High voltage problem with lead acid battery charger

[Qaisar Azeemi]

Hi all;

Following is the design of my pb-acid battery charger that i've designed to charge 24v, 100AH pb-battries with a current of 10A approx. first i used 10A, 24v
transformer with the circuit but it could deliver 6.3A hardly and the voltages at the transformer o/p terminals also dropped down to 20V so it was unable to charge.
After that i used a 36V, 32A mighty transformer. In the circuit after rectification the voltages boosted up to 50V DC that is quite problemous and
blown out the MOSFET IRF3205, i thought it is blown out due to the reason that it can withstand with a maximum voltages of 55v dc so i used IRF 540 that can
withstand with 100v dc but it is also burn out the same way. the current through the circuit is 10A approx: now my questions are that:
1) why voltages are boosted after rectification?
2) what is the role of power of transformer?
3) why MOSFETS are burning out? as i was fully confident by looking at data sheet that it can withstand with 50v dc? ( i am using mosfet to cutoff after battery
is fully charged, the cutoff circuit is not shown here).

View attachment 67885

 

[alec_t]

1) The 36V AC, being a sine wave, has a peak voltage of 36 x 1.414 = 51V. The bridge rectifier diodes drop that down by 1.2V.
2) Input power must not exceed the rated power or the core will saturate and the tranny will release the magic smoke.
3) The FET needs several volts between gate and source to turn it on. Those volts are also across drain and source. Volts times 10A = lots of Watts, so the FET will get hot and burn out unless it has an adequate (= large) heat sink.

 

[panic mode]

1) lookup how AC voltage is defined (hint - it is not peak). when you rectify AC, capacitors get charged to a peak (minus drop on rectifier diodes).
2) assuming you are talking about voltage transformers - they transform voltage. this way you can have output voltage that is different than supply...
3) because they are not used properly. most important things to consider are:
- transistor current
- transistor voltage
- transistor dissipated power

all need to be within design limits.

for example you mosfet is rated for 50V and 110A. but not at the same time... because P=V*I=50*110=5500W which is way over rated 200W. even if current is 10A and your power supply current protection kicks in, you get (32*1.4 -2*0.8)*10 or some 435W which is more than double that this transistor can handle (of course, when mounted on massive heatsink).

few things to consider:
- why built regulated DC supply to charge lead acid batteries (did you do any research, did you check any commercial lead-acid chargers)?
- what is the purpose of mosfet in this circuit? you seem to use it as a series resistor to limit current. why? all it does is dissipate heat.
- what is the purpose of D1, D2 other than cause voltage drop and dissipate heat?
- why are you paralleling three LM317L? each is capable of only 100mA. LM317T or LM317K can do 1.5A (still far cry from 3.5A if outputs were to be shared perfectly).

 

[Qaisar Azeemi]

1) lookup how AC voltage is defined (hint - it is not peak). when you rectify AC, capacitors get charged to a peak (minus drop on rectifier diodes).
2) assuming you are talking about voltage transformers - they transform voltage. this way you can have output voltage that is different than supply...
3) because they are not used properly. most important things to consider are:
- transistor current
- transistor voltage
- transistor dissipated power

all need to be within design limits.

for example you mosfet is rated for 50V and 110A. but not at the same time... because P=V*I=50*110=5500W which is way over rated 200W. even if current is 10A and your power supply current protection kicks in, you get (32*1.4 -2*0.8)*10 or some 435W which is more than double that this transistor can handle (of course, when mounted on massive heatsink).

few things to consider:
- why built regulated DC supply to charge lead acid batteries (did you do any research, did you check any commercial lead-acid chargers)?
- what is the purpose of mosfet in this circuit? you seem to use it as a series resistor to limit current. why? all it does is dissipate heat.
- what is the purpose of D1, D2 other than cause voltage drop and dissipate heat?
- why are you paralleling three LM317L? each is capable of only 100mA. LM317T or LM317K can do 1.5A (still far cry from 3.5A if outputs were to be shared perfectly).

Thanks for your kind reply panic mode. Acctually i am new to the field of power electronics so all i am doing is a group of experiments in an R&D deptt:.
1) No i didn't check or more accurately say i cant't get any schemetic or hardware; i just study the material available on net and decided to use regulators because i wanted to develop a digitally variable charger for 6v, 12v and 24v batteries using atmel 89c51 microcontroller. can you please guide me about how they make commertial chargers.

2) i am using mosfet to cutoff the charger when battery gets fully charged. it is use as a series switch but now i am thinking i have to use a transistor with the adjustment pin of the regulator. but some times such design also don't work due to analog behavior of the transistor switch, its cutoff isn't sharp enough to pull the voltages of the charge down to 1.2v of the regulator and turn the charger off.

3) i am using D1 and D2 to block the DC voltage of the battery to hit the output of the regulator directly.

4) sorry i am using LM338, which is 5A regulator, so three regulators in parallel can conduct 15A current. i used lm317 in diagram because lm338 is not available in proteus.

 

[Qaisar Azeemi]

2) Input power must not exceed the rated power or the core will saturate and the tranny will release the magic smoke.

thank you alec_t , but i can't understand. kindly explain it a bit more. because my 10A, 24v transformer could deliver 6.3A hardly and the voltages at the transformer o/p terminals was dropped down to 20V also it provided dc 20v after rectification. why the volts drop down from 24v to 20v ac?

 

[panic mode]

ac has characteristic form. after rectification you get same shape but all positive (abs value of sin). this is pulsing DC where pulses are halfperiods of sin. to get smooth DC, one can filter it. if there is no load, filter capacitor charges to peak voltage (Vac * 1.41 - 2*0.8V). the 0.8V is drop on diode. full bridge rectifier has two diodes conducting at the same time so we need to subtract two drops. 1.41 is sqrt(2), recall what RMS stands for and how it was defined.
so if your AC after transformer is 20Vac, then after rectifier you get 26.5Vdc.
but when there is load connected, capacitor does not stay fully charged, it starts discharging when AC input is not charging.
https://tymkrs.tumblr.com/post/12794006130/bridge-rectifiers

this means that with load, your DC voltage will be lower (actual value is function of load).

for electronics one normally needs stable voltage so this filtration is not enough, one uses regulators to produce stable voltage.

for charging lead acid batteries you do not need this, all it is needed is rectified juice (can be pulsed, no need to filter it). you have internet so use it. do some research. i just typed "lead acid battery charger" and there was tons of hits including circuits. next one was to type "lead acid battery" in wiki. and there is more, much more.

 

[audioguru]

Your original transformer was rated at 24V AC, 10A AC which is 240VA.
But the rectification produced a peak voltage of 34V which was used by the charger circuit. But you wanted a continuous output of 10A DC so the transformer must be rated to provide 34V x 10A= 340VA, not 240VA. Your transformer was overloaded!

 

[jpanhalt]

On other thing about the mosfet, it is an N-channel and you have it configured with the load connected to the source. Since there is very little voltage drop across a mosfet when it is fully turned on, the gate will need to be at your supply voltage plus about 10V. That you can't do with the circuit you have, so the mosfet never turns on completely. Rds is quite high, which will lead to considerable I^2 R heating.

If you sourcing current, you need a P-channel mosfet or reconfigure your circuit so the mosfet sinks current. That is, its source should be connected to ground.

John

 

[Qaisar Azeemi]

what if i use a tip3055 or 2SD1047 transistors to provide reduced voltage at the input of lm338. the output of the rectifier is 50V DC and i need to provide not more than 35v at the input of the lm338. and how should i calculate that how many transistors will be needed to pass a maximum of 14A current and 50v dc input voltage at collector as shown in the figure below. and how many volts will be needed at its base to totally cut it off to stop the charging current?

View attachment 68121

And how should i measure the power dissipated in the transistor (BJT or MOSFET)? is it (input V - output V ) * current passing through it ????

 

[Qaisar Azeemi]

What if i use a tip3055 or 2SD1047 transistors to provide reduced voltage at the input of lm338. the output of the rectifier is 50V DC and i need to provide not more than 35v at the input of the lm338. and how should i calculate that how many transistors will be needed to pass a maximum of 14A current and 50v dc input voltage at collector as shown in the figure below. and how many volts will be needed at its base to totally cut it off to stop the charging current? and how to calculate if i use IRF3205? i searched the net but can't get any clear formula or like situation.

View attachment 68121

And how should i measure the power dissipated in the transistor (BJT or MOSFET)? is it (input V - output V ) * current passing through it ????

 

[dr pepper]

Chargers designed to be left connected to battery have different modes, bulk charge which is a constant current, taper charge which is a constant voltage and then when the current drops to a certain level the charger remains in constant voltage but at a lower maintenance level, this can help prolong the life of a battery. There are also variations on this scheme.
If your going to the extent of a microntroller controlled charger then you might as well lookup a switching regulator to control the juice to the battery, this will increase effeciency and therefore decrease the size of the heatsink, what you've talked about so far requires a big sink.
One possibility for gate drive on your mosfet having read what I've read so far would be to use a gate drive transformer, then you can use the regulating mosfet as a high or low side regulator without any special power supply for the gate drive (this idea is for a switching reg not a linear reg).

Heres a 8 amp 3 stage charger I made that keeps my boat battery healthy, its a linear regulator.

 

[circuitspecialists]

You are correct .... to determine the power dissipated in the transistor (BJT or MOSFET)? is it (input V - output V ) * current passing through it. Look at the data sheet for the particular device you choose and make sure you are operating it in the "safe area of operation". Also be aware of theheat being generated and use the correct heat sink required to disipate the heat from the transistor.

 

[ronv]

If your big transformer has a centertap you might be able to switch between a bridge rectifier and a center tap rectifier for the 6 volt batteries. This would help.

 

[crutschow]

....................................
2) Input power must not exceed the rated power or the core will saturate and the tranny will release the magic smoke.
....................................

That's a common misconception about saturation. A transformer core saturates due to operation above the rated voltage and/or below the rated frequency, which causes a large increase in the magnetizing current. They do not saturate due to a high current load.

The magic smoke will come out from over-current though, due to excess power dissipated in the winding resistance.

 

[crutschow]

If your big transformer has a centertap you might be able to switch between a bridge rectifier and a center tap rectifier for the 6 volt batteries. This would help.

Actually, you don't need extra rectifiers to get 1/2 the voltage. The bridge rectifier will act as a full-wave rectifier for the centertap, giving 1/2 the rectified DC voltage directly at the transformer centertap as well as the full-rectified voltage at the bridge output (ignoring the diode drops).

 

[ronv]

Sure, Why didn't I think of that. Even better. He can probably get 12 and 6 volt from the 18 and the 24 from the 36 volts. Down to 120 watts.

 

[dr pepper]

I agree with cruts, transformer cores saturate with volatge over time, too much volatge or too much time, excessive current causes the windings to overheat.
A core in a not so well designed smpsu might saturate with overcurrent, as the inductor on time varies with the load (by various methods).