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[b]Power Supply Design[/b]

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I need your help to build a power supply working on 230VAC mains with a battery of 6V/4Ah alkaline type for supplying the power in case mains fail. The output should be 5V/750mA and 24V/100mA , both regulated and DC and Isolated. The circuit should be capable of charging the battery as fast as possible when the mains supply is on AND free from noise as well !!


I am wondering if you got the right battery type there. AFAIK, alkaline batteries are not rechargeable.
Another thing, your requirements are for at least 7 watts of DC power, how long do you wish to run your equipment on standby power? You will have to size your battery accordingly.
Sorry ! My mistake regarding the battery type. The battery in my mind is Globe make 6V/4Ah rechargeable type, used typically in emergency lamps.

Ur point regarding wattage is well received. But isn't it that if I want to increase back-up time , I need to use battery of higher ampere rating ? keeping other things same !! But what would be the charging time then???

if you chrage a 4Ahr battery with a 1 amp current it takes 4 hours for a full charge. Alternatively if you charge with a 4 amp current it takes 1 hour theoretically. Some manufactures do not recommend booster charging as the internal plates may get damaged. (I have some experience on this). However, looking at the data you have given, I do not understand how you are going to charge a battery of 6V/4ahr with an out put of 100mA

Nevertheless,you may obtain an unregulated low-voltage dc supply using transformer-rectifier arrangement.Rectifier diodes could be connected in bi-phase arrangement or most common full wave ckt configuration based on four diode bridge. The following formulas apply:

Output voltage(Vdc) 0.71 x Vac
Output Current (Idc) Iac
Input voltage (Vac) 1.41 x Vdc
Input current (Iac) Idc
Reservoir Cap.
recommended min 2200 x Idc
(micro F. if Idc in A)

Vdc-1.41 XVac
Idc-0.62 X Iac
Res.Cap. 2200 x dc
In each case the reservoir capacitor should be rated at around 2 times of Vdc in order to provide with adequate safty margin.

You may use fixed voltage regulators to get the out put. 7805 & 7824 with 1amp out put current are recommended. Care should be taken to ensure that the inregulated dc input voltage is within the range specified by the manufactures (e.g +7V to +25V max for a 7805) Where the input voltage is very much higher than the output voltage, power dissipation within the regulator may be excessive. Conversely, where the input voltage is only marginally greater than the output voltage, regulation may be very poor and ripple may be present at high current levels. The optimum input voltage (on load) must normally be at least 3V greater than the regulator's out put voltage. In order to avoid excessive power dissipation, the input voltage (off load) should normally be no more than about 6V greater than the output voltage.

Cheers !

Thanks for ur reply. I read it carefully. However some of the unanswered questions are as follows

1. How the battery is charged ( schematic plz)
2. Are the outputs isolated ( different grounds)
3. The circuit should provide output till the battery voltage drops down till 4V. How this is possible ? Plz note that in order to charge the battery I can increase voltage above 7.6VDC( Best charging voltage I believe for this battery).

What I have in mind is converting mains 230VAC to 7.6VDC output of LM317 -TO3 pack. Using this volatage to charge the battery and also the convert this DC voltage to 22VAC & 6VAC ( appx) using a transformer for isolation purpose ( Xformer say 4-0-4 Input , 0-22 & 0-6 isolated outputs). Then on the output side of transformer using the type of regulator u r talking about.

In brief this is DC-DC converter with Isolated Outputs with Battery for uninterrupted power supply. (UPS if I am not wrong )

I hope u got this. Or should I post a block diagram ?!

Keep up the good work !! Thanks again !!!!
The maintanence-free lead-acid batteries designed for continous charge mode. You must keep the voltage on 6,9V.
Do You thinking about circuit like this?


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This will not work if the battery voltage falls to 4V. Isn't it? For this we have to include both boost/buck arrangement !!

Secondly grounds of 5V & 24V are not isolated. I will be using 24V for field sensors which may be prone to noise due to lot of cable connections.
Hence the need of transformer in between.

LDO !! Hmmm . Well I would prefer to stick to cheap LM78XX series as LDOs are generally costly. And will it source 750mA ? I am not sure

Thanks for your effort the sketch and all that ......
For 6V lead-acid battery the 4V deep-discharge is end of life.I recommend the discarge about 5,4...5,3V, so the LDO can work in 5V section.(this is cheaper as boost regulator). The 24V is floating, because the quad with four legs symbolize a DC-DC isolated converter. I mean, no cheaper solution with Your parameters...
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