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# Thread: Using battery in line with circuit

1. Ham Member
Join Date
May 2012
Location
MN
Posts
41
Originally Posted by KC9AXZ
Take a look at this website.

This may give you a few ideas.

Jon KC9AXZ
I was going to reply with that also. The manual even shows two ways to hook it up (the one at the end seems more what you are looking for).

2. Originally Posted by KF5LJW
Only one problem doing that. Vout drop of 3 volts minimum according to data sheets. Battery or output voltage will be set to 13.6 to 13.8 volts, than means the DC power supply voltage on the input must be equal to or greater than 16.6 to 16.8 volts. That presents two problems:

1. Your 12 volt power supply must be able to go that high.
2. Your connected 12 volt equipment must not let the magic smoke out at 16.6 to 16.8 volts.

If you connect the LM317 as a constant current battery charger it will charge to within 1.2 volts of its input voltage.

1.2 volts drop across it when charging. After it stops charging it would still pass a very small amount of current, until the input voltage was reached. You can test that theory by charging a capacitor with a LM317 connected as a constant current battery charger.

Given the supply is 13.8v or about it would work just fine to protect the battery.
It would protect the battery even if the supply was 12v, just would not charge very well.

3. Ham Member
Join Date
Jun 2011
Location
Sherman TX
Posts
1,168
Originally Posted by KA9JLM
Given the supply is 13.8v or about it would work just fine to protect the battery.
It would protect the battery even if the supply was 12v, just would not charge very well.
Well I respectively disagree. A battery with an at rest voltage of 12 volts is only 30% capacity and would sulfate in a very short time if left at that low of a voltage. You have to get up to the battery specified float voltage to achieve 100% capacity, and the float voltage varies depending on battery chemistry of 13.2 to 13.8 volts.

But I do not quite understand what the point of what the project is to accomplish. What ever the device is uses 10 ma in standby from what I can ascertain. But what does it use when in use? A 7 AH battery can only supply a load current of a couple of amps and only for a couple of hours.

The easiest way is just leave the DC power supply on and call it done. My Astron has been turned on for about 20 years with a 200 AH AGM battery used as emergency power when commercial power fails. It does not take any interface between a lead acid battery and the power supply assuming the battery AH capacity is matched to the Power Supply max current, and set to the Power Supply voltage to the battery float voltage. Commercial repeater operators have been doing that for 50 or more years, and over 75 years in the Telecom sector.

4. Originally Posted by KF5LJW

But I do not quite understand what the point of what the project is to accomplish. What ever the device is uses 10 ma in standby from what I can ascertain. But what does it use when in use? A 7 AH battery can only supply a load current of a couple of amps and only for a couple of hours.
My reading of the OP is that the circuit that needs to be running only draws 10ma. Says nothing about that being a standby current. So, a 10ma current is not very large.

The question is how long the power supply will be turned off, and if the circuit actually needs back up power, or simply another supply that runs 24/7. If this is the case, and backup is not required, I would simply install a wall wart to run the circuit. Those are cheap to buy, or you can find them at any second hand store for a buck or two.

There may be a reason to turn off the large supply, we don't know. In any case, the wall wart would save electricity since it will use far less than the bigger supply, when powering a 10ma load.

If back up is needed, the wall wart could charge a small NiCd or NiMH battery that would last for a day or two. I think people here are over thinking what is needed to solve a very simple problem.

Joe

5. Ham Member
Join Date
Jun 2011
Location
Sherman TX
Posts
1,168
Originally Posted by K7JEM
My reading of the OP is that the circuit that needs to be running only draws 10ma. Says nothing about that being a standby current. So, a 10ma current is not very large.

The question is how long the power supply will be turned off, and if the circuit actually needs back up power, or simply another supply that runs 24/7. If this is the case, and backup is not required, I would simply install a wall wart to run the circuit. Those are cheap to buy, or you can find them at any second hand store for a buck or two.
Joe that is why I was asking. It is not clear, at least to me anyway, what the person is trying to achieve and the conditions.

6. Originally Posted by K7JEM
My reading of the OP is that the circuit that needs to be running only draws 10ma. Says nothing about that being a standby current. So, a 10ma current is not very large.

The question is how long the power supply will be turned off, and if the circuit actually needs back up power, or simply another supply that runs 24/7. If this is the case, and backup is not required, I would simply install a wall wart to run the circuit. Those are cheap to buy, or you can find them at any second hand store for a buck or two.

There may be a reason to turn off the large supply, we don't know. In any case, the wall wart would save electricity since it will use far less than the bigger supply, when powering a 10ma load.

If back up is needed, the wall wart could charge a small NiCd or NiMH battery that would last for a day or two. I think people here are over thinking what is needed to solve a very simple problem.

Joe

That would be a good approach.

I never did see the voltage requirement for the monitoring device. I think there is a lot of assuming that it is 12V.

For 10MA it would not take much of a battery pack, as long as the charger charged every now and then.

7. ## Spam

Originally Posted by PKMAHESHWARI
Hi all,

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75th Anniversary.

8. Ham Member
Join Date
Feb 2009
Location
North Wales
Posts
34
Hi,

Thanks very much for all the detailed replies. I've learned a lot from reading them.

I'm looking for a really simple solution, so am going to go with the battery in parallel with the supply and using diodes. The load needs a constant voltage, so I shall use two 12V batteries in series and a 12V regulator. I'll keep an eye on the batteries regularly and charge them if necessary.

The battery is being used as a backup for when the power supply goes off, which is unlikely to be more than twenty times a year, but could be up to a day at a time when it does go off.

With many thanks again,

73 de Matt
MW3YMY

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