# Battery backup circuitry

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#### pd9

##### Newbie level 4
Hi everyone,

I need help with this battery backup. I am using a DPDT relay here to detect the on-off of power supply. When there is power, the LM340 steps down the voltage to 5V and the battery is getting charged. I am using 4X 1.2 V NiMH battery (2600mAH) cells. I want the output to be always between or equal to 12V and 4.8 V even when there is power or no power. When I test this circuit, the voltage drop across R2 is around 9V and I am getting output voltage of around 2.66 V. So, I removed resistor R2 and shorted the connection. Now i get voltage drop for R1 = 5.4 V and the output voltage to be 11.21 V but the batteries start getting overcharged. So can anyone please guide me here? thank you very much for the help.

Last edited:
pd9

### pd9

points: 2

#### pd9

##### Newbie level 4
Thank you Kak111. So, for the 2nd circuit diagram, I should use the same battery charger circuit that I have and then use two relays..Right?

Thanks once again!

#### kak111

Let me explain.............

Circuit 1.
(Battery charger is your LM340 circuit )
- when +12Vdc is available it is charging battery via diode and output is Ucharger
minus voltage drop in diode ( 0.7...0.8V). ie. output voltage is Ucharge.
- 12V not available , output voltage is Ubattery.

Circuit 2.
- when +12Vdc is available, LM340 is charging battery via diode and relay is ON
and connect output to +12Vdc
- 12V not available , relay is OFF and connect output to battery

( Circuit is drawn when there is no 12V input .)

Or do I have some misunderstanding with your needs

Regards KAK

pd9

### pd9

points: 2

#### pd9

##### Newbie level 4
No..you are not misunderstanding anything. Only thing is that I want to use a 5V relay. So I will have to step down power somehow..so the relay can take it?

Thank you.

#### pd9

##### Newbie level 4
And also, a diode needs to be added in parallel to the relay coil to dissipate the current

#### Syncopator

##### Full Member level 6
And also, a diode needs to be added in parallel to the relay coil to dissipate the current
No. It's to clamp the voltage.