Automatic Lead Acid Battery Charger Circuit Using IC 555
The following article explains a simple, versatile automatic lead acid battery charger circuit. The circuit will allow you to charge all types of lead acid battery right from a 1 AH to a 1000 AH battery.
The IC 555 is so versatile, it can be considered the single chip solution for all circuit application needs. No doubt it's been utilized here too for yet another useful application.
A single IC 555, a handful of passive component is all that's needed for making this outstanding, fully automatic lead acid battery charger circuit.
The proposed design will automatically sense and keep the attached battery up to date.
The battery which is required to be charged may be kept connected to the circuit permanently, the circuit will continuously monitor the charge level, if the charge level exceeds the upper threshold, the circuit will cut off the charging voltage to it, and in case the charge falls below the lower set threshold, the circuit willconnect, and initiate the charging process.
The circuit may be understood with the following points:
Here the IC 555 is configured as a comparator for comparing the battery low and high voltage conditions at pin#2 and pin#6 respectively.
As per the internal circuit arrangement, a 555 IC will make its output pin#3 high when the potential at pin#2 goes below 1/3 of supply voltage.
The above position sustains even if the voltage at pin#2 tends to drift a little higher. This happens due to the internal set hysteresis level of the IC.
However if the voltage continues to drift higher, pin#6 gets hold of the situation and the moment it senses a potential difference higher than 2/3rd of supply voltage, it instantly reverts the output from high to low at pin#3.
In the proposed lead acid battery charger circuit design, it simply means that, the presets R2 and R5 should be set such that the relay just deactivates when the battery voltage goes below say 11.3V (for 12V batts) and activates when the battery voltage reaches above 14.2V.
Nothing can be as simple as this.
The power supply section is an ordinary bridge/capacitor network.
The diode rating will depend on the charging current rate of the battery. As a rule of thumb the diode current rating should be twice that of the battery charging rate, while the battery charging rate should be 1/10th of the battery AH rating.
It implies that TR1 should be around 1/10th of the connected battery AH rating.
The relay contact rating should be also selected as per the ampere rating of TR1.
How to set the battery cut off threshold
Initially keep the power to the circuit switched OFF.
Connect a variable power supply source across the battery points of the circuit.
Apply a voltage that may be exactly equal to the desired low voltage threshold level of the battery, then adjust R2, such that the relay just deactivates.
Next, slowly increase the voltage up to the desired higher voltage threshold of the battery, adjust R5 such that the relay just activates back.
The setting up of the circuit is now done.
Remove the external variable source, replace it with any battery which needs to be charged, connect the input of TR1 to mains, and switch ON.
The IC 555 is so versatile, it can be considered the single chip solution for all circuit application needs. No doubt it's been utilized here too for yet another useful application.
A single IC 555, a handful of passive component is all that's needed for making this outstanding, fully automatic lead acid battery charger circuit.
The proposed design will automatically sense and keep the attached battery up to date.
The battery which is required to be charged may be kept connected to the circuit permanently, the circuit will continuously monitor the charge level, if the charge level exceeds the upper threshold, the circuit will cut off the charging voltage to it, and in case the charge falls below the lower set threshold, the circuit willconnect, and initiate the charging process.
The circuit may be understood with the following points:
Here the IC 555 is configured as a comparator for comparing the battery low and high voltage conditions at pin#2 and pin#6 respectively.
As per the internal circuit arrangement, a 555 IC will make its output pin#3 high when the potential at pin#2 goes below 1/3 of supply voltage.
The above position sustains even if the voltage at pin#2 tends to drift a little higher. This happens due to the internal set hysteresis level of the IC.
However if the voltage continues to drift higher, pin#6 gets hold of the situation and the moment it senses a potential difference higher than 2/3rd of supply voltage, it instantly reverts the output from high to low at pin#3.
In the proposed lead acid battery charger circuit design, it simply means that, the presets R2 and R5 should be set such that the relay just deactivates when the battery voltage goes below say 11.3V (for 12V batts) and activates when the battery voltage reaches above 14.2V.
Nothing can be as simple as this.
The power supply section is an ordinary bridge/capacitor network.
The diode rating will depend on the charging current rate of the battery. As a rule of thumb the diode current rating should be twice that of the battery charging rate, while the battery charging rate should be 1/10th of the battery AH rating.
It implies that TR1 should be around 1/10th of the connected battery AH rating.
The relay contact rating should be also selected as per the ampere rating of TR1.
How to set the battery cut off threshold
Initially keep the power to the circuit switched OFF.
Connect a variable power supply source across the battery points of the circuit.
Apply a voltage that may be exactly equal to the desired low voltage threshold level of the battery, then adjust R2, such that the relay just deactivates.
Next, slowly increase the voltage up to the desired higher voltage threshold of the battery, adjust R5 such that the relay just activates back.
The setting up of the circuit is now done.
Remove the external variable source, replace it with any battery which needs to be charged, connect the input of TR1 to mains, and switch ON.
Good Day.
Sir, I have a 12v 500 watts inverter which is of new kind (using chopper instead of a big transformer). If Iconnect a solar panel as a power source to it instead of a lead acid battery, will it work ?. I just want two 25w energy saver lights and a fan to run on it. And what should be the specs of solar panel I use ?
Thank you.
Rashid
Yes it can be used, but the voltage will need to be first regulated at the specified input voltage level of the inverter by using a high current voltage regulator controller circuit.
In this post Ansari told that he has a 12v 500w chopper based inverter...
Will you explain about that chopper based inverter and can you drav a circuit for 12v 500w chopper based inverter with solar panel powered(to charge 150Ah battery) and with full automatic operation like battery overcharge protection and automatic shift on mains failure
A chopper inverter is actually much easier to make than an iron core type, however the only difficult part is the ferrite transformer which needs proper calculations, and presently I do not have any practical experience with them.
I have discussed one such idea in this blog, you may refer to this article:
http://homemadecircuitsandschematics.blogspot.in/2012/09/making-200-watt-compact-pwm-inverter.html
Rashid.
You would require a 200 watt, 20V panel approximately, and a charger controller as shown in the following article:
http://homemadecircuitsandschematics.blogspot.in/2012/11/high-current-transistor-tip36-datasheet.html
Great help indeed.
God bless you always.
Rashid
It does seem to me that this circuit should be more efficient than the one we discussed earlier. I'm I right? Please let me know.
Thank you.
yes you may be right,because the above circuit has a built in hysteresis control which makes the operations smoother and hassle free.
Are C1 - C4 polarized electrolytic or, what are their voltage ratings. What is the wattage for the diodes (D1-D5) can I use 1N4007 for all of the diodes?
Please let me know. I like to try this one.
Thanks
D1---D5 should be as per the rating of the battery, it's current should be near about the AH rating of the battery
But in this case I'm using a generator producing 32VAC which I use an LM2596 DC-DC Buck Converter Step Down Module to convert to 14.2VDC after rectification with a bridge diode. Can I go ahead and use the said rectified and step down supply to charge the battery as it is?
One more please, I want to add led indicator for charging process and led indicator for full charge condition in this circuit, where that two led can be placed?
thanks
what component needs to be replaced if I want to modify this schematics to be a 6v and 24v battery charger?
only the transformer and the relay should be changed to 6V
I'm sorry, but this battery charger circuits is not work, the R2 nor R5 didn't respond when I trying to setting low and higher voltage. I doubt about pin configurations of the IC NE555 in this circuit. Are the pin configurations of the IC NE555 in this circuit are correct? 'cos I found NE555 pin configurations datasheet like this :
http://t3.gstatic.com/images?q=tbn:ANd9GcRgOr-tmDbJjRyEsukn43WC00Ud35Uf_P8lvlBXiWh1PlFbdU-4
pin 1 : ground pin 8 : +vcc
pin 2 : trigger pin 7 : discharge
pin 3 : output pin 6 : threshold
pin 4 : reset pin 5 : control voltage
Thanks Mr. Majumdar...
the pins of the IC are all correctly configured.
For testing you will have to connect a variable 0-14V supply across the battery terminals and vary the voltage to see the results, then you can adjusts the presets to set thresholds.
http://homemadecircuitsandschematics.blogspot.in/2011/12/high-current-10-to-20-amp-automatic.html
what is the transformer rating for charging 100Ah/12V battery?
Primary = 220V
Secondary = ?
Regards,
Mike L
It should be in between 10 to 20 amps, @14V
i have 12amp/24v (12-0-12) transformer... can i use it?
yes, I'm using a variable 1.2 - 20v supply. when I trying adjust voltage level, yes the relay is deactivates when voltage level is about 7.5v, and then activates again when voltage level reach 13.5v (this is a default setting I think, 'cos I dont adjust R2 or R5). so with this setting, low voltage level is 7.5v and cut off level is 13.5v.
But, when I'm using this circuit to charge 45ah lead acid battery, the charging process still continous above 13.5v and still continous when voltage level is above 13.8v (remember the circuit setting for cut off level is 13.5v) why this could happen?
I'm using a 15v on 5A transformer.
any idea Mr. Majumdar? Or maybe you have another battery charger circuit that have low and high voltage level system?
The circuit now is working, because the low voltage setting is 7.5v the circuit won't charging the battery when the battery voltage is about 11.6v . If I set the low voltage about 11v, the highest voltage for cut off level is 16v and I can't change this voltage (for 12v battery 13.5v)
I want setting the low voltage is 12.5 - 12.9v, so under this voltage the charging process starting, and cut off voltage is about 13.5 - 13.8v.
any help and idea Mr. Majumdar?
Thanks.
I think if the preset adjustments are done correctly as given in the article, you wouldn't face much problems.
I don't know where is my mistake, and yes I do exactly like your tutorial in this article.
wihout connecting the circuit to the transformer and using 1.2 - 20v supply this is the result :
- when switch on the variable supply the relay is activates, right?
- I adjusting voltage to 11.4 like you say, then adjusting R2 (10K PVC trimpot) and nothing is happen still no respond
- I try adjusting voltage to 14.4, then adjusting R5 (same 10K PVC trimpot) and same result nothing is happen
- then I adjust voltage between 5 - 15v and the relay is respond, deactivates when the voltage reach 7 - 8v, and activates again when the voltage reach 13.5 - 14v
maybe, maybe you can diagnose where is exactly my problem or mistakes?
thanks Mr. Majumdar
I connect 7.5v zener diode with the pin 8 and pin 4 of the NE555 exacly like the schematic in this article.
- the catode of the zener diode and C 22u (-) are connected than connect to ground
- the anode of the zener diode and C 22u (+) are connected than connect to the R 270 ohm 1/2 Watt, pin 4 and pin 8 of the NE555 IC
anything wrong?
alright I'll try to replace 7.5v zener diode with 12v zener diode soon. I'll reporting the progress to you Mr. Majumdar.
Thanks
After building the above circuit the relay stays activated regardless of the adjustment of the R2 or R5.
Now, here is my part of the troubleshooting/testing the circuit, I connect an LED in series with a 1K resistor to the +ve terminal of the supply end and then the -ve to ground.on supplying the voltage (in this case, 11.4VDC)the relay is activated and stays close, so the LED goes OFF (meaning it is not charging).
However, when I unplug the 555IC from the circuit, the relay deactivates and the LED turns ON. Do you have any idea why it is doing that? What do I need to do?
The confusing results may be happening due to the presence of C1 and C2, just remove them and then check.
This circuit uses a 7.5v zener diode which is the supply voltage for the 555 timer.
What is the relay coil voltage?
Thanks,
Kerry
The transformer and the relay voltage ratings should match the battery voltage rating
emitter to bridge positive supply
collector to circuit positive.
base to pin#3 via a 1K resistor.
jose from Argentina
you can try this circuit:
http://2.bp.blogspot.com/-GbL69BMaL5c/UUvoqjC2yKI/AAAAAAAADus/nTPGvNzfc7Q/s1600/6v%204%20ah%20charger%20circuit.png
ariel.
thank you....can you please specify the battery voltage and AH that needs to be charged?
Pease do exactly as it's given at the end of the article.
You will need a variable power supply and follow the steps as mentioned at the bottom of the article.
Also replace the shown zener diode with a 4.7V across the supply pin of the IC.
I don´t have a 7.5v zener, could I put a lm317 and make it 7.5v instead?
Keep up the good work!
5.1V zener should also work.
provide all the details of the procedures that you have done with this circuit, then I'll try to troubleshoot.
I have a question for you. I have a 12V, 130AH lead battery and an inverter (not smps). I run a pedestal fan on it which is of 150W 220V. It runs only for about 2.5 (two and an half) hours.
According to formula W/V=A which is here 150/12= 12.5
and Time = Battery Amp/Amp of appliance which is 130/12.5= 10.5
Question is it should run for atleast 10 hours, why it is only run for about 2.5 hours ??
Hopefully you will help in detail.
Thank you.
Rashid
Your result is as per ideal conditions and nothing in this world will generate ideal results, even a new battery here will not give 10.5 hours of back up.
Therefore either your battery has become old and inefficient or it's not being charged correctly and fully. Try using a step charger circuit for charging it optimally....
In fact my battery is just brand new and I use a PWM charger to charge it. And when it becomes full, indication LED emits on its panel.
Any way thank you again for your reply.
Rashid
Only a step charger circuit ensures that much possibility.
Yes you are right, i am not sure about the battery either it is fully charged or not. I only assume that by the indication LED on the charge controller.
Sir, can you explain what is the step charger, how it works. And if possible can we have it's circuit diagram.
Thank you.
Rashid
You can go through one example circuit given below, however it is not meant for high current batteries, I will hopefully try to produce a similar circuit for higher current applications soon:
http://homemadecircuitsandschematics.blogspot.in/2012/10/make-this-3-step-automatic-battery.html
you can charge any type of battery with this circuit.
I tried this circuit with the present modification, but it didn't work. i used 7.5v zener instead of 6. still the circuit activates at around 7.6v. i again removed bc547, c1, c2 and used a 12v zener still the same, i.e., it activated at around 7.6v regardless of variation in pot. please help. i used a variable dc power source controlled by lm317.
one more que, disconnecting transformer section means to remove only transformer or along with bridge rectifier????????
I have modified the diagram by replacing the zener with a 7805 IC, please check it once gain with the modifications, I am sure it would work now.
If still it doesn't work then you could probably opt for a 741 IC circuit which is a thoroughly tested design.
Thanks for this great circuit. I am looking forward to start putting one together for hibernating my RVs battery over the winter. However, can I exchange the transformer + diode bridge with the +15V DC power output from an old PC power supply, i.e a switched power supply? I don't see any reasons why not, but don't know too much about the charging restrictions for 12V Lead Acid Batteries. Any Comment is appreciated.
Thanks again
Superbender
yes an smps can be used here as the power supply, no issues, just make sure it's current is not above 1/10th of the battery AH which is being charged.
Yes a single common power supply source is just what is normally required to charge two batteries or even more numbers or for any other application where more number of recipients are needed to be operated.
I would be addressing your above question in the form of a new post soon where I would provide all the required circuit details.
I have published your design here:
http://homemadecircuitsandschematics.blogspot.in/2013/09/automatic-charging-of-two-batteries.html
what are 22u and 0.1u cap sir it would be helpful if you give the exact volts with those cap. Can i use 16v 220u instead of 22u and 50v 2.2u instead of 0.1u. thanks
I have removed the capacitors from the design, please check the new diagram.
See the LAST design in this post:
http://homemadecircuitsandschematics.blogspot.in/2012/07/making-simple-smart-automatic-battery.html
Pls, suggest me the changes in circuit to charge 24 V. 7.2 Ah two battery pack to be charged with 24 V. solar panels.
http://homemadecircuitsandschematics.blogspot.in/2012/02/how-to-build-automatic-6-volt-12-volt.html
Is there any problem using this circuit in co-operation with a medium power ( about 350 W )inverter ? I will have to draw more currents from the battery while discharging .
And another serious doubt to clear is, whether this circuit can be used to charge all types of deep cycle lead accid batteries with normal charging cycle fulfilled effectively ? Plz reply me soon......... Sry 4 the bad english
The circuit can be used in conjunction with all types of inverters because the charging procedure has no connection with discharging procedure.
The circuit can be used with any AH battery, provided the trafo, diodes and the relay are dimensioned as per the load current.
the 7805 is only for powering the IC and has no connection with the battery, so does not need to be changed.
My 555 ic takes too much time to respond.......
I was tuning the pot at terminal #2 for lower threshold after sustaining the variable voltage to 11.4 V. But initially no response was there although the pot undergone complete rotation. I tried one more time and realized that when pot is rotated at the maximum position such that the resistance measured between the ground and its adjusting terminal becomes 8.2 ohm ( in this situation i think that pot has contributed complelte resistance towards its Vcc side ) and when the power supply ( 11.4 V ) is directly connects, then the relay activates at the same moment. In brief i couldn't efficiently adjust the pot since the relay took too much time to respond.
I just disconnected the relay from the circuit and put a piezo buzzer at the pin#3 of IC to know whether the ic responds to the adjustments made to the pot, but the result made me so much unhappy. I found that the pin#3 becomes active only after 35 seconds when the pot adjusted maximum. Because of this i can' t adjust the pot properly..
Sir what is the reason ? How could it Be solved ?
I used a 0.1 uF CERAMIC CAPACITOR at pin#5 as in the diagram. A TRIM POT was used as R2 and a normal one as R5. I had to use a SERIES combination of resistances to substitute a single higher one in the circuit due to lack of the exact one specified.
Will these effect the overall working.
Anyhow expecting a quick suggestion from your side
http://2.bp.blogspot.com/-nwMfNvxGD5g/ULSDWeSQ3qI/AAAAAAAABpU/ZLh413MLJwQ/s1600/40%20watt%20led%20emergency%20light%20circuit.png
ignore the lower relay, it's not required. this circuit will work perfectly....adjust the 10k preset for the high cutoff and the 100k preset for the lower cut off.....remember to disconnect the 100k preset temporarily while setting up the 10k preset
I want to Clear something. I have noticed from the comments given that there are 4 capacitors in this circuit also a zener diode too. But i couldn' t see any such components except c4. That means you renewed the old one composing all these components with a later edition circuit consisting of only C4 compared with the other ?????
Yes, I have removed the caps as they were of no relevance with the design
But a bit of confusion i have, with the image you given.
Can i use this circuit to be used for both lower and higher thresholds?
How can i set the threshold as in the previous case. I am asking you this, coz it seems constant power supply from SMPS is used here.
Can i use 4k7 and 47k pots instead of the ones given here, coz i don't have a 100k pot with me( it is possible to get a range upto 100 K by periodically adjusting the 10 k pot with a series of resistances in multiples of ten upto 90 , but will it turn the behaviour of the circuit ? )
smps can be replaced with any 15V DC source, but the current should be 1/10th of batt AH rating.
you can try 47K preset, I think it should work.
the circuit link, for others to follow:
http://homemadecircuitsandschematics.blogspot.in/2012/11/making-40-watt-led-emergency-tubelight.html
But the problem is that, when i am adjusting the 100k pot for lower threshold 11.4 V ( KEEPING 10K POT CONNECTED ), nothings seems happening, no LED glow even the relay sound couldn't heared.
Actually what should be the indication for lower threshold setting ?
What may have occured to ma circuit ?
Sir i have to say onething that, I HAVE NO 100K POT WITH ME. So i HAD USED THE 49.2 K RESISTOR ( 47k+ 2k2 ) in series with the adjusting terminal of the 50K POT which gave a variation of 50 to 100 k at one full rotation, and the remaining was taken as usual.
IS THERE ANY PROBLEM IN REPLACING THE ORIGINAL 100 K POT with this configuration. Will it affect the result ?
Anyway i am waiting for your valuable suggestions about this problem to keep ma project go ahead
Everything is centered around the reference voltage set by the zener voltage at pin#2.
While setting the higher threshold, the LED lights up because the pin#3 voltage exceeds the zener reference voltage.
Now in order to switch OFF the LED the 100K must be adjusted to a point where pin#3 voltage is again dragged below the zener voltage of pin#2.
Measure this with a meter while adjusting 100K pot.
If this is not happening then probably either we'll have to increase the value of the 100K preset to some higher level or connect the 100k preset with reference to ground.
You may confirm this first then we can proceed further.
Where could i connect a Green LED in the circuit of opamp to indicate charging progress ?
If i am using a 12-0-12 V inverter transformer as the power supply, then if i only want maximum of 6 A for safe charging, What will be the value of current limiting resistor which is to be used prior to the bridge rectfier section ? Will it be 2 ohms, since by ohm's law r= v/i ( 12/6 ) ???????
positive--------->I------^^^^----------pin6----------->I---------^^^^------------ground
the resistor shown are 4k7k each
yes your current resistor calculation is OK.
THANK U VERY MUCH.....
Now i just connected the power supply section. As i said earlier, an inverter transformer rated 25 A at secondary is being used. I had followed the transformer with a bridge rectifier ( formed using four 1N5402 diodes ) and a 1000 uF/25V capacitor in parallel. But when i had just finished the circuit and powered up, a bursting occured in the center tap or the transformer seconrary, then i suddently disconnected the device. I have only tested the power section ( without the charging portion ) to avoid any awaiting damages.
I am using a 25 A rated 230/12-0-12 inverter transformer with 5 inputs( primary ) and 4 outputs ( secondary ). The center tap had been taken BY PAIRING THE TWO CENTER WIRES IN THE SECONDARY TOGETHER and fed this tap to the -ve terminal of the capacitor forming ground. The other two terminal wires have been connected to the respective junctions in the bridge......n but the system FAILED.
I have noticed several circuits in which this bridge configuration is employed, but in all these, the transformers used were not having a center tap, having only two outputs..........
THen, was my connection wrong ?
How can i solve this issue sir ????'??.....
http://homemadecircuitsandschematics.blogspot.in/2012/01/how-to-make-hi-fi-100-watt-amplifier.html:
It's better to use the half winding, that is 0-12V winding and connect the bridge to it in the conventional manner, this method would create full wave rectification and also eliminate confusions.
On noticing the envelope of the battery, it has been written there that , the battery has to be operated by a charger with the feature of constant current and constant voltage charging. So any problem will my battery face operating with the opamp circuit sir ? Will this simple circuit be able to provide the enlisted features here ?????
I have found one of your schematic named ' automatic 3 step charger '.
But as one of your followers said in that blog, i am also not able to collect such low value resistors r1, r2, r3 ( of the order of 0.01 ohm ) from the local shops in our area. So what will i do to charge my battery with the safety features and without the lack of such triffle components ????
Give me a better suggestion sir....
With a higher current trafo, yo would certainly require a current control circuit along with the opamp circuit.
I have tried the 0-12 tap, eventhough spark and the bursting sound along with smell comes again. I am very much afraid of this and not so much confident to connect this into the opamp circuit. I think the spark comes due to the sudden high current flow towards the rectifier section. I have used a 2 ohm resistor to reduce the current to 6 A. But nothing happens as i wish. I should have to use this transformer to follow the inverter design along with this. Can you please suggest me a better modification which can be applied to the secondary terminal with 4 NO. OF OUTPUT LEADS ???????
No problem yet.
But i don't wanna this, coz i think it will not be efficient for my purpose.
Then what may have occured to ma bridge config. '????'''
Using a meter set at AC range, check the voltage of the taps, and find the one which gives the required 12V, and connect these taps to the bridge.
I think the diodes might have got shorted and burnt....you can check them to confirm or use fresh diodes to make a new bridge rectifier with correct polarity and connections.
I have a small doubt. Will you clarify it??
I have three sealed lead accid batteries of 12 V each. All the 3 having same Ah rating. I had been using the batteries for 3 or4 months before. But since then it hasn't been ever used. But no problem..they are zero maintenance batteries.
Now i have to charge all of them. But the problem is that each batteries shows different open circuit voltages on checking with a voltmeter. My doubt is whether they can be used in parallel to charge the whole setup or not?
I prefer a parallel set up, becoz it suits my purpose ot getting the required higher Ah keeping voltage constant..
I am very sorry for the large describtions which may have wasted your time much more. Again sorry for the bad english.
Hoping a better suggestion from your side
You can connect them in parallel with each individual high current rectifier diodes in series with their positives. The diodes will automatically balance outthe difference in voltage levels of the batteries and charge them uniformly, you mayu refer to the following post:
http://homemadecircuitsandschematics.blogspot.in/2013/08/connecting-batteries-in-parallel.html
Be sure to use diodes rated with current twice that of the charging rate of the batteries.
What about 6A4 diode ?
What word i have to search for the maximum current rate of a diode in it's datasheet ? Is it 'maximum forward current ? '
You can use two 6A4 diodes in parallel for each battery positive.
6A4 can handle 6amps so two would provide a safety margin of 12 amps
I haven't update diode datasheets here yet so at present you won' t be able to find any relevant info in this blog
How do i measure the current drawn from my high power transformer using multimeter???
I want to ensure that the current limiting resistors are working properly to avoid any awailing damages.
I have heard that it is possible to measure the current only under LOAD. But except ma batteries i don't have any other devices functioning at that much current with me. I don't wanna to take risk on ma batteries too.
If you check your meter features you will find a slot for 10Amp or 20 amp range, you will have to plug the red prod into that slot, set the selection knob to 10Amp position and atsrt checking current by inserting the prod terminals in series with the transformer output. Use DC range if you are connecting them after the bridge, and AC range if it's before the bridge.
I have 5 no.s of 20 Ah batteries connected in parallel forming a 12 V 100 Ah battery bank. What will be the required charging rate of the whole setup ?
Is it 2 A max ( 1/10 th of the individual battery ) or 10 A max ( 1/10 th of the whole battery capacity ) ?????
It'll be 10amps and not 2 amps because parallel connection adds up current.
1N4508 is enough ??''
When i am turning on my dc power supply the resistors connected in series with the +ve secondary of the 300 W transformer is getting burned. I have tried a 2 W resistor, eventhough the burning effext didn't change. what to do sir ?
My power supply connections are everything perfect
Secondly the resistor will need to be dimensioned according to the current limit that's being used at the output...if you can tell me the charging rate of your system then I'll tell you the correct resistor dimensions
I have noticed a battery charger circuit using 741 opamp which you had given as a reference circuit to one of your readers. In this circuit when the battery voltage exceeds a higher cut off ,opamp output becomes high due the increase in voltage at terminal 3 compared to zener reference voltage at pin#2.
But i want an extremely opposite action to this. I mean, when the voltage rises above the zener voltage, output of opamp should go low and vice versa. ie., opamp in -ve config.
Can you please help me sir...n
Just reverse the pin arrangement, meaning connect the zener to pin3 and the preset to pin2
I have a 300 W inverter transformer with me ( bought in order to make a medium power inverter ). From basic calculations it could be understood that anyhow it draws maximum 25A current under full load at secondary.
I have connected the rectifier section along with the transformer for the first time, but a fuzzing sound could be heard. Therefore, i decided to add a calculated resistor of value 2 ohm ( 12V/ 6 A) in series with the +ve tap of the secondary to limit the current. But it also got failed. Resistor burned.
Here i am using a 60Ah battery. Therefore it should draw maximum of only 6 A from the transformer secondary, but the thing is that the secondary terminal supplying very high current.
I am employing your automatic single transformer inverter/charger/changeover circuit with the charging circuit using opamp which i got from this blog. So all the above problems came from this single transformer design such that current limiting and power handling problems.
BUT I DON'T EVEN WANT TO PAY ANY MORE FOR ANOTHER TRANSFORMER and i have have to implement the specified charger block only since i prefer the charging of my lead accid batteries with complete safeguarding
So in brief ( talking from a beginners level ) i want to clarify the below things-
1). What should i do to limit the current for charging section???? Will it be a single resistor alone with high power rating????
2). Any modification to be carried out for remaining sections ( such as oscillation section, relay coils ) which deal directly with transformer secondary ???
3). Can the limiting resistor be used in series with transformer secondary??? Will it be able to limit AC current???
And the last thing to ask to you ' Whether i have to use a LM338 IC or Traic circuit as a current limiter???????'''''''
Now i think you got the exact thing that i meant before and asking you several times. So if you can i want your valuable suggestions and help on this. Please don't neglect ma request as a triffle. I am expecting a satisfactory reply
If it can, i need help in modifying other sections of the inverter handling with high current delivered by high power transformer
That would be the right approach, since you are new in the field you shouldn't try adventurous circuit concepts, it could be risky.
I would also suggest that you spend some more money and buy another transformer and make a separate charger,using IC LM196 in place of IC338.
By the way which inverter design are you trying right now?
NTC will not work as current limiter, because it's function is to limit surge current and not continuous current
Sir i want to clarify something on that inverter circuit.
I need the circuit to be function as follows.......
I am connecting the output of the inverter directly to an ac wall outlet with a plug in a room. My purpose is that during power failure, suitable instruments (to the inverter power) can be turned on which is situated at any room in the house powering from the ac fed by the inverter at the other end through that plug.
One of my cousin had done a similar inverter and he plugged the inverter directly to the wall socket and swithed on it after turning the mains switch off during power failure. It worked well. But all the thing he had done was changing a normal PC UPS into an inverter.
On searching the internet, i am confused about this output connection, because all of them saying that the instruments to be powered during power failure are directly connected to the inverter output. So will there be any problem in my strategy.
And also I want the inverter to be connected permanently to the ac wall outlet and ni need to TURN OFF THE MAINS SWITCH during power failure to make it operating.....mTry giving me best reply considering my cousin's situation
But you should first confirm the operations using smaller transformer and without wall socket connection, otherwise you may end with a possible fire hazard.
http://2.bp.blogspot.com/-nwMfNvxGD5g/ULSDWeSQ3qI/AAAAAAAABpU/ZLh413MLJwQ/s1600/40%20watt%20led%20emergency%20light%20circuit.png
http://1.bp.blogspot.com/-VMaUoV6aqLQ/UcsSdjCa6TI/AAAAAAAAEqo/NufzQhrsZeE/s1600/48v%20battery%20charger%20circuit%20automatic.png
For your circuit use 10k resistors in place of 22k.
ignore the msfet stage and use a relay as given in your circuit
I want to use the series config. for my need. i.e, as 24V bank always.
Can you tell me which are the lower and higher threshold levels for this combination and how can they be set?
I also want to know whether any damage will occur to the individual batteries with this charging method?????
Parallel charging would be better using the above explained method, series charging will not give optimal charging results compared to parallel so it should be avoided.
I need a little info about tabular acid battery.
I purchased a new 100 AH battery(not branded). I am using the battery @ of 10Amps load using a mosfet inverter battery is lasting 5hrs and 10min.
according to my knowledge It should give backup of 9+ hrs @10 Amps discharge.
dealer claims he sold the correct item of 100AH but I feel being cheated.
I want to know how to determine the correct AH of the battery. It's only 15 days old .
Possibly your battery might not be fully charged....to charge it fully keep it connected with a 14V 10amp charger for about 14 to 18 hours and then you can expect the desired results from it.
1. battery has nothing printed as it is not branded dealer build it at his shop assembling the battery plates.
2. Battery gets charged using the same transformer ( inverter + charger) @10 -15 Amps automatically it does not get over 15 Amps during charging. and has 14V at the battery terminals. After switching off the charger battery has 13.93 V at the terminals.
(Purchased this ready made kit from market based on Mosfet and SG35xx series)
3. I had fully charged the battery like you are saying for 14-18hrs before testing , and as the charging is automatic I don't need to switch off the charging as it gets charged to it fullest level charging automatically kept @ 14V.
4. Tested the 100AH battery 2-3 times in past days after fully charging but it never gives more than 5Hrs backup @10 Amps discharge.
5. Lastly I have another 32AH battery (1 year old) which gives a backup off 2Hrs 40min at 10Amps discharge rate using the same inverter for charging . This is the reason I highly doubt there is something wrong with the 100AH one.
A hand assembled battery can never produce results equivalent to branded ones due to the low tech manufacturing procedures involved.
It's better you try the first circuit from the following link, which is much easier:
http://homemadecircuitsandschematics.blogspot.in/2012/07/make-6v-4ah-automatic-battery-charger.html