Discharge constant current from battery 3

[johnwqtan]

Currently, i am making a simple discharge circut to discharge batteries at a constant current. Using a resistor won't work because the close circuit voltage will be different depending on the health of the battery. The question is how can i achieve this constant current draw from the battery for the whole lifespan. If i need to, i might have to do it using microcontrollers to control the load resistance while observing the voltage of the battery. but really don't want to go there. Is there a simpler solution to this? Thanks in advance.

 

[melone]

Buy a current load. It's similar to a power supply, but will sink / source current at a constant rate.

 

[Skogsgurra]

What size is the battery?

The simplest way of doing ot to use a transistor with an emitter resistor and then feed the base of the transistor with a voltage that corresponds to voltage drop over resistor plus Vbe.

An example: Say that you want to discharge with 2 A constant. Select a transistor that can dissipate current times voltage, say 12 V. That will be 24 W. Use a heat sink that corresponds to that. The base resistor can be something like 1 ohm. That will give you a 2 V voltage at the emitter. Then feed the base with 2.8 (approximately) V to get the constant discharge current. Use a power rheostat (more than a few hundred mW) to adjust base voltage to get the current you want.

Gunnar Englund

http://www.gke.org

 

[itsmoked]

Add a resistor to skogs plan to spread around the power dissipation.

But the big question is still; What size battery. ?


Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[nbucska]

What is the min/max of the voltage and desired current ?


Plesae read FAQ240-1032
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[BrianR]

Many batteries do not respond well to complete discharge and definately not when they are in a pack where one cell can go flat first and then be reverse charged by the others which damages it. So your constant current discharger may need a voltage cutoff as well. Eg NiCad/NiMh should cutoff at 1.1V/cell.

 

[johnwqtan]

3.6 volt lithium ion battery
Desired current: 50mA to 200mA
cutoff: 2 volts
Thanks for all the response

 

[itsmoked]

Study this data sheet for the robust LM317. It will explain the heat-sinking and all other details. Note the simple current source near the bottom. You can make a current source of any current which will reflect back to what your batteries will have to supply, that is, a constant current load.

http://cache.national.com/ds/LM/LM117.pdf

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[jimkirk]

An opamp, transistor (low Vgs MOSFET if you can find one), voltage reference and a couple resistors will give you a suitable current sink. Another opamp or comparator will allow you to detect the battery voltage and turn the current off when you reach the end of life to avoid damage to the battery.

If you use low power parts (example, the LT6000 can be powered by 1.8 volts at 13 uA with rail to rail input and output), you can have the whole thing powered by the battery itself.

Add an opamp to sense the voltage and a reference and you shouls be able to do the whole thing under about 50 uA.

Just adjust the current to account for the circuit drain, and be aware that the small current will still be loading the battery once you've reached the cutoff voltage. But even that should drop to even closer to zero below 1.4 volts or so.

Of course, heat sink the transistor adequately for the watt or so you're dissipating.

 

[OperaHouse]

You have two problems; low voltage and low current. If you use a transistor, the drive current wil be added to the load current. A FET would be better and you wil have to power the op amp from an external source. Because of the low voltage, a three terminal regulator will not work unless a constant voltage source is placed in series. The timing required makes using computer control a reasonable choice.

 

[jimkirk]

You can get MOSFETs with very low Gate Threshold Voltage. IRF7410 from International Rectifier is a P-channel with Vgs(th) specified at -0.4 to -0.9 volts. It will typically conduct 1 amp of drain current with only -1 volt gate source.

A bandgap refernce will get you 1.24 volts with 10 uA or so of bias. Use a couple high value resistors to divide this down to 100 mV or so as the reference to the current sink and comparator to do your end of life voltage shut-down. That will leave plenty of voltage to drive the MOSFET to conduct 200 mA or more.

With this, you won't need any external source.

As OperaHouse says, if you go the bipolar junction transistor route, you'll need to account for the base current. Probably about 1% or so with a decent transistor. A Darlington might help, and you'll probably have enough voltage to drive the two Vbe drops even at end of life.

 

[nbucska]

you want 50 to 200 mA -- let's set the goal to sqr(50*200)=
100 at medium voltage i.e 25 ohm == 60 mA @ 2 V and 114 mA @ 3.6 V.

If you want, for 5 mA more you can add a comparator+LED
to indicate low voltage.

If you want to disconnect, use a saturated NPN
and recalculate the R -- if you are fussy -- to make
up \for the Vsat.


Plesae read FAQ240-1032
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http://geocities.com/nbucska/>

 

[nbucska]

sorry 4 typo: it is 80 mA at 2 V.

Plesae read FAQ240-1032
My WEB: <

http://geocities.com/nbucska/>

 

[johnwqtan]

Thanks for all the help. I will try a few and let you guys know how it worked out. Again, really appreciate the help.

 

[AllorNothing]

The LM317 is really the way to go. Cheap, reliable, easy.