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# Car battery circuit design - comments, feedback?

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Hi,

I have designed, based on another car battery charger (http://www.aaroncake.net/circuits/charger1.asp?showcomments=all), a circuit to charge a lead-acid battery to about 14.4V then auto-shut off and light an LED.

I think it may need some work but will the concept work?? Am I crazy? The supply voltage will be from a computer PSU regulated at 15V

Please forgive the diagram, KTechLab doesn't have relays as an option.

Version 0.2 is here:

I am stuck on how to get it to switch off the power to the battery, but still monitor it so it float charges the battery, so when the supply voltage drops again it begins to charge, and so on..

How it works:

The HIGH output on the comparator keeps the relay ON, then as the supply voltage increases due to the increased resistance in the battery (decreased current flow), which once gets higher than the reference voltage, the voltage comparator output is then LOW, switching on the LED indicator

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With the capacitor in series with the two resistors then your circuit does not have a reference voltage. The input is at +15V.

Usually a reference voltage is a regulated voltage.

The LM311 will not work with its inputs so close to its positive supply voltage because its inputs have a typical max input of 13.5V when it has a 15V supply.

The LM311 must have a pullup resistor on its output.

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Thanks Audioguru,

I have switched LM311 with LM339 which has a voltage range 2-36V: http://www.national.com/mpf/LM/LM339.html#Overview

The 15V is regulated by a computer PSU. Is that what you mean?

I have added a pullup resistor on LM339 output - value as per datasheet

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The "input common mode voltage range" of an opamp or comparator is the range of voltages that the inputs still work. The max common mode input voltage for an LM339 quad comparator or an LM393 dual comparator is 1.5V less than its positive supply voltage (the same as the LM311 comparator) so again your circuit will not work. Reduce the voltage at the (-) input of the comparator with a voltage divider made with two resistors and reduce the voltage at the (+) input.

The voltage divider you are using has very low value resistors that will dissipate almost 1W. The input current of the comparator is 244,000 times less so you can use resistor values at least 1000 times higher.

The datasheets for all comparators recommend adding a little hysteresis (positive feedback) with one resistor so that the output has a fast "snap action" when the inputs are at the threshold voltage so that the output does not oscillate.

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Thank you audioguru,

I have gone back and done some calculations with the inputs, and have dramatically reduced the current.

I don't understand the part about positive feedback but I am reading the datasheets now and wikipedia.

my interpretation of that is:

and I'm not sure on what value that resistor should have, doing more reading - Schmitt trigger?

edit: Ok so I did more reading and so for positive feedback I need to make the INPUT more (1 - 10mV more) so that it acts fast and sharp to prevent oscillation.... now I currently thinking how I can do that

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Now your circuit will work properly. The hysteresis resistor is a high value. Try 1M.

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thank you very much audioguru, I have bought the LM339 IC and almost ready to test.

but first I want to add a feature to auto stop charging battery once the LED goes on. it doesn't have to keep monitoring it would be satisfactory for it to completely shut off until reactivated manually via a push button

my thoughts there are 2 ways to do this

1. tap into the computer supply voltage regulator and connect the output to the adjuster

2. have it modular design: have the output swich on a N/C relay. and add a diode to prevent battery from powering the circuit once switched off

i prefer 2. as then PSU can supply function as generic PSU when not being used as a charger

drawing up a schematic now for 2. will post shortly

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As promised! I'm really pretty good about it but I'm sure I've missed something

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Your latest circuit will work but the transistors should not be "emitter-followers". Instead they should be common-emitter for less voltage loss.

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I've fixed up so they are common emitter, thanks for the tip:

How would I find out the forward voltage drop over the lead acid battery while charging? I suppose that may not apply it's not an active device. Would it just a matter of measuring the battery's internal resistance and using Ohm's law? (edit: it is the charging voltage of the battery, the excess voltage will drop over the resistor)

It appears I may be charging this battery at about 16V (edit: no, I'm not -> the excess voltage drops across the resistor in series)

Maybe I'd be better off using 14V and supplying the IC chip 16V from another power out in the PSU (common ground). (edit, no it's fine)

I may have designed this whole thing badly :) (edit: maybe, but it appears to work)

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Just in case anyone was following this, here's a pic of the prototype:

It appears to functioning correctly. I have limited the current more than normal as I am using thin aligator cables.

Oh and the power is coming from the 5V and -12V rail, which is giving about 16V. I tried tapping into the regulator and changing one of the resistors, but that didn't work and it's really crammed and I can't see or move anything! But this works for this purpose.

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• 1 month later...

A follow-up, I have soldered this to perf board, and hooked it all up but there's a problem...

It charges at 700mA then gradually drops to say 500mA, but it never seems to get over 12.8V in the Vout at voltage divider at R8-R9. I can go up to 1100mA but I don't see it'll make a difference.... It's a 12Ah motorcycle battery, I have left it for a week charging I think at once, and still not really getting in 13V range.

And I topped up the fluid levels to just under MAX, but after a couple of days of charging some of them are almost below the LOWEST level! And it's not been overcharged, as the voltage was only 12.8V when I removed it.

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The battery won't charge fully.

A lead acid will slowly charge if the voltage is connected to a 13.8V constant voltage source, it will fast charge if the voltage is charged until the voltage reaches 14.4V then is disconnected or floated at 13.8V to prevent gassing. If the voltage never reaches 13.8V, the battery will never fully charge.

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Oh I think I probably confused people with all the different diagrams. The battery is getting > 14V across the terminals when charging, as measured with multimeter as per the design, but the "supply voltage" comparison voltage for the voltage comparator is not getting above 12.8V. And in fact when I disconnect the battery, and measure open circuit voltage straight away it is still only around that 12.8V mark.

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The original circuit posted seems to operate by switching on/off rapidly. Is that how it's operating?

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Hi KevinIV,

That is not how I intended it to operate. I don't really know what I'm talking about, but this is my idea behind the design:

The 16V regulator comes from a computer PSU. The first voltage divider (20k/100k) acts as the Reference voltage - which is about 13.56V in real life - to the Voltage Comparator IC. The second voltage divider (15k/100k) acts as the Battery voltage reference, which is fed into the Voltage Comparator IC as well. When the Battery voltage goes higher than the Reference voltage, the Output from the Voltage comparator IC switches on an LED and switches off power to the battery via that relay - apart from a very weak trickle charge (via the 1k resistor which bridges across the relay).

The 3.3 Ohm resistor acts a voltage shunt and the voltage across it is the supply voltage minus the battery voltage. As the battery voltage increases, the voltage across the Battery voltage reference increases as well.

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I misread the original circuit. But I don't think a 200mohm resistor would see a voltage drop, unless it's a high capacity battery.

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I'm not sure what you're referring to?

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I'm sorry, the circuit looks good. The 3.3 ohm resistor isn't power efficient, but maybe it can be replaced with a circuit.

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Another way to design the circuit might be to use a 3 amp NPN in series with the battery, and a 200mA PNP transistor in series with it's base to get a constant source of current to the battery.

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Thanks, I do think constant source of current to the battery would be very useful to charge the battery faster

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