was to plug the supply and it happened to be set at a higher voltage?

KE4NYV
10-29-2008, 10:43 PM
...Could a person put a diode on the supply to prevent reverse polarity issues?

My board has a reverse protection diode for this very reason.

shaggybirdman
10-30-2008, 12:36 AM
well as long as the wall wort is set to 6v, and the center pin is negative your good to go. i'm using a 5.8v wall wort, so i'm safe, but i'd put a piece of tape over the voltage selector if your worried about it.

KE4NYV's board is nice, but then i'd need to carry a brick sized power supply with me if i don't want to use the internal batteries. plus my charger now has a $24 upgrade to it. i really don't need it to be more expansive. stupid stupid me. oh well. for $4 the simple wall wort mod rocks :YeaBaby:

the other mod i'm using is a plug to charge the internal batteries while i'm on the wall wort. i mite get fancy, and wire it, so i can use my triton charger as well. i'm not sure yet though.

aristo
11-03-2008, 05:31 PM
shaggy did you finally git r done?:thumbup:

shaggybirdman
11-04-2008, 06:42 PM
nope. gotta get my butt in gear, and do it.

Steve Joblin
11-04-2008, 07:59 PM
I got a 6v 1A wallwart that I ordered off ebay for a whopping $3 (shipping included!). Great deal huh? Well I used my meter to confirm which wire was the positive, and to my surprise, the voltage was 9.75 volts! I'm gonna fry the little sucker or expode my battery with that kind of voltage! I was thinking that I could either use a voltage divider by putting two resistors in parallel or using a bunch of diodes in series to drop the voltage to 6v.

1. Am I correct in that 9.75 volts is too much and I need to drop the voltage?

2. What is the best method to drop the voltage?
a) a voltage divider using two 1/8 watt resistors
b) a bunch of diodes in series
c) something else I am unaware of

aristo
11-05-2008, 12:04 AM
Personally Steve I would be scared to use that walwart.

aceisback
11-05-2008, 03:53 AM
I am no electronics expert, but I bought (2) 6v 800ma wallwarts from All Electronics. When the voltage is measured with no load attached, the voltage measurement was 10v, but with the charger hooked up, the voltage dropped to it's 6v output. I have set up 2 chargers using these wallwarts and have had no problems whatsoever with them.

Duane

kamtsa
11-05-2008, 05:45 AM
I got a 6v 1A wallwart that I ordered off ebay for a whopping $3 (shipping included!). Great deal huh? Well I used my meter to confirm which wire was the positive, and to my surprise, the voltage was 9.75 volts! I'm gonna fry the little sucker or expode my battery with that kind of voltage![/INDENT]

Try to connect a resistor of let's say 1k to 10K between the (+) and (-) wires of the wall wart and measure the voltage across the resistor. Possibly the 9.75V that you see is due to some leakage and the resistor will take care of that. Some power supplies need a minimal load to provide the specified voltage.

Kam

olooki
11-05-2008, 06:50 AM
I got a 6v 1A wallwart that I ordered off ebay for a whopping $3 (shipping included!). Great deal huh? Well I used my meter to confirm which wire was the positive, and to my surprise, the voltage was 9.75 volts! I'm gonna fry the little sucker or expode my battery with that kind of voltage! I was thinking that I could either use a voltage divider by putting two resistors in parallel or using a bunch of diodes in series to drop the voltage to 6v.

1. Am I correct in that 9.75 volts is too much and I need to drop the voltage?

2. What is the best method to drop the voltage?
a) a voltage divider using two 1/8 watt resistors
b) a bunch of diodes in series
c) something else I am unaware of


The cheaper wall warts use poor wire in the internal transformer. This wire has a higher resistance and thus your open voltage reading will be high, but when placed under a load as rated on the wall wart, you should get close to the rated voltage.

Dumb Thumbs
11-05-2008, 03:13 PM
I'll stick with using rechargeable nickel metal hydride's in the charger. "Energizer" has a pritty good charger out. Charges 4 AA's in about 15 min & I think it's a balanced charge!

shaggybirdman
11-05-2008, 10:03 PM
put them on charge. then pull 1 battery out. if it keeps charging each battery is being charged by itself. if it stops it's charging all 4 as a pack. i highly doubt it's a balance charger though. only li po, li on, and a123 cells use a balance charger as far as i know.

15 min to charge 4 AA cells? i'd expect them to discharge quick. my venom cells take forever to charge, but say on the packaging to only charge at .5 amp. not sure how many packs it takes to deplete there charge, but it's alot. i found that a nice slow rate of charge gives a much better charge than a fast charge, and a dc power source is even better yet. meaning a battery, not a power supply.

Dr. Memory
11-06-2008, 11:32 AM
I got a 6v 1A wallwart that I ordered off ebay for a whopping $3 (shipping included!). Great deal huh? Well I used my meter to confirm which wire was the positive, and to my surprise, the voltage was 9.75 volts! I'm gonna fry the little sucker or expode my battery with that kind of voltage! I was thinking that I could either use a voltage divider by putting two resistors in parallel or using a bunch of diodes in series to drop the voltage to 6v.

1. Am I correct in that 9.75 volts is too much and I need to drop the voltage?

2. What is the best method to drop the voltage?
a) a voltage divider using two 1/8 watt resistors
b) a bunch of diodes in series
c) something else I am unaware of
Steve asked me to comment on this.

Apologies in advance, Steve, but I'm just too lazy to read through the prior 86 posts to get fully up to speed on what you all are trying to do here. I think I can answer your questions without it, but be patient in that I won't know everything that was discussed prior.

I couldn't tell for sure from the OP what the intent is here. It looks to me like you guys are trying to adapt a cheap AC-source charger to 6V to charge batteries. I'm also going to assume we're not talking lipos here.

Given you have 9.25V, want 6V, and assuming a significant load (which would be the case for charging batteries), a voltage divider is not the way to go. The desired current range of operation makes this impractical -- load current will be large for much of the cycle, and then fall off toward zero near the end of charging. With only 1A total available, the design constraints rule out this approach -- there's no design solution within those constraints.

A series diode network is a viable solution, although there are some considerations here as well. First, how well is the 9.25V regulated? Remember that a step-down regulator depends on a well-regulated source voltage. I suspect that if you are getting 9.25V without a load on a 6V power supply, it's gonna drop -- a lot -- when a load is applied. The first thing I'd do is get a 9.25V x 1A = ~10Watt resistor, value = 9.25V / 1A = ~10ohms, place this across the supply output, and measure the voltage. If it's still within 8.5-9.25V, then the supply is probably good, just somehow maladjusted.

I the voltage drops more than that, toss the thing -- it's bad.

If it's good: You want to calculate the voltage drop according to the end-state charge voltage. If the target voltage of a charged battery is 6V, then simply (9.25-6)/0.7 = 4.64 diodes.

How to get a half a diode drop? Use a specialized diode called a Schottky, which has about a 0.3V drop. So, 4 general purpose diodes + 1 Schottky gives you 2.8V + 0.3V = 3.1V, dropping 9.25 to 6.15, close enough.

Keep in mind that at peak current, 3.1V x 1A = 3.1W of power will be dissipated in the diodes -- they will get warm. When you get that 10ohm test resistor, I'd also pick up a 6ohm, 10watt test resistor to test out the step down setup before putting a battery in as a load. Use the 6ohm as a load, check voltages, let it run for an hour and check voltages again. The supply will be heating up, being run at capacity, and will start to drop in voltage. You need to see how this affects everything.

I'd also let it go overnight and see how it settles -- it may drop enough that this won't work.

These tests are important to perform with a cheap resistor, where the worst thing that can happen is that you burn out the resistor (not if you get one with enough power rating, as I've outlined above). You'll also generate a bunch of heat.

Regardless, I wouldn't get near a lipo with this setup. As I'm sure you know, lipos are fussy animals, and have no tolerance for overvoltage or overcharging without presenting very real danger.

This setup can be used for nicads, and even be left unattended, once you've ensured that the starting and ending voltages are correct. For nicads you can safely charge at 2C, which would mean a 5cell, 500mah pack with a 1A max current. However, at 6V the actual current through the pack should be down to C/10, which is 50mA. So, you may have to put a low-value resistor in series with the diodes to get the current right at exactly 6V.

You can see how complicated this gets. For an application like this, an active regulator (as opposed to diodes or resistor networks) is called for. The reason is that supply voltages are likely varying quite a bit (I'll bet as much as 30%) with load, the load current range is big (varying from 50mA to 1A), and the voltage tolerance small (ideally, 6V +/- 0.1V -- off the top of my head).

Battery charging applications are demanding from the standpoint of electronic design. This is why unattended charges are basically microcontroller operated switching power supplies, and therefore expensive. Safe, unattended charging algorithms, whether lipo, nicad, or NiMH, require more controllable electronic designs, with a sophisticated controller (i.e. software program) to manage the currents and voltages.

kamtsa
11-06-2008, 12:07 PM
...I couldn't tell for sure from the OP what the intent is here. It looks to me like you guys are trying to adapt a cheap AC-source charger to 6V to charge batteries. I'm also going to assume we're not talking lipos here....

The intent is to power from 110VAC a 1S 300ma LIPO charger that is currently powered by 4 AA batteries. The charger is used to charge small 100mah LIPO's.

Kam

Dr. Memory
11-06-2008, 01:19 PM
The intent is to power from 110VAC a 1S 300ma LIPO charger that is currently powered by 4 AA batteries. The charger is used to charge small 100mah LIPO's.Ahhh...

Then, Steve, don't worry about most of what I said (although it's good info to understand). Since there is a lipo charger doing the charge management, you're okay.

So, the answer is, hook 4 general purpose diodes in series between the PS and the Charger, and you're good to go, provided it holds 9V under load.

As I said above, I suspect it won't, and that it's blown, because it's showing 9.25V open circuit, and it's a 6V supply.

So, the first step is just to put a suitable load on it and check the voltage. You might get lucky -- as soon as there is any load, it may drop to 6V, and hold pretty steady there all the way up to the rated load current of 1A. If that happens, you don't need any adjustment -- just use it as is.

Dr. Memory
11-06-2008, 03:07 PM
Here's a little very cheap circuit I just whipped up (and spent a long time in MS Paint drawing it :arggg:) that you can use for this project. All you need are soldering skills, and something to contain it.

If you already have a good, heavy-duty doorbell transformer lying around, this is less than ten bucks of parts at Radio Shack:
http://www.helifreak.com/attachment.php?attachmentid=68145&stc=1&d=1225999945

Values for R1 and R2 are calculated as follows: Junction bias voltage for T1 is about 0.7V higher than the target output voltage, so we want our voltage divider to give us 6.7V.

For a voltage divider, Vout/Vcc = R2/(R1+R2). In our case this ratio is 6.7/12 = 0.56. For an 18V doorbell transformer (another common part), the ratio is 6.7/18 = 0.37.

Based on the divider formula, R1=R2*(1-r)/r, where r is the voltage ratio.

We want the current in the divider to be small compared to the current supplied to the load. This will maximize efficiency. We also must account for the forward bias current of T1, ensuring that at 1A collector current, the necessary bias current on the base of T1 doesn't significantly alter the voltage at the divider. We account for this by suitable selection of resistor values that meet the above relationship.

A reasonable "rule of thumb" figure for the gain of T1 is 100. This means that, at 1A through the CE junction, we need a current of about 10mA through BE. We'd like the output voltage at the emitter of T1 to be +/- 0.3V (5%).

From these specs we can calculate the R values. With a 100ma current through the divider, we meet the 5% error tolerance. So, we have 110ma through R1 with a 5.3V drop ==> 5.3/0.11 = 48ohm. R2 has 100ma flowing through it with 6.7V across it, => 67ohm. These calculations are under full load, so the load will see 6V while drawing 900mA (100mA are being lost to heat through the voltage divider).

For 18V, the numbers are (18-6.7)/0.11 = 103ohm, R2 is the same at 67ohm (R2 doesn't change with different supply voltages.

Dumb Thumbs
11-06-2008, 04:00 PM
put them on charge. then pull 1 battery out. if it keeps charging each battery is being charged by itself. if it stops it's charging all 4 as a pack. i highly doubt it's a balance charger though.

15 min to charge 4 AA cells? i'd expect them to discharge quick.
It'll charge 1 cell at a time, 3 cells, 2,or 4. Energizer (as in the Bunny) NiMH charger
I've been using same 4 AA's in the Tx for over a year now and am using them in the LiPo charger now.
As a test I pulled the four AA's out of the LiPo charger. Voltage's were
1.206
1.203
1.211
1.211
Charged just one cell till full - 1.393
Put all four cells into the charger, including the full charged one. 15 min. later & waiting 1/2 hour to stabiles & cool (Yea they do get very warm at that rate of charge) Volts read...
1.438
1.431
1.421
1.428
Not bad for batts that are over a year old & have been cycled I don't know how many times, in different devices.
Just my 2 cents...

olooki
11-06-2008, 05:59 PM
put them on charge. then pull 1 battery out. if it keeps charging each battery is being charged by itself. if it stops it's charging all 4 as a pack. i highly doubt it's a balance charger though. only li po, li on, and a123 cells use a balance charger as far as i know.

15 min to charge 4 AA cells? i'd expect them to discharge quick. my venom cells take forever to charge, but say on the packaging to only charge at .5 amp. not sure how many packs it takes to deplete there charge, but it's alot. i found that a nice slow rate of charge gives a much better charge than a fast charge, and a dc power source is even better yet. meaning a battery, not a power supply.

I actually own on eof those Energizer 15 min chargers. It DOES charge each cell independently. And the cells hold a charge as well as NiMh batts can. It works quite well.

shaggybirdman
11-06-2008, 07:52 PM
oh don't get me wrong. i was just worried about heating up the cells. nimi cells don't like to get hot. i'd be curious to know the charge rate of the charger. if we were talking ni cad i wouldn't have said a word, but seeing as they are nimi's it would seem to me to be charging at like 4C plus, or such. 1C gets nimi's hotter than i like.

how many charges do you get for your 15 min charge?

Steve Joblin
11-06-2008, 08:19 PM
Thanks much good Dr.!

I tested the voltage under load by checking the voltage across a DC motor I hooked up to it, and the voltage only went down to 9 volts. I think I will stay away from using this charger as I just don't trust it!

I do have a multitude of other 9v AC Adapters that say 9V on the lable and actually measure 9 volts with my DMM... I think I will use one of these instead.

As long as I have to start adding electronic components to reduce the voltage to 6 volts, I probably should use the component that was actually designed for the task... the venerable V-Reg! Jameco sells the 7806 in a T-220 package for $0.43... I just need to filter caps on the input and output.