simple precharging circuit will this work?

[Roff]

What's hysteresis? cutoff voltage range is 3.2V - 3.4V constant current is 100mA but a selectable range of 50mA to 200mA would be great for experimenting, on what does the 0V cells better in terms of revival rates.
terramir

If the current cuts off when the voltage is 3.3V (or whatever), and turns back on when the voltage drops to a lower (definable) level, that's hysteresis.

 

[terramir]

.2V below the set predefined voltage should be fine if the cell doesn't need an extra jolt it shouldn't drop more than .1V as long as it isn't above 3.3 even charged to 3.6 it will drop to 3.3-3.4 in a day or so. so it will do for the trickle charging to have a .2V hysteresis
terramir

 

[terramir]

Anyone have an idea what a good but chaep replacement circuit would look like because this one has a bit too much play
terramir

 

[Roff]

Anyone have an idea what a good but chaep replacement circuit would look like because this one has a bit too much play
terramir

I or one of the other guys can design it, but I'm going gold prospecting tomorrow, so if it's me, you'll have to wait.

 

[terramir]

Well I can wait a few days as long as I'm still charging cells that have retained some charge these circuits will completly do it's when I get to the 0V ones that I need definetly more control to ensure that I'm using a consistant protocol. But while your designing please keep cost and availiblilty of dip components cause I am miserable at soldering smd components. Have fun prospecting, hope you find enough at least to pay for the expenses Good luck .
terramir

 

[dougy83]

Well, I guess that leaves it to the 'jealous one'...

Have a look at the attached cct. There is a point marked 'I SET', which you need to put a voltage on to set the current; to work out the current into the LI cell, divide this voltage by R2 (which is suggested as 4.7 ohms). So 0.47 V to 'I SET' will provide ~100mA.

The other point is marked 'V SET', this is nominal voltage that the Li cell will go below 5V. So for 3.3V, 'V SET' should be ~1.7V.

There's some hysteresis on the voltage, so for 'V SET' of 1.7V, the turn off point will be at VBATT=3.42, on at VBATT=3.2V (assuming LM324 output goes to 4V when high). Decreasing 'V SET' will increase both these values; increasing it does the opposite.

The 2 voltages should be provided using pots, I suggest 220ohms - 1K. The voltages may be shared between all the circuits you use, and will make them all act in the same manner.

Good Luck!

EDIT: I forgot to add the 'done' LED. The LED can go from the output of the voltage comparator (pin 7 on the diagram), through an appropriate resistor then to ground.

 

[terramir]

Ok I'm a little confused but let's see if I got it right the complete schematic as you wrote it is attached. But let me see if I understand this right. one OpAmp U1A drives Q1 to regulate the constant current while U1B doesn't do anything till the proper cell voltage is reached. then it saturates Q2 instantly which turns Q1 off.
Where I'm confused it this How does this circuit work in this circuit exactly work? (cause I want to learn) I see that it's a quad op amp. (what does an opamp do?)

I see each chip can make two chargers that much I already see.
But lets see something else the voltage regulators in this case are just simple pots (voltage divider) I should use multiturn ones for precision I assume. Using pots also means that the circuit won't be very acurate if my PSU fluctuates in voltage too much (because the divider will be off then). I could use an lm317 to provide a sturdy 1.7V and use one pot to provide the 0.47 V to make it more accurate if need be Have to test my PSu first to find that out.
Did I get this right so far?
terramir

 

[dougy83]

I'm sure there are some opamp tutorials around, but I couldn't find a good one atm; hopefully someone can locate one for you.

Basically, for an opamp with negative feedback (resistor going from output to -ve input), assume the following rules (these are simplifications that allow you to work out what's going on):
1. the -ve & +ve input will be at the same voltage (the opamp will change its output voltage to cause this to happen)
2. the -ve & +ve input draw no current (they have very high impedance)

So the top opamp U1A has 0.47V applied to it's +ve input. It also has negative feedback through R1 & Q1, so the above 2 rules apply. So the -ve input is held at 0.47V too; this voltage is also across R2, which we can calculate to be passing 0.47V/4.7ohms = 100mA.
So, this opamp circuit will ensure 100mA is flowing through R2. A similar current (ignoring Q1 base current) will be flowing through the Li Cell.

The bottom opamp has positive feedback, which causes the output to be either high (~4V) or low (~0V); you can't use the above 2 rules for this one and it works like a comparator (i.e. when -ve input is less than +ve input, output is high; else output is low). When the output is low, the voltage divider R4 & R5 set a level of 1.58V (assuming 1.7V at V SET) on the +ve input of U1B. As the Li Cell charges, the voltage on the -ve input decreases until it passes 1.58V which causes output to go high. This turns on Q2, which turns off the constant current source. As the output is high, the voltage on the +ve input is increased to 1.8V (due to resistor divider).

As the Li Cell voltage discharges (for whatever reason), the voltage on the U1B -ve input increases. When it passes 1.8V, the output flips to become low again. Q2 is turned off, and the constant current is reapplied.

 

[dougy83]

But lets see something else the voltage regulators in this case are just simple pots (voltage divider) I should use multiturn ones for precision I assume. Using pots also means that the circuit won't be very acurate if my PSU fluctuates in voltage too much (because the divider will be off then). I could use an lm317 to provide a sturdy 1.7V and use one pot to provide the 0.47 V to make it more accurate if need be Have to test my PSu first to find that out.
Did I get this right so far?
terramir

Yep. Sounds good.

 

[terramir]

so what would happen if I made I set 0.23V or .96V I assume that then 50 mA or 200mA would be the current respectively. how about V set? if I set it to 1.6V or 1.8V (think I'm gonna avoid 1.8 LOL) would that then set the battery cut off voltage 3.32 and 3.52 respectively or would that mess the whole thing up? I bought 6 of those chips at this time that should be enough for 12 cells precharging which should be more than enough atm cause I only got 9 chargers atm. (well finishing 5 of the 9 right now two should be up and running with-in the hour. another 3 by tommorrow afternoon. If I can repair the paper one I built (something's wrong havn't had the time yet to trouble shoot I'll have ten. at a constant current of 100 mA the precharging should complete with-in 2 to 5 hours at the most. The full charging is what takes the longest because of the CCCV schema.
the current drops to 20mA @ about 3.63V then they are done.
Ty for the circuit I'll be breadboarding them tommorow.
terramir

 

[dougy83]

Yes. Current is set as you say.

With V SET=1.6, switch points are: 3.36/3.54
With V SET=1.8, switch points are: 3.08/3.31

You will see I made a mistake in my above explanation; increasing V SET increases the switching point of the Li Cell negative terminal, which means it decreases the Li Cell voltage switching point.

 

[Roff]

If you want better precision on your pot, put resistors in series with each of the end terminals, so that it only covers the range you want (plus a little on each end to account for resistor and pot tolerances). Use Ohm's law to calculate the values.

 

[dougy83]

Catch any gold?

T, I just realised that if your 5V suply is not stable, then the charge termination voltage is not stable.

So, you'll need to provide the 0.47V referenced to ground for I SET (you can use an LM317 and resistor divider); and you'll need a ~-3.2V referenced to the 5V rail for the V SET (you can use a LM337 for this).

Or just use a stable 5V supply and leave it as you had it.

 

[Roff]

Catch any gold?

The skunk followed me around all day (in a word, NO), but my buddy Mike found a nice 4.6 gram nugget, predrilled for a chain by mother nature. My other buddy Freddie found a little dink.
Prospecting with a metal detector is kinda like fishing. If you find (catch) something, great! If you don't find anything, you still have a great time.

 

[dougy83]

Cool! I'm surprised that there's gold just lying around; I guess you have to know a likely place to start looking, somewhat uninhabited - otherwise you just end up finding either nothing or a bunch of bottlecaps and other useless junk.

 

[Roff]

Cool! I'm surprised that there's gold just lying around; I guess you have to know a likely place to start looking, somewhat uninhabited - otherwise you just end up finding either nothing or a bunch of bottlecaps and other useless junk.

That's exactly what I found - useless junk, that is. I was in the mountains of eastern Oregon. I found bullets, .22 shells, a tiny shotgun pellet, a BB, a square nail, a boot tack, a rusty little tin can with a hand-made wire bail (handle), and some foil.
I do find gold on most of my prospecting trips.

Australia has some of the best gold prospecting on the planet.

 

[terramir]

Catch any gold?

T, I just realised that if your 5V suply is not stable, then the charge termination voltage is not stable.

So, you'll need to provide the 0.47V referenced to ground for I SET (you can use an LM317 and resistor divider); and you'll need a ~-3.2V referenced to the 5V rail for the V SET (you can use a LM337 for this).

Or just use a stable 5V supply and leave it as you had it.

Well I'll be using a pc-psu with a fairly stable 5V rail Once I have most of these circuits up and running I'll be using the 5V car lamp trick to provide a constant load so I can use the 7V trick for my other chargers. That should be enough so the lm317 in my otherchargers are stable (6V won't do at 3.65V 7 is the bare minimum. Anyways thanks alot I will be doing some experiments and as long at the charger turns off before 3.5 I'd be fine (would like it to be 3.3 exactly but heck I'll see what I'd get. PC-PSU's that are capable of more than 400W should be able to stay stable at these low loads ( 12 precharging 20 something charging (that's the plan)) will be less than 100Watts total drain it should be fine I'll run some tests. I got a spare 317 and well I got a 337 laying around somewhere as well.
terramir

 

[dougy83]

I don't follow what you said and I've no idea of what the car lamp trick &/| 7V trick are.

If the 5V is stable:
* you don't need the LM337
* V SET should be 1.8V for 3.3V turn off.

 

[terramir]

If you don't have a load on the 5 volt rail of a PC-PSu the other voltages get totally out of whack. (that's the car lamp trick) 7Volts is the differential between the 5V positive and 12V positive rail (effective 7V)
Got a question though I measure the V-SET on the voltage divider between ground and the "middle" of the pot right? cause I got a bit confused looking at something you said between positive etc.
terramir

 

[dougy83]

Oh right. I've seen people use a wire-wound resistor to load the 5V line.

Yes, V SET = 1.8 has been measured from ground. Just to confuse: if you measure from 5V, it will be 3.2V.