This is the circuit I have designed. I want to be sure whether I have designed it properly or not.

Here is the Purpose:
I am trying to power a Network device 24 hours which requires 9v (600ma). I have a 12 Volts 9AH (SLA) battery and a 12-0-12 (2 ampere) center tapped transformer. When there is mains available this circuit should charge the battery, auto cut off when full charge state is achieved and protect the battery from over discharging.

Circuit: enter image description here When battery voltage reaches 13.8 V then the T1 (Transistor BC547) goes on, making relay K1 on so the default connection between point 'a' and 'b' breaks and the charging stops. Similarly when battery voltage drops below 10.8v then T2 goes off so is the relay, Hence the over discharge is prevented.

Problem: This circuit is working to some extent, but as mains go off this circuit will provide power for only 50-55 minutes. I don't have any idea why this battery drains so fast. Also the relay bounces back and forth for a while near to the cut-off voltages.

So my point is how do I stop this battery from draining so fast. I am using a brand new battery and there is no problem with the battery. Is relay drawing so many power or may be the linear regulator 7809. Also how do I make it more efficient. Any suggestion would be appreciated.


This is what is written on battery:

Cycle use: 14.4-15 Volts

Stand By use: 11.5-13.8 Volts

Initial Current not less than 2.7 ampere


1. Moreover, At what current and voltage should I charge this battery?

2. Suppose I am charging the battery at 2 ampere current and lets say, I do not apply any current regulator (meaning, Once I have started charging I do not regulate the current there after) , does current gradually decrease as the battery charges up or do I have to reduce the charging current after 70% charge of state?

3. How important is the float charge in my case? Battery is frequently drained lets say once in a two day period.

4. How can I cut down the components and ultimately the price?

  • \$\begingroup\$ Part of the problem is that 13V is not enough to fully charge the battery. It would be normal to switch from fast charging to float charging at 13.8V, and stop charging at 14.4V, but that still doesn't entirely explain the short life. How many amp-hours is your charging cycle? \$\endgroup\$ – user_1818839 Aug 3 '15 at 10:46
  • \$\begingroup\$ Thanks for your response, I am not getting what you mean by amp-hours charging cycle. Battery is 9 AH capacity. \$\endgroup\$ – Arjun Aug 3 '15 at 10:48
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    \$\begingroup\$ The inductive spikes from the relays may have destroyed the transistors. Google "relay protection diodes" \$\endgroup\$ – user_1818839 Aug 3 '15 at 11:39
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    \$\begingroup\$ Charge to AT LEAST 13.7V - maybe more. DO NOT discharge under 11.5V. IF the mains is off often and this drains often then you are in cycle mode and 14.4V should be used and still do not discharge under 11.5V. A LA battery floated at under 13.7V will die quickly. \$\endgroup\$ – Russell McMahon Aug 3 '15 at 13:40
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    \$\begingroup\$ You can save some battery power by arranging the circuit so that the relay is operated when AC power is available, and released when operating from battery. \$\endgroup\$ – Peter Bennett Aug 3 '15 at 15:52

Charge to AT LEAST 13.7V - maybe more.

DO NOT discharge under 11.5V.

IF the mains is off often and this drains often then you are in cycle mode and 14.4V should be used and still do not discharge under 11.5V. A LA battery floated at under 13.7V will die quickly.

If floated at 13.7V the battery can be left connected indefinitely - the current will fall to zero in time.

If doing frequent discharges you should ideally charge to say 14.5 volts, hold it there "for a while" then set the voltage to 13.7V. "A while" above depends on how deep you discharged it. See Battery University - read all the lead-acid material.

Also see these prior answers of mine some of which are relevant, and skim these for relevant material.

Modem: At 15V in to the 7809, if the modem draws 600 mA, then the 7809 will dissipate
Power = V x I = (15-9) x 0.6 = 3.6 Watts. It will need a substantial heatsink. You can reduce 7809 dissipation by placing a series resistor between supply and 7809 (NOT in battery circuit). Measure supply when supplying modem.
Rseries ~= (Vsupply -9V + 3V_headroom) / 0.6
If Vsupply = 15V then Rseries ~= (15-9+3)/.6 = 3/.6 = 5 Ohms.
Use say 4R7.

Power in R = I^2 x R ~= 1.8W.
Use a 5W resistor, air cooled.

Measure Vin at 7809.
Probably needs 11 to 12V - see datasheet.
Adjust R at max modem draw to still have enough Vin to 7809.
Add capacitor at 7809 input - 100 uF at least - bigger is better.


The voltage detection is imprecise and varies with ambient and part temperature, an extreme exampr of this is "T2". As it begins to turn off it drops more voltage C-E, burns more watts and heats up, this reduces the Vbe needed to turn it on and it stays partially on well below the desired set-point. Only when it's well baked and the battery is almost ruined does the relay turn off, even then the battery is still required to run the two voltage dividers.

T1 has a similar problem during the charging of the battery, but this time the feedback is partially positive, so it probably doesn't fare as badly as T2

to make it cheaper, loose T1 , K1 , and the Shottkey diode, and instead charge the battery from SCRs connected to the transformer secondary. this will probably also allow a smaller capacitor to be used.

for voltage level detection a TL431 offers a temperature compensated set-point and needs only a small bias current (meaning you can also use larger resistances in the divider) the TL431 may be able to drive K2 directly, but add a freewheeling diode ob K2,

drive the resistive divider for K2 voltage sense from the appliance input, this way it will turn off when the battery is depleted.

the the charging voltage sensor is not as easily isolated from the battery, but you could add a diode between the SCR and the battery ads sense voltage there,

swapping the 7809 for a switched mode regulator will result in improved power efficiency and thus allow savings by using a smaller battery but getting the same life.


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