Battery Charger help! (sch included)

[Cloud9]

Oooh more brains! Thanks all.

Yes, Lead-Acid or Gel-Cell. specs for the battery are 12VDC 33mAh

 

[RadioRon]

Oooh more brains! Thanks all.

Yes, Lead-Acid or Gel-Cell. specs for the battery are 12VDC 33mAh

Well, 33mAh can't be right, that's way too little. In any case, the reason I ask is that I've had to repair a few chargers in my time and found that with many lead-acid charger circuits (especially automotive ones) you have to attach the battery or its equivalent circuit in order to test or understand the charger operation properly. In other words, it doesn't make sense to look at the circuit solely as a voltage source or current source feeding a load. You may have to imagine a battery attached before you begin to analyze. I don't know if that applies to this one or not, just something to keep in mind.

 

[Cloud9]

Yea thanks, I learned that one the hard way unfortunately lol. I am testing with a load connected.

Also, the 33mAh is the battery specs.

The charger will not exceed 4Amps.

 

[RadioRon]

OK, now I've gone through the schematic and read Roff's observations and have these of my own:

- voltage regulation is done via the feedback path of CR2, Cr1, D2 around to Q2 and thus to R18. This is a negative feedback system. The fact that Q4 gets hot indicates that it is operating linearily, ie. not saturated.

- for analysis purposes, it might be ok to ignore the fast charge detect circuit as its main function is to light an LED. There is some positive feedback to the regulator via R1 but I suspect this is simply to add a little bit of hysterisis into the system.
- also for analysis purposes, I suggest that you imagine removing CR3 and ignoring the over-current protection circuit for now.

Now, if VR1 is adjusted slowly CCW, there would be more current flowing into the base of Q3 which in turn attempts to increase the voltage at the collector of Q4. As this voltage goes up, so then does the voltage to the base of Q2 go up and then the emitter of Q2 voltage increases. This tends to cause Q3 to turn off to some degree and the system finds a new balance. So, then, what is the purpose of VR1? Does it simply set the output voltage of the linear regulator? In other words, does it set finish voltage of the battery?


So, back to the problem. By the way, can you restate the problem as you see it right now?

EDIT: Another way to view Q2 and Q3 is that they are working as a differential pair, sharing the same emitter voltage. When one is more on, the other must be more off. So, it would seem that turning VR1 more clockwise means Q3 gets to be "more on" than Q2 and the so the voltage output goes down. Turn VR1 CW and the output voltage would go up.

 

[Cloud9]

Great observations, thanks.

In simplest terms, the problem appears to be that the charger never wants to cut off once it is activated. Of course, taking into consideration the Fast-Charge Detect only lights a LED, means this may not necessarily indicate it will not stop charging, just that the LED will not cutoff once lit.
However, what is known, is that when charging is complete the fast-charge LED shuts off when charging is complete on a good unit.

I suppose the first question is exactly what conditions cause the charger stop? Also, a slow rolloff or all at once? Seems the rolloff makes more sense with the integrator ramping up on the fast detect for some reason (surely the circuit didnt need an integrator just to light the led - does it?) Once I know this, I need to develop a good way to test this function (cut-off).

I've been abit concerned about just leaving it on (with my multimeter setup as ammeter to monitor charge current) because I do not want to damage neither my meter nor the battery especially. I fear the battery might explode!

I ran it for probably an hour with the average current being 1.2A before I felt how hot the heatsink mount was. I may be overreacting though since the boards (with their heatsink) get mounted to a humogous aluminum sink on the chassis - obviously we expected it to get very hot.

On a known good unit:
Today, I tried charging a battery that was around 6.8V starting at about 9am. It wasnt until 2 pm that the charge finally finished. I measured the batt voltage at different times and found that it exceeded 12.5V for at least an hour before finishing. After charging finshed the batt voltage settled to around 11.58V, that was a surprise.

In general this is the process I take when I first connect a board:
Set the VR1 full CW. This makes it so when I connect a slightly low battery (11.8V or so) it does NOT start charging.

Next I will turn the pot CCW slowly until the charge light kicks on. I was assuming this set my trip voltage thinking that anything below this will activate the charge.

Now heres the point where I am. I sit waiting for it to cut-off at some specific voltage but that doesnt happen. I turn the pot in vain trying to make it cutoff. This I why I suspect the function of the integrator in the fast-charge is there ramp off the voltage over time.

Hope this helps, I will have more specific info in the AM.

Thanks a ton for your time,
Cloud

PS:
Could that 10K at the top of the circuit be used as a trickle charge when the main charge is off - or is just the bias for VR1?

 

[Roff]

RadioRon, you are very astute. That path totally escaped me. I just thought it was an indicator circuit. I've been running some sims, which are very revealing. The maintenance voltage that the circuit settles at is dependent on load resistance.

Cloud9, is the battery loaded while it is charging? If so, what is the load resistance or current?

 

[Cloud9]

The problem seems to have appeared when we changed battery types. Similar specs but a different manufacturer. I wonder if its just coincidence.

Roff, what do you mean? It is a valid battery usually between 10-11V. As far as current and load resistance, I dont know. I wish I had one of those fancy batt analyzers right about now.

 

[RadioRon]

Cloud9, you keep mentioning a shut-off function and you know, to be honest, I don't see anything that I would call shut-off. What I see is a charger that applies a relatively constant voltage (subject to Roffs analysis which I suspect will tell us more about how the charger is dependent on load) and just keeps pushing current into a load until the load's voltage fights back and matches the charger's voltage. The fact of life with lead acid batteries is that it is quite OK to trickle charge them all the time and that is what happens with this charger I think. It just brings the battery up to its set-point (say, 13.6 V) and as the battery gets near full charge and reaches this voltage, the charger just keeps it there.

Lead acid batteries, like SLA types, don't suffer from this as long as the trickle charge current is fairly low, which it will be if you set the voltage correctly. Such batteries, while charging, typically will go up to about 14.5 or 14.6 volts if the charger puts out a voltage higher than this. It is not wise to charge them above this voltage. If you then remove the charger, their terminal voltage (even with no load) will drop back down to the natural cell voltage which is much closer to 12.0 volts I think. For example, in a car, if the motor is running (and the alternator is charging the battery) you should read 13.6 volts at the battery terminals. The moment you shut the car off (and the lights and everything else in the car) your battery voltage should drop to about 12 V.

I recommend that you set your charger output voltage at about 14.0 to 14.3 volts open circuit. Then, if you connect a battery thats a bit flat, with say a terminal voltage of 11 volts, then you can expect a current of roughly 3 volts divided by all the circuit and battery internal resistances to flow.

You might also want to read some application notes from a major lead-acid battery maker on how to charge their batteries. One good maker is Enersys who I believe also own Genesis, Optima, and Hawker batteries, all very good batteries.

Getting back to that cut-off function for a moment, it is possible that the little bit of hysterisis that I mentioned before is their way of slightly shifting the end-point as the battery charge current ramps down slowly so that when the charge current drops below 560 mA the effect at VR1 of the fast charge detector changing state is a little bit like somebody adjusted VR1 slightly CCW which would force the endpoint voltage slightly lower and drop the charge suddenly down to a lower trickle rate, lower than 560 mA. So I think this is their simplified version of shut-off, the charge current drops down to a trickle rate and just keeps trickle charging.

You are too focussed on that C3. The time constant that it brings is roughly 100 microseconds, so it doesn't do all that much, just slows down the rate of change of U1B to avoid glitching I think. It is important in maintaining stability of the system I think, but for just testing if the system is working or not, don't worry about it.

Now, I'd like to hear what Roff has learned about the effect of the battery being attached in his simulation.

 

[RadioRon]

I went back and reread your statement of the problem again. You seem to have the problem that the charger goes into fast charge OK when a battery is attached, but does not "shut off" as expected. This "shut-off" which is actually just a drop down to trickle charging condition, occurs when the fast charge LED turns off. So, the red led is not turning off ever, eh?

This would indicate that you have your set point turned up too high. If you set point is at, say, 13.7 volts, then the charge current MUST decrease slowly as the battery's charge voltage approaches 13.7 volts. When the charge current drops below 560 mA (according to the schematic) the red LED must turn off. When this happens the charging current will fall to some value considerably lower than 560 mA (because of the hysterisis feedback through R1) and then continue to decrease to some very low value depending on the battery construction.

You will probably have to experiment a bit by:
- adjust the open circuit output of the charger to some value like 13.8 volts using VR1.
- attach a flat battery and continously measure the current flowing into it (best to use a 0.1 ohm 1 watt resistor in series and measuring the dC voltage across that resistor with a battery operated meter. 1 amp of current will read as 100 mV on the meter). Attach another meter to measure the battery voltage and watch it increase.
- note that the battery voltage should increase to 13.8 volts or some value near that and then after it switchs the red led off note what the current is.
- continue to measure the battery voltage. It should not go higher than your initial set voltage (13.8 volts in this example)

They are playing "as good as it gets", a movie, on the tv now and its one of my favorites, so I'm gonna go watch. See ya later.

 

[Brian Hoskins]

You might also want to read some application notes from a major lead-acid battery maker on how to charge their batteries.

EPE did a good project on lead acid battery chargers. In the write up they explain quite comprehensively the optimum way to charge the battery. I can't remember which month is was now, but I reckon it was within the last 12 months. Might be worth a scout on their website.

Brian

 

[Cloud9]

Well I am back at the shop again this morning and I see you've left me lots of info to play with. Your thoughts sound pretty promising.

I will adjust as you suggest, and also give it more time to finish charging. Like I said before, I've been abit timid about letting it run until it stops. I will post my results later.

Thanks

 

[Roff]

The problem seems to have appeared when we changed battery types. Similar specs but a different manufacturer. I wonder if its just coincidence.

Roff, what do you mean? It is a valid battery usually between 10-11V. As far as current and load resistance, I dont know. I wish I had one of those fancy batt analyzers right about now.

Is the battery powering something while you are trying to charge it, or do you remove the battery from the load and connect it to the charger? If it is still connected to the load while you are charging it, what is the load?

 

[Cloud9]

Oh sorry Roff, I see. No load, in any case we will disconnect our device from the battery.

Also just to clarify, where I should be measuring my 13.8ish volts when adjusting the open circuit? I used the bottom of R17 but I fear that wasnt the appropriate place.

I appreciate the time you guys have taken to help me out!

 

[Roff]

Oh sorry Roff, I see. No load, in any case we will disconnect our device from the battery.

Also just to clarify, where I should be measuring my 13.8ish volts when adjusting the open circuit? I used the bottom of R17 but I fear that wasnt the appropriate place.

I appreciate the time you guys have taken to help me out!

I think RadioRon is talking about 13.8V across the battery. I don't think we know enough to tell you how you can set the charger without a battery in place. I realize that puts you in a difficult position. Maybe we can figure it out when I do some more simulations.
I'm running some interesting sims, but I am finding that the charge cutoff voltage (for a fixed pot setting) is very dependent on the voltage from your rectifier (the voltage across C1). Can you measure that when there is no battery connected to the charger, and let us know what you get?

 

[Brian Hoskins]

Oh sorry Roff, I see. No load, in any case we will disconnect our device from the battery.

Also just to clarify, where I should be measuring my 13.8ish volts when adjusting the open circuit? I used the bottom of R17 but I fear that wasnt the appropriate place.

I appreciate the time you guys have taken to help me out!

Con4 pins 3-4 would seem as good a place as any to measure it. I presume these points connect to your large crocodile clips for direct connection to the battery. I think what Ron is suggesting is that you measure the output with no load attached and adjust the pot if necessary to achieve 13.7V at the output. Then, when your battery is connected, the current supplied will drop right off as the battery voltage approaches that of the charger output. (no potential difference, no current flow)

For now it might be an idea to measure the output off load and report your findings back here before you adjust the pot. If it's way off, then there's probably a problem elsewhere that needs to be investigated first - BEFORE making any adjustments.

Brian

 

[Brian Hoskins]

Ron - does the off load output need to be set at a slightly higher voltage than that required to charge the battery fully in order to take into account the headroom of the series pass regulator circuit?

I see a pnp transistor is used as the series pass device, which I think gives it low-dropout functionality but I'm not quite sure what this drop-out figure would be.

Brian

 

[RadioRon]

On reflection, I agree that we can't be sure of the best output setting without a bit of experimentation. Obviously it is necessary for there to be some resistance between the rectifiers and the battery in order to limit current and that must be contributed by Q4. So we can't assume that this circuit is an ideal constant voltage source such as we see in a DC power supply, but rather one with significant series resistance by design. I still think that the best starting point is, however, to choose an open-circuit voltage set point and work from there, charging a battery and watching how the circuit behaves.

Thinking back, didn't Cloud9 also say many posts ago that he measured 13 volts on all terminals of Q4. If this is true, I wonder why. One explanation might be that Q4 is blown and shorted. That would certainly explain why the unit won't come out of fast charge mode. Another explanation is that Q3 is completely off and the battery is charged up as much as the charger will go. But if Q3 is completely off, then Q4 must also be off and there can be no current flow into the battery except through R19, not enough to keep the Red LED on. Yet, C9 did say the red LED was on in this state.

Puzzled.

 

[Roff]

Ron - does the off load output need to be set at a slightly higher voltage than that required to charge the battery fully in order to take into account the headroom of the series pass regulator circuit?

I see a pnp transistor is used as the series pass device, which I think gives it low-dropout functionality but I'm not quite sure what this drop-out figure would be.

Brian

Brian, we've got two Rons in this thread. maybe you could be more specific.
The PNP is a Darlington, so it will always have at least ~0.7V Vce(sat).

 

[Brian Hoskins]

Thinking back, didn't Cloud9 also say many posts ago that he measured 13 volts on all terminals of Q4. If this is true, I wonder why. One explanation might be that Q4 is blown and shorted.
Puzzled.

Agreed - that might be worth a cold check. If it is shorted, it's a pretty simple test to find out.

Brian

 

[Brian Hoskins]

Brian, we've got two Rons in this thread. maybe you could be more specific.
The PNP is a Darlington, so it will always have at least ~0.7V Vce(sat).

Sorry about that I completely neglected to consider which Ron I was reffering to, but actually I'm happy responding to either of you (or anyone else who wanted to chip in) so I guess it doesn't matter.

Regarding your 0.7V headroom - agreed. I think with npn designs the base-emitter voltage is derived differently and means that a greater headroom is required.

So does the output voltage need to be set 0.7V above the required voltage to ensure the regulator only drops out when the required battery voltage is achieved?

Brian