Diode to reduce voltage - questions...

[malow]

hi, im trying to reduce the voltage of a li-ion cell to replace 2 AAA bats.

as i look around, ive found that some batteries use a diode to reduce from 3.6v/3.7v nominal voltage to 3.0v (with maximun voltage around 3.6v)

as i live in a city far away from big cities, my only shop from electronics is dealextreme

my first idea, was using a 10440 li-ion in one "slot" of AAA battery, and a dummy AAA with the diode in the other.

BUT, direct measutring the voltage with a "simple" (diode removed from small AC>DC transformer) the diode, read 3.8v. after consuming about 30ma, it fell to 3.5v (my device will use about 35mAh)

so, my questions:

1- its normal the voltage be higher when there nothing consuming power from it? (only batt and diode)

2 - consuming more ma, the voltage should drop?

3 - i was thinking in buying a "Soshine 3.0V CR123A" battery, and use the diode found on this battery (does the same as i want)

4 - the diode from Soshine battery will have a better "stable" output, or all of them have this variation depending of the mAh consumed?

my device can have MAXIMUM 3.6v input. diodes reduce about 0.7v right? this is a constant reduction?

tnks for patience. :thumbsup:

 

[VidPro]

a breif blurb. not an answer.

its not constant, i dont know why and all the variations, but it varies with the current , and can vary with temperatures too, read the voltage drop across the diode. therein lies your answers.

chances are good that the "regulation" in a 3V li-ion battery that the cell internally is a 3.6v cell having a 4.2V high, that it uses 2 diodes
|---|>----|
|---<|----|

that way it gets a "drop" when pushing current through, and voltage flows in both directions.
so with 2 cheap diodes i would guess you could simulate the curcuit used.

 

[alpg88]

1 yes voltage with no load will be higher
2 yes
3 diode does have resistanse, but it isn't something you wanna to control voltage with, to keep steady v with varuable draw you use voltage regulator, lm350, lm 317..ect.

 

[NextLight]

SNIP

my device can have MAXIMUM 3.6v input. diodes reduce about 0.7v right? this is a constant reduction?
:thumbsup:

A silicon diode (1N4001 or similar) will drop about 0.7V. It is not EXACTLY constant, stable, or linear, but it is pretty close at near zero to modest currents. It will drop a 4.25V maxed out LiIon to about 3.55V, with light or no load. With a 1A maximum load, the diode might drop 0.9V, plus you would experience battery voltage droop.

I would not hesitate to use a single LiIon and a silicon diode in your application, or I might choose two series connected germanium diodes, (1N4148 or similar) for a drop of 2x 0.5 V or ~1.0V, if I was worried more about overvoltage risk, than power efficiency on my 35 mA device.

It sounds as though you are on the hairy edge of your understanding. Reading up a bit more about diodes would be useful to reduce the chance of any mistakes in what you are doing. Radio Shack's Basic Electronics, or the ARRL Radio Amateur's Handbook may help.

Welcome to CPF. There is much for all of us to learn here.

 

[malow]

tnks for the attention guys. :twothumbs

i found a 1N4001, i will put inside a AAA dummy bat and test.

:thumbsup:

 

[malow]

yeap, works fine!

tnks again. after a lot of digging, ive found that a minimal current consumption cause the higher voltage to drop to expected "voltage drop" from a diode. so a 5k resistor will drain 0.7mah and "fix" this.

appears that logically, i dont need the resistor to reduce, cause the consumption from the device will do the same. but i belive im paranoic :ironic:

at least i know the device will have a voltage in limits when powering on (3.6v without powering on, 3.4v after powering it) :thumbsup:

perfect 3.6v on a just-charged li-ion at 4.2v