Hi

I wish to run a device that draws 3.5Amps and it was designed to run from 4.5v but the battery is no longer avalable and i can find no other substitute so i have a 6v sla 4.5Ah battery that fits inside very well but as this device is of a 1964 vintage i dont want to dammage it and so i would like to ask for some help / advice in how i can reduce this batterys dc voltage to 4.5v but still retain the full current capability of the battery?

Thank you for any help you can offer.

svdsniper

 

Hello,

There are sevetral possibilities.
All will lead to powerloss.
1) 2 diodes in series with the powerline (each will drop about 0.7 Volts, loss about 25 %).
2) a Series resistor, the voltage difference will be lost as heat. (about 25 %).
3 A DC/DC converter (Buck converter), this will have a loss of 5 - 10 %.

Greetings,
Bertus

 

Hello Bertus

Thanks for your reply.
I want to avoid as much heat generation as possible so with this in mind what option would you suggest as best?
can you tell me what components to buy?
i am a total newbie with this electronics thing.

thanks

svdsniper

 

Hello,

National has some buck converter controllers.
Take a look at the LM3475, see the datasheet for more info.
With that you can build a small conveter.

Greetings,
Bertus

 

Hello

The space inside the battery compartment is very limited so i would like to try the two diodes in series method, could you advise what type of diodes i should buy?

thanks

svdsniper

 

Hello,

You can take standard 5 Amp or more diodes.
The diodes will get a bit warm.

Greetings,
Bertus

 

Hi Bertus

Thank you for your advice
i will try the diodes in series and see how that effects things.
thank you very much
svdsniper

 

does it matter if i use zener diodes or rectifier or switching diodes?

thanks

svdsniper

 

Hello,

You can use standard rectifying diodes of 5A or more.

Greetings,
Bertus

 

thank you very much
i will go out to buy some diodes tomorow and hoefully test this out soon.
Thank you for your help
svdsniper

 

Note that the two diode will produce exactly the same amount of heat as a resistor or as a linear regulator like an LM317. You will be dropping 1.5V at the load current, so since P=IE, the P is the same because the I is the same...

 

Hi All

Today i bought 3 rectifier diodes rated for 6A , i soldered all three in series and got down to around 5v direct from the battery and it was still too high but when i powered up the device through the diodes i found the voltage reaching the lamp was now too low. I removed one diode at a time and retested and found that by using a single diode i was able to achieve exactly 4.5v at the lamp and this is the voltage that the lamp is rated for, so i am guessing this device is now running at the correct voltages as the image intensifier tube may have a higher voltage circuit inside but as i now have the correct output from the control module to the lamp i expect the internal voltages to be correct too.

Thanks to everyone for all the help, now i just need to wrap the diode in a fibreglass sleeve and heat shrink as it does get quite hot.

svdsniper

 

Don't wrap the diode in fiberglass and a heat shrink sleeve; there will be no place for the heat to go, and the diode will eventually melt.

You need some way to get rid of the heat. You might try epoxying the body of the diode to the enclosure using J-B Weld. If the enclosure is not metal, you will still have a problem getting rid of the heat.

 

Another problem with using a diode to drop the voltage is that as the temperature of the diode increases (due to the power dissipation across it), the voltage across the diode will decrease. The hotter it gets, the less voltage that it will drop.

You say your load requires 3.5A @ 4.5v.
A 6v lead-acid battery has nominally 6.3v to 6.4v across the terminals when it is charged.
(6.35v-4.5v)/3.5A = 1.85/3.5 = 0.529 Ohms (rounded off)
Power dissipation = 1.85v x 3.5a = 6.475 Watts. Doubled for reliability, that's 13 Watts.

Radio Shack carries 1 Ohm 10 Watt 10% tolerance power resistors.
Link: http://www.radioshack.com/product/i...0&filterName=Type&filterValue=Power+resistors
If you connected two of them in parallel, they would measure roughly 0.5 Ohms, drop about the right amount of voltage, and be able to dissipate 20 Watts of power. Your load would have roughly 4.5v across it. The power resistors won't change their resistance as they heat up.