Backup Internet Access through Secondary Router

[haseebhm]

Hello,
I have made this 12V adapter for my wireless router to keep it running even when the power goes off. Please check the attached schematic.


I connected the battery in series with a digital multimeter to determine the current in both the cases of power ON and OFF. The problem is when the input voltage is available, the current still flows through the battery in the same direction. However it is reduced from 0.45A to 0.35A.

Is it alright ? The router should take current from the power supply rather than the battery when the input voltage is available. And as far as i think, the DMM should show a current reading with the opposite sign (-) in that case because of charging. Please suggest how i need to modify the circuit to avoid this problem. Thanks.

 

[poor mystic]

It is not crystal clear to me exactly what is going on here, but the colour of the problem suggests to me that you have made an error in construction, which you may shortly feel sheepish about. Don't let it discourage you - sheepishness is probably a necessary step in learning electronics.

 

[Laplace]

What is the voltage across C2 when the battery is charging? What kind of battery is it? Does the battery require special circuitry to keep from being overcharged?

 

[(*steve*)]

Or more likely, is the battery voltage in excess of the charger voltage (it won't be for long).

At 12V, a lead acid battery is well on its way to being discharged. You may want to increase the voltage of the regulator to just under the recommended float voltage.

I'd also recommend placing a resistor between 0.47 and 1 ohm between the output of the LM338 and R2, in series with the load.

 

[poor mystic]

Steve's remark has stimulated me to think! (Thanks, Steve!)
My new opinion is that you'd be better off with a proper continuous-use charger for your gel-cell (a presumed detail there, maybe it isn't a gel cell) which would keep the battery in good condition.
Then, I'd regulate the battery voltage to provide power for the router.
Mark

 

[duke37]

Maybe the regulator voltage is a bit low. The battery should have 14V on it when charging. At a fairly high current the diode will drop about a volt, so the regulator will need to supply about about 15V.

I think the regulator output will be a little low Am I wrong when I calculate 11.6V?

 

[haseebhm]

I'm sorry I didn't mention complete details. I'm using 2.5AH Lead Acid battery(image attached). And the power goes off after every 4 hours for a duration of 1 hour. So the battery should charge enough in 4-hours time. I think this float charge mechanism would charge it too slow, wouldn't it ? And how can I regulate the voltage for the router ? It cannot be done using LM317 because input to output voltage difference would be too low for most of the cases.

 

[haseebhm]

I'm sorry I didn't mention complete details. I'm using 2.5AH Lead Acid battery(image attached). And the power goes off after every 4 hours for a duration of 1 hour. So the battery should charge enough in 4-hours time. I think this float charge mechanism would charge it too slow, wouldn't it ? And how can I regulate the voltage for the router ? It cannot be done using LM317 because input to output voltage difference would be too low for most of the cases.

View attachment 7158

 

[haseebhm]

Further details,
I'm using a 220V Primary to 12-0-12V Transformer which can supply upto 1.5 amperes, Half wave rectifier is used with two capacitors in parallel with a value of (4700 uF)

 

[duke37]

Why use a half wave rectifier, put another diode in and make it full wave.

How fussy is the router. The battery should be sufficient to control the voltage to 14V on charge and 12V on discharge.

Measure the input and output voltages of the regulator and the battery when the mains is on and off.

 

[(*steve*)]

12V with full wave rectification will give you a little over 16V. With 2V overhead for the regulator, you can go as high as 14V.

Check the specs for float charge voltage on this battery. Float charge does not mean low current, it means that the battery is charged up to this voltage.

The extra resistor I recommended is to limit the current if the battery is dead flat.

What current does the router require?

 

[haseebhm]

Why use a half wave rectifier, put another diode in and make it full wave.

How fussy is the router. The battery should be sufficient to control the voltage to 14V on charge and 12V on discharge.

Measure the input and output voltages of the regulator and the battery when the mains is on and off.

I have set the float charging voltage to 13.8V. When the battery is connected, the voltage is always below 13.5V. But the problem is that the battery is not charged enough the next power breakdown(within 4 hours).

 

[duke37]

You have not said whether you have added a diode to convert to full wave rectification.
You should measure the voltage on the input to the stabiliser, if this drops too low between charging pulses, then the output will be low.

 

[haseebhm]

You have not said whether you have added a diode to convert to full wave rectification.
You should measure the voltage on the input to the stabiliser, if this drops too low between charging pulses, then the output will be low.

The input voltage is 15-16V. I used half wave rectifier because I had center tapper transformer. I'll change it to full wave now.

 

[duke37]

I am not sure what you have, a centre tapped winding and two diodes will give full wave rectification. If you use half the winding and a bridge recitfier you will also get full wave rectification but a lower voltage since there will be an extra diode voltage drop.

There will be considerable ripple on the input to the stabiliser, a change from half wave rectification (if that is what you have) to full wave will halve this ripple.

15V average means that, with ripple, the minimum voltage is unlikely to be sufficient to drive the stabiliser throughout the cycle. Increasing the reservoir capacitor may help a little. A low drop-out regulator will help also.

 

[(*steve*)]

...unlikely to be sufficient to drive the stabiliser...

Let's call it a regulator eh.