[SOLVED] 9 V and 5 V UPS + battery monitoring

Sir,

I agree with you, but since im on short interms of money right now i can't buy DMM asap but the DMM that i will borrow from a friend is also my groupmates, so i think ill borrow it from him until the project will successful...

I have just noticed, while im charging on sir, the transformer getting hot. ive doubled checked the connection on your circuit sir before charging the battery and testing the circuit itself, ill stay the circuit on for maybe 12 hours or more for me to see if the battery can back up enough the circuit. Ill send some progress report tomorrow sir regarding the UPS, hopefully no troubles will be face.

---------- Post added at 15:15 ---------- Previous post was at 13:46 ----------

Question again sir, if the mains are off and the battery will back up the circuit/ system, will the LED indicator on the circuit will still be on?

dabby21 said:

I have just noticed, while im charging on sir, the transformer getting hot. ive doubled checked the connection on your circuit sir before charging the battery and testing the circuit itself, ill stay the circuit on for maybe 12 hours or more for me to see if the battery can back up enough the circuit. Ill send some progress report tomorrow sir regarding the UPS, hopefully no troubles will be face.

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Just dropped in on this thread. If the transformer gets too hot to keep your hand on it, then the battery is drawing too much current. In that case it may help to install a higher resistance than 22 ohms. It will limit charging current further, and it will take a longer time to charge the battery.

The hotter the transformer gets, the more danger of varnish breaking down between windings, possibly creating shorts and ruining the transformer.

Question again sir, if the mains are off and the battery will back up the circuit/ system, will the LED indicator on the circuit will still be on?

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Yes. It will only draw a little current, as compared to the much greater current needed to power the inverter.

Pjdd has been leading you right. Good luck.

Sir brad,

Thanks for the information, i have touched the transformer just now, its not that too hot, my hand still manage to keep on touch with it, so if the transformer getting hot,the circuit is ok, i mean , ive correctly mounted the components plus the PCB layout. So its a good sign for being on the right track how i mounted them?

Regardless with that issue sir brad, all had been cleared and noted. So ill charge the battery overnight and see tomorrow if the transformer will get really really hot. Again, ill post the progress and problems again tomorrow sirs. thanks

---------- Post added at 16:35 ---------- Previous post was at 16:28 ----------

Just by now sir, the wire wound 22ohms resistor getting hot too, way more hotter than the transformer. Is it natural? a good sign? or the other?

Try to think of some way to give us an idea of how hot the transformer and the resistor get.

First, the transformer: Is it just a bit warm or quite hot? After it has been on for about half an hour, can you press your fingers firmly against it and keep them there without becoming uncomfortable? If you can do that, it's still safe for the transformer.

Second, the resistor: It's normal for it to get a bit hot, but not too hot. Can you grip it with your thumb and forefinger without danger of burning your fingers? If you can do that, it's OK.

Brad, this is how I guesstimated the proper value of the charging resistor without knowing the precise characteristics of the transformer. I estimated the no-load secondary voltage at about 13V AC. With the rest of the load turned off and the transformer being loaded mainly by the battery, the rectified and filtered DC output would be around 16V. Subtract a drop of ~0.7V by D5 and we get about 3.3V/(22 ohms) or 0.15A charging current. If the load is switched on, the DC voltage would drop and the charging current would also be less.

In the absence of a proper charging control circuit, this was a compromise to avoid both too feeble a charging current and rapid overcharging. Even if the DC output were as much as 17.5V, the charging current would still be less than 0.25A.

Question again sir, if the mains are off and the battery will back up the circuit/ system, will the LED indicator on the circuit will still be on?

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If you mean the single LED in series with the 1.5k resistor (not the battery monitor LEDs), the answer is no. Current flow from the battery to that LED is blocked by D5.

I agree with you, but since im on short interms of money right now i can't buy DMM asap but the DMM that i will borrow from a friend is also my groupmates, so i think ill borrow it from him until the project will successful...

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I understand. I know what it's like to be just starting out in electronics and being short of funds to buy what we need. I used to be in that position myself for a long time.

However, just to make sure that you're not overestimating the cost of an entry-level DMM, let me give you an idea of what they cost in terms of US dollars. (I'm not in the USA myself but the US $ is universally known). A basic Chinese DMM starts from about $2-3 and $10-20 models are quite good although they are not as rugged as premium brands like Fluke.

Pjdd said:

Brad, this is how I guesstimated the proper value of the charging resistor without knowing the precise characteristics of the transformer. I estimated the no-load secondary voltage at about 13V AC. With the rest of the load turned off and the transformer being loaded mainly by the battery, the rectified and filtered DC output would be around 16V. Subtract a drop of ~0.7V by D5 and we get about 3.3V/(22 ohms) or 0.15A charging current. If the load is switched on, the DC voltage would drop and the charging current would also be less.

In the absence of a proper charging control circuit, this was a compromise to avoid both too feeble a charging current and rapid overcharging. Even if the DC output were as much as 17.5V, the charging current would still be less than 0.25A.

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This is sound. We don't know whether the UPS switches off the charge when the battery reaches an upper limit. So .25 A should not be too severe of a trickle charge (even if it is doing so 24 hrs a day to a charged 5AH Lead-Acid battery).

This was in my mind when I said 'automatic charger' a few posts back, instead of simply charger.

Pjdd said:

Try to think of some way to give us an idea of how hot the transformer and the resistor get.

First, the transformer: Is it just a bit warm or quite hot? After it has been on for about half an hour, can you press your fingers firmly against it and keep them there without becoming uncomfortable? If you can do that, it's still safe for the transformer.

Second, the resistor: It's normal for it to get a bit hot, but not too hot. Can you grip it with your thumb and forefinger without danger of burning your fingers? If you can do that, it's OK.

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Sir, i think the transformer and the resistor are all safe, last night i was charging the battery and monitor it for 6 hours, yet i can still touched the transformer even pressing my fingers on without any burns at all, its not that too hot, its just getting hotter. For the resistor, i can still manage to pressed my fingers on it with same results, no burns but not that comfortable compare to the transformer. I think the heat produced by the resistor is more than the transformer.

Pjdd said:

If you mean the single LED in series with the 1.5k resistor (not the battery monitor LEDs), the answer is no. Current flow from the battery to that LED is blocked by D5

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Sir,
If i want to have an Battery Indicator if the mains were off, it will be like a load after D5? i mean just like putting up the battery with same polarities? Or maybe it will be a good idea to put that Battery Indicator on the battery monitoring?


By the way sirs,

I am still charging the battery right now, the total hours is about 15 hours, ill still go for it for 18-19 hours, and test the output same goes with the battery.

Sirs,

Question again, the resistor cool down and not getting hot sir. what does it mean? the battery is full charge?

Question again, the resistor cool down and not getting hot sir. what does it mean? the battery is full charge?

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Yes, it sounds as though the battery is charged.

It's expected for the 22 ohm resistor to get hot while charging. If just .22 amp is going through it, then that is over 1W of power which is turning to heat. It can make a component too hot to hold between the fingers.

If i want to have an Battery Indicator if the mains were off, it will be like a load after D5? i mean just like putting up the battery with same polarities? Or maybe it will be a good idea to put that Battery Indicator on the battery monitoring?

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It is okay to leave the led monitor connected at all times, whether it is charging or powering a load. Those are the times you want to keep track of the battery state of charge.

Notice which is the topmost led lit up. Can you check the battery volt level with a meter now? It is probably around 13 or 13.5 V, because it has been charging.

If it goes over 15 volts or so, then it is overcharging.

However when the battery sits idle, disconnected from everything, you need not have the led monitor attached. It will drain the battery slowly if attached. Even if nothing is attached, the battery will self-discharge slowly.

Sirs,

Ive been charging the battery right now and have some reading, before the battery was charged, it reads 12 V and after 2 hours of charging it has a reading of 13V. ive tried to turn off the mains and see if it can output 9V and 5V, unfortunately it did not. After turning the mains the circuit outputted on LM323K 5.1V and for 7809 it outputted 9 V.

Questions sir,

Based on the observation, 13V cant supply the circuit, what is the range of the battery that it can supply the circuit if the mains are off?

Is it safe for our micro controller to supply 5.1 sir? and how can i measure the Amperes on the 5V output. since we are in need of 5V and minimum of 1.5 A for our GSM module.

Thanks

Sorry for having been silent for some time. I've been concentrating on something else.

It can't be the battery voltage that's causing it to fail to supply power with the mains off. Anything from 11V upwards should work. There must be some other reason. Please check the actual construction again. For example, make sure D7 is fitted with the correct polarity. If it's reversed, it will not work on battery power.

Regarding the 5.1V output, it's within the specs of the LM323. Practical devices always have a +/- tolerance of some %. 5.1V should be OK for the uC as most 5V devices are specified to work within a range of at least 4.75-5.25V. You can check the datasheet to make sure.

https://obrazki.elektroda.pl/80_1326795827.jpg
Sir, is the above circuit same as yours? we made it as a full wave since our transformer is a center tap.

https://obrazki.elektroda.pl/87_1326795827.jpg
here is our PCB layout, i have in doubt with it sir, ive labeled it on white.

Just by now sir the circuit was in my bag together with the battery and transformer, when im trying to get it out, it smoked, coming from the battery wires through the circuit, i dont know what happened sir but i guess it shorted. i did measure immediately the battery still its 12V, from 13 to 12V. the battery wires melt on some part of it. It was my fault then, since its been open circuit and the feet of the components were not cut for i am trying to observe it. Sir, any idea of what components may be affected and will result to serious damage? like it will malfunction. Thanks sir

---------- Post added at 11:42 ---------- Previous post was at 11:21 ----------

https://obrazki.elektroda.pl/81_1326796948.jpg

Here's our transformer sir.

---------- Post added at 12:59 ---------- Previous post was at 11:42 ----------

Sirs,

what number of stranded wires will i used to connect the battery to the circuit? same goes with the LM323K. I've tested the circuit if its functioning, fortunately sir it is, the LED indicator still lit up. But any idea sir for serious damage on the components? Really have no idea why the stranded wire of battery smoked. all through out the circuit.

---------- Post added at 13:07 ---------- Previous post was at 12:59 ----------

Sirs,

Just remembered something, a while ago sir while im testing the output of the circuit,around 5 hours from now, when the main is on, the output of the circuit to the battery is about 16 V i think. but I'm sure its more than 15 V with no battery attached on it, I made the VOM as a substitute load for the battery for me to know the output voltage, what does it mean sir? does the circuit have a correct output to charge the battery? more than 15V can charge up the battery? thanks

The wire size is not critical and almost any wire of reasonable size will do. I don't know what wire standard you use over there. As an example using the British standard wire gauge (SWG), one common size is 14x36 swg which is good enough. It means that there are 14 strands of No.36 swg wires. 36 swg is 0.0076" (0.193mm) in diameter. And then there are other sizes like 23x36swg, etc. Some flexible wires also use a larger number of smaller strands and heavy-duty wires may use larger strands.

For the moment, don't concern yourself too much with the wire size. The reason your wire got burned must have been a short-circuit, probably the positive and negative lines of the battery must have accidentally got shorted somehow.

One practical point is that a larger wire is mechanically stronger. When a wire breaks from mechanical stress, it is usually at the point where the sleeving (insulation) was removed to make a connection and the bare copper just emerges from the insulation. So, a larger wire has the advantage of mechanical reliability even if the higher current carrying capacity is not needed.



One thing you should keep in mind is not to make the wires longer than necessary. It is particularly important that the wires to the LM323 should be as short as is practicable.

It's quite normal for the charger output to be about 15-16V when the battery is not attached. If it wasn't higher then the battery voltage, it would not be able to charge it. When the battery is attached, the voltage difference will be absorbed by the 22-ohm resistor. It is the voltage difference divided by the resistance that determines the charging current.

Regarding possible damage caused by the accidental short circuit, it's hard to even make a guess from a distance. What you can do is to power up the unit again and check the various voltages.

Sir thanks for a detailed explanation about wires, i was worried about the exact wires to be used as to correct amperes and voltages, that was my outlook then, but now its clear for me and corrected.

I've powered up the circuit again sir and still ok, the LED still light up, i cant measure yet the voltage, i am not very lucky this day. The probe of the VOM was detached from the pins soldered on the steel probe. I have the VOM here sir but will test it all again tomorrow, will buy another test probe for this.

Sir about the bridge type diode that we have modified. We change it to full wave. is it ok sir? what output will i expect on each capacitor sir?

I forgot to answer some of the points raised in the first part of your post.

Your rectifier configuration is OK if you use a centre-tapped transformer. You don't have to parallel two 1N5408s on each side as one diode each has more than enough current rating for your load. On the other hand, paralleling them does no harm.

The track marked "WRONG?" is OK, it's not wrong. It should be there to join the two rectifier sections together.

Copy Sir,

Ill report again tomorrow regarding the circuit, hope its good news. and will expect to charge up the battery and will back up the circuit.Will be more careful now regarding short circuit, just hoping the other components did not take any damage at all.

what output will i expect on each capacitor sir?

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Without knowing the characteristics of your transformer such as the no-load voltage at the secondary, and the winding resistances, it's not possible to make accurate calculations. My estimate is that when there's no load at the outputs and the battery is disconnected, the voltage at the 2200uF cap will be about 17V, and about 0.6V less at the 100uF capacitor.

When the battery is being charged, the voltages should drop a bit from those values, and drop still more when the load is turned on.

That reminds me of something I intended to say earlier today but forgot to. The 100uF cap should have a voltage rating higher than 16V. 25V is fine. This is because the unloaded voltage at that point can sometimes be higher than 16V.

The voltages at the 0.1uF caps should be the output voltages of their respective regulators, that is, 9V and 5V.

dabby21 said:

The probe of the VOM was detached from the pins soldered on the steel probe. I have the VOM here sir but will test it all again tomorrow, will buy another test probe for this.

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Happens to my meters incessantly.

See if you can unscrew the plastic piece from the metal pin piece. If so then you can solder the probe wire to the pin.

There may be a bowl filled with the broken-off wire. Heat it and extract it.

Scrape and tin the probe wire. Solder it in the bowl and you've got your new meter probe.

Abiso ng Redirect

Above link is the test probes of my classmates VOM, its kinda glued on the handle, i cant loosen it sir. or maybe its china made. Im going to the electronics shop right now and try to ask some help about this. Ill report immediately about the circuit.

Sirs,

Goodnews, the incident last night were it has some shortcircuit, did not done serious damage to the circuit and the battery. Im testing out right now sir. Sir Pjdd was right, the battery can back up the circuit with 11V and up if the mains are off.

https://obrazki.elektroda.pl/31_1326857600.jpg

the above picture shows that the battery back up the system with the mains off. It reads 9V output from LM7809. I'll post later for the 5V from LM323K. Need to go to school. Thanks sirs

dabby21 said:

Above link is the test probes of my classmates VOM, its kinda glued on the handle, i cant loosen it sir. or maybe its china made. Im going to the electronics shop right now and try to ask some help about this. Ill report immediately about the circuit.

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Yes, one of my meters has a probe like that. Molded around the wire. The connection broke after a while.

It's not easy to fix like the type in my previous post (#57).

I fixed it but first I had to to cut/file away plastic to expose 1/4 inch of the inner conductor. It took some care.

After soldering on the probe wire, I applied hot melt glue around the exposed connection. The glue forms a flexible insulation which bends with the wire.

The wire used to bend sharply when new. Now it bends gradually. It is less prone to break again.