Reducing battery voltage

[andygo]

I need to replace the batteries on my Cateye light system on my Mountain bike.

Currently it uses two sets of 5 metal hydride batteries with a rating of each set of 6.ov, 2200mAh.

This gives the lights on full power a life of abot 50 mins. They are about 4 years old now and the life isn't as good as it was plus my mates and I seem to be going on longer rides in the dark. (last nights was 3 1/2 hours)

Anyway, what I would like to do is buy some li lion batteries to extend the length of riding tiem.

The main problem is that li lions seem to be rated at 7.4v rather than 6v. Will this cause the bulbs to blow or will I need a gizmo to reduce the volts to 6ish?

I could buy a whole new system, but they are nearly £300, so a battery pack seems a better bet.

Any ideas folks?

Thanks,

Andy

[Plotloss]

1.4v resistor.

Under the saddle.

Mmmmm heated seats

[The Excession]

Not familliar with that make, does it use high power LEDs or is it a set of incandescant bulbs?

If it's using LEDs the chances are that there is an electrical circuit in there that is actually switching the LEDs on and off very rapidly. This allows the unit to overload the LEDs to the point where they would burn out if left on constantly. This creates the huge amount of light. Because there is also an off period the LEDs can cool down without lasting damage.

The unit does this hundreds of times a second which is way faster than your eye and brain perceive so you 'see' a continuos beam.

Potentially the unit could run quite happily at a higher voltage, the only way to check would be to hook it up to a variable power supply (bench PSU etc) and slowly increase the voltage whilst monitoring the current consumed - (or observing all the LEDs burning out).

[andygo]

The lights are two halogen bulbs 10w and 15w I think.

The whole setup is really good - just need to extend the battery life.

[Spaceframe]

Hi Andy,

I'd go with the resistor idea too.

We need to know the current to calculate the required value and power dissipation so...

Assuming your NiMH cells are a bit old now, I'd guess they are supplying around 2A at 6V for the 50mins - which suggests they have a useful capacity of 1666 mAh. This is fairly typical for some well used cells.

To drop 1.4V across the resistor we apply ohm's law: R = V/I, so 1.4/2 = 0.7

The closest standard resistor value to this is 0.68 ohms.

The resistor will need to dissipate 2.8 watts (V x I) so given they are pretty cheap, go for a 5W one.

A 0.68 ohm 5 Watt resistor should be no more than 2 quid.


Have a read about LiPoly cells though. If you over discharge them (very easy with high power bulbs) they'll become unstable and can catch fire in a really nasty way (remember burning lithium in chemistry lessons) . I don't store mine in the house.

Some newer cells like the 18650 have built in PCBs which cut the supply once the minimum discharge voltage has been met.

Hope this helps

Jon

[Spaceframe]

Andy,

I'd assumed ONE set of cells gave you 50 mins but have just seen your additional post which says you have a 10W and a 15W bulb.

Are you using both sets of cells in parallel through a balancing arrangement? (NiMH cells shouldn't be used in parallel without this)

If so, at 6V this puts current draw at around 4.2A with both bulbs on, so the resistor calculation should be...

R = 1.4/4.2 = 0.33 ohms

Dissipation now needs to be 5.8 Watts

I'd go for a 0.3 ohm 10 Watt resistor


If you can switch each bulb on independently, then the resistor arrangement won't work, since it requires the current draw to stay constant in order to drop the correct voltage. You'll need a voltage regulator circuit in this case.


Having said all of this, the fully charged voltage of a NiMH cells is 1.4V. This means a fresh pack will supply 7V for a short while instead of the nominal 6V. I think I'd be tempted to bung on a 7.4V LiPoly and see what happens...

[Pigeon]

Don't use a resistor... use two diodes (ordinary diodes, not Schottky ones) in series... they will drop 0.7V each, they give an essentially constant drop independent of the current.

Pick ones rated at 5W or more.

[Ordinary Bloke]

Pigeon said:

Don't use a resistor... use two diodes (ordinary diodes, not Schottky ones) in series... they will drop 0.7V each, they give an essentially constant drop independent of the current.

Pick ones rated at 5W or more.

http://www.maplin.co.uk/Module.aspx?ModuleNo=46413

Edited by Ordinary Bloke on Friday 14th November 00:25

[Pigeon]

Yep, they'll do.

[GreenV8S]

I don't know how many cells you have in that pack or how many volts per cell, but could you just get rid of one cell?

[Zad]

Anything that drops X Volts at Y Amps will produce (X*Y) Watts in heat, whether diode or resistor! To be honest, I'd be tempted to go with an LED system, they produce more light for less input power. Is there room to wire up 5* 1.2v LiIon cells instead?

[Spaceframe]

Pigeon said:

Don't use a resistor... use two diodes (ordinary diodes, not Schottky ones) in series... they will drop 0.7V each, they give an essentially constant drop independent of the current.

Pick ones rated at 5W or more.

Good call! Hadn't thought of that

[tamore]

higher voltage bulbs with the same fitting?

[vrooom]

Ah if you are using lithum polymer batteries, you can destory the batteries by draining it all way down. you need to limit battery down to 3.7v per cell, or else it swell up badly.

Jay

[tom g]

bulbs should be fine on 7.2V - well they'll be brighter! it's common to run 12v bulbs at 14.4v (um, like cars )

how would you charge your li-ion batt?

you might find it easier to use two lots of 4xAA nimh aranged in parallel. that would give you 5.6Ah if you use 2800mah batteries

Edited by tom g on Friday 14th November 15:30

[Hyperion]

tom g said:

how would you charge your li-ion batt?

This is an important question.
Take care with Li-ion batteries - incorrect charging can cause them to burst into flames.
Don't brush the warnings off as scare mongering - they really are very volatile if not handled correctly.

Edited by Hyperion on Friday 14th November 15:39

[tom g]

tom g said:

bulbs should be fine on 7.2V - well they'll be brighter! it's common to run 12v bulbs at 14.4v (um, like cars )

how would you charge your li-ion batt?

you might find it easier to use two lots of 4xAA nimh aranged in parallel. that would give you 5.6Ah if you use 2800mah batteries

Edited by tom g on Friday 14th November 15:30

sorry, nimh is 1.2V, so you;d need 2x(5xAA). Still, they're cheap and easy to charge.

Oh, an R/C car battery pack & charger would be perfect. Problem solved.

[andygo]

If i got a couple of remote control battery packs rated at 7.2v, surely when they are freshly charged they will be shoving out 8v or so. Surely that will blow the 6v halogen bulbs?

[Zad]

That might well be 8v off load though. Measure the voltage on load. Even so though, running at a higher voltage will make them draw more current. For a linear load, power increases as the square of the voltage (E^2/R) so a 7.2V battery driving a 6V bulb would draw proportionately more current, and generate 44% more light and heat. Definitely not good for bulb life!

How much are 7.2V replacement bulbs?