Reducing 14v to 10-12v for lipo charger, Experts needed!

[cb1rocket]

Hi all

A bit non bike related but still relates to bike batteries and the elecrtrical system!

My mate has a lipo charger that specificly states not to have the car/bike running when using the charger to charge lipo heli batteries and the like. So obviously I want to be able to use the charger when the bike is running.

My memory have practically gone blank on ohm's law and I need to know how (using resistors obviously) to reduce the voltage down to under 10 - 12 volts max from the running 14.5v that we usually encounter on a bike's charging system.

Advice is much appricated and i'm familar with most things electrical! ie i know parallel and series circuits but as said I have gone blank in knowing what value reisistor i need and how to wire it in.

cheers

[The Shaggy D.A.]

You want a voltage divider circuit, where R1 and R2 is a percentage of the voltage you desire in comparison to the input voltage, i.e. if the input voltage is 10v and R1 & R2 are the same value, then the output would be half that, or 5V.

This means that if you want 10v out of a 14.5v input, that would be roughly 65% (100/14.5*10), so R1 would need to be about half the value of R2.

[edit] Knew I had it somewhere :-

R2 = O * R1 / (I - O)
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[carvell]

The Shaggy is correct.

However, you need to make sure your resistors are beefy enough. You need to know how much current is going to be drawn by the target circuit, and ensure that the power dissipated in each resistor won't exceed its power rating.

If I were you, I'd use a voltage regulator rather than a potential divider.

https://cgi.ebay.co.uk/L7810-7810-Voltage-Regulator-10-Volts-1-5A-/270533103195?pt=UK_BOI_Electrical_Components_Supplies_ET&hash=item3efd078e5b
That's what you want. You'll need to heatsink it if you're pulling over 500mA odd through it.

You can get 12V ones too. Search ebay for "7812 regulator".
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Yamaha TDM 850

[The Shaggy D.A.]

True, having read up on LiPo charging rates, it says a 5000mAh battery would charge at 5 Amps, so I'd go down the regulator with a decent heatsink route too.
____________________
Chances are quite high you are not in my Monkeysphere, and I don't care about you. Don't take it personally.
Currently : Royal Enfield 350 Meteor
Previously : CB100N > CB250RS > XJ900F > GT550 > GPZ750R/1000RX > AJS M16 > R100RT > Bullet 500 > CB500 > LS650P > Bullet Electra X & YBR125 > Bullet 350 "Superstar" & YBR125 Custom > Royal Enfield Classic 500 Despatch Limited Edition (28 of 200) & CB Two-Fifty Nighthawk > ER5

[Codemonkey]

Another option might be to get 4 big silicon diodes rated at greater than whatever the maximum charge current is and connect them in series in line with one of the leads. Each diode should drop about 0.6V, 4 should get you about 2.4 volts drop. Make sure they're not Shottky type though cos they only drop about 0.2V each.

Resistor divider isn't really much good with loads that vary like a charger since you need to know the currents to size the resistors, and since the currents vary, you would need resistors that do automatically, or as the posters above suggested, use a regulator.

Dropping just 2 volts might be a tall order for lots of regulators though, often they need a bigger in/out voltage difference to be able to function correctly, so if you do down that route, look for a "low drop out" regulator, (means it doesn't need as big a differential).

If you go for the easy diode option, connect them up like this:

+14V in -------|>|---|>|---|>|---|>|-------- +11.6V to charger

[gfiandy]

If this battery is a large lithium cell as I suspect, please stop this project now unless you can find someone who really knows what they are doing.

Lithium cells are very sensitive to their charging voltage they don't like spike or noise on the charging system and need carefull voltage and current regulation.

If you don't do this you will damage your batteries and there is a very real possibility they will explode.

If you want help with this I would recomend going to an electronics site where you can find someoen who knows how to charge this type of cell. They are not like lead acid that will put up with just about anything.

At the very least we need to know the exact battery pack you want to charge and the charger you plan to use. Then we might be able to find out how robust the charger is and what voltage and current characterisitcs the battery needs.

Regards,
Andrew
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ANDY

[Codemonkey]

[finpos]

If you want to do this you really need a proper voltage converter which will probably be quite costly - 14V is a nominal value with the engine running, in fact it will be all over the place and full of spikes, the bike's regulator isn't all that fussy about it's output (aka. it's built to a price)

f.

[cb1rocket]

[cb1rocket]

[cb1rocket]

[Codemonkey]

Just be aware that if you're putting 5 amps through anything, diodes or resistor, they're gonna get hot! If you opt for diodes, I'd go for stud types that can be bolted to a heatsink. In this case beware that the case of the diode may be connected to one of the connections, so you'd have to insulate them from each other (seperate heatsinks).

In the case of 4 diodes, at 5 amps you'd be dissipating 2.4 x 5 = 12W

[cb1rocket]

[Codemonkey]

Something like this https://uk.farnell.com/ixys-semiconductor/ds17-12a/diode-recti-1200v-25a-stud-anode/dp/1841858?in_merch=New%20Products&in_merch=Featured%20New%20Products&MER=i-9b10-00002068

[cb1rocket]

these are quite bulky but cheers for the link. I'm surprised its looking more complicated than I thought. Although the 7810 and 7812 voltage regulator looks more promising despite they are rated at 1.5 amps max each so I would need ideally 3 - 4 of these in a parallel circuit somehow?

Can someone provide a circuit diagram for these 7812 semiconductors? cheers

[The Shaggy D.A.]

[carvell]

The diode idea is a good one just for knocking the voltage down, I'd go with that as the easiest. The only disadvantage with that is that it won't condition the voltage in any way.

To answer your question though, the three legs on the regulator are for input, output (both positive) and a common 0v. Note that if you really are going to be pulling 5 amps, then the one I sent a link to isn't beefy enough.

All a bit of a faff! The problem that you've got is that your charger could be pulling so much current. If you have a multimeter handy you could measure the actual current pulled by the target in order to get an accurate idea of how beefy your circuit needs to be.
____________________
Yamaha TDM 850

[Codemonkey]

[cb1rocket]

[Codemonkey]

You're going to struggle to do this with a resistor divider network. You need to calculate the values twice.

Case 1: For the circuit when there is no load on it (no battery being charged),

Case 2: for full load (flat battery being charged).

The voltage dropped across resistor R1 is proportional to the current flowing through it (V = I x R), hence it will drop a different amount of volts in both cases.

Resistor R2 helps in Case 1 by providing a path for current to flow through R1 when there is no load (or very little load). When the load is increased it will effectively be in parrallel with the load and conducting very much less current than is flowing through the load. You can effectively discount it and use the current flowing through the load instead.

You can do this with resistors, but you will be losing more power dissipated as heat than you are going to be using to charge the battery. Resistor dividors work well when there is very little load (eg, when you need to create a certain reference voltage within a circuit, or to scale a voltage to a lower level), but they are pretty much useless when you need to create a certain voltage out for a varying load.

To work out how much power a resistor dissipates, it is I squared R

so, Current squared times the ohms value of the resistor. so eg for a 10 ohm resistor and 5 amps flowing through it, that would be 5 x 5 x 10 = 250W. At that current it would drop V = I x R so 5A x 10ohms = 50V.

You need to drop 2V at 5A, so R1 can be found by R = V/I = 2V / 5A = 0.4ohms

It would dissipate 5A x 5A x 0.4ohms = 10Watts when the battery is charging.

However, when the battery is not charging, lets be generous and say the charger would still draw 1A, R1 would now only drop 1A x 0.4ohms = 0.4V


You can size the resistors for each case, but not both.

[The Shaggy D.A.]

[cb1rocket]

Many thanks so far for all replies.

After much thought about everything on all the circuits it seems the only way forward is to just sodding buy one rather than bother making one from scratch.

The voltage divider was my favourite, along with the basic of using diodes but saw the pitfalls when the loads differ from each state of the charging lipo with regards to current draw having an affect on input voltage, esp. when on standby with load being light that the voltage drops as much as 6v! When under load it can be just right at being 12v. Sodding lipo charger! Wish I just got him to get a charger with variable input voltage, so much for the cost savings!

I saw this and thought thats great if only they had a 3 amp version so to speak:

https://cgi.ebay.co.uk/12V-DC-Voltage-Regulator-Module-Board-Based-7812-/150448615786?pt=UK_ConsumerElectronics_SpecialistRadioEquipment_SM&hash=item23076fb16a

Some of you have said using regulators so please post all your regulator ideas. Does some of you not think it would consist of another bog standard bike reg/rec that could be used? cheers to all and karma for all

[kramdra]

It sounds like your lipo charger will take some variation - certinally at 5A, if it doesnt have its own wall PSU, it must be designed for some - at minimum to work from a part discharged 12v battery..

possibly upgrade its heat sink, so that it is capable without any additional regulator.

[Robby]

I presume the reason you want to have the engine running is to charge the batteries while you're out playing with the helicopter. You would need to spend a lot of time on charge to get the bike battery low enough to fail to start the bike, and even then just leaving it for 20 minutes should give you enough charge to start the bike - chemistry in action there.

If you want to charge the battery whilst riding, then I would be concerned about the heat from various parts directly under your arse of behind, depending on where the storage is on your bike. You can't smell something catching fire on a bike until it's too late.

I would go for one of the following options:

1. Build/buy the circuit and use it to charge the lipo batteries with the bike turned off.

2. Buy a 240v inverter and use it to charge the batteries with the bike engine running at a standstill, or with the bike engine turned off just using the battery.

A bike battery holds a good chunk of charge, around 12Ah. That gives you over 2 hours or charging at 5 amps with enough left over to start the bike afterwards.

[cb1rocket]

hi all

yes as stated in the manual the bike/car must not be running when using the charger. Hence why I don't want to blow the charger when the car or bike is running as we all know its roughly regulated at 14.5 volts.

I want to do the charging while running whether in the car or bike so to save time justr charging at the field. Its a good 30 mins drive so its also a good time to get the batteries on charge during that time.

cheers