Designing a 9 V Battery Backup, Need some help.

[Renardak]

Hello everyone, I'm doing a project and could use some advice or help, I'm trying to designed a 9 V backup battery with a 9 V primary source coming in.

I'm trying to design it so that when the primary source gets disconnected or turns off the backup battery will power the rest of the circuit. I'm having troubles as well with the LED's because the first LED works as it should but the second LED showing that the backup battery has taken over doesn't illuminate properly.

Any help would be greatly appreciated.

 

[MikeMl]

What are you really trying to accomplish with this circuit??

If the goal is to extend the run time 2X over what a single battery would supply, just connect two batteries and two diodes (to isolate the batteries from each other) to one load.

Your LEDs are wasting about as much power as your load consumes, making the run time with two batteries about half of what a single battery would have supplied without any other circuitry.

btw- LED2 is connected backwards and has no current limiting resistor in series with it... not that that would make the circuit useful...

The more usual requirement is that you have an AC powered power-supply, and you want to keep the load running in the event of a line power failure. The circuit below is the way this is usually accomplished:

Since the resting voltage of a new Alkaline 9V battery is about 9.2V, you must start with a power supply that has a higher output voltage, say 9.5V. Use two 1N400x Silicon Rectifiers as current steering diodes in the way shown. If you want a LED indicator, put it across the DC supply output, so it is powered when the AC is ok, otherwise a LED across the battery will greatly hasten its demise.

Look at the simulation where the AC-powered 9.5V DC supply goes off at 1 second. Prior to that, the DC supply is carrying the entire load, and battery current is ~zero. The voltage across the load is 8.75V (9.5V-1 diode drop). The LED current is ~10mA,

After the AC failure, the load current (25.6mA) comes from the battery, while D1 prevents current backflow into the LED or DC supply. Note that the load voltage dropped from 8.75V to 8.46V.

 

[cowboybob]

Hello everyone, I'm doing a project and could use some advice or help, I'm trying to designed a 9 V backup battery with a 9 V primary source coming in.

I'm trying to design it so that when the primary source gets disconnected or turns off the backup battery will power the rest of the circuit. I'm having troubles as well with the LED's because the first LED works as it should but the second LED showing that the backup battery has taken over doesn't illuminate properly.

Any help would be greatly appreciated.

Welcome, renardak!

Interesting circuit. Good for you for at least taking a shot at it. As you (and MikeMI) point out, it does have some problems.

You might consider a circuit for the backup supply such that it is "isolated" (i.e., is not in the circuit until called for). This type of arrangement is ordinarily handled by a dedicated chip that monitors the condition of the primary supply (battery in this case) and switches to the backup source as needed, for instance:

https://www.electro-tech-online.com/custompdfs/2013/01/slls917.pdf

As you can see, not necessarily a simple circuit.

Ordinarily, for a battery backed radio or clock (for instance), the primary source of power is "robust". That is to say, powered by mains (or a much larger battery). With the device in a "tickle-charger" arrangement that keeps a battery backup fully charged such that upon the loss of the primary the power source, the battery takes over (until depleted or primary power is restored).

Otherwise, a device, such as noted above, is required.

As to your circuit (other than the noted, minor LED orientation), with a suffiiciently "robust" primary power source, a switch-over relay at the primary battery end and some rewiring of the backup battery and its LED circuit, it should work as you expected.

The relay switch-over circuit is key to a simple (albeit, energy intensive) addition to your concept.

 

[Renardak]

Thanks for the really quick replies guys and for the warm welcome.

This is actually a project for a backup battery designed for an ECG machine. We're doing an add-on component for the later year graduates and their machine and they handed us this back up power supply. They essentially want it so that there are indicator lights both Green, Yellow and Red each to come on as the power decreases in the 9 V battery to show the battery losing voltage. I'm sorry I kind of posted a rough draft this morning of that power supply I got it working and I'll post it when I get home at least i think it's working properly.

The schematic I have for the indicator lights is drawing a lot of current and voltage and not really sure how to handle it if I can run a voltage regulator into the circuit's output to make sure that I get 9V at the load, only problem is I need about 18 V to run the indicator lights and they drop at different values than the battery itself. I'll post a schematic of what I have when I get back home just wanted to give those that helped a little update and a thank you.

 

[MikeMl]

If you are adding LED indicators to show battery condition, you need to have a press-to-test button so that LEDs are not on except for a few seconds...

 

[Renardak]

I agree that would make a lot more sense but unfortunately that's how their group wants it to be constantly displayed which just seems like waste of what battery life is left.

 

[MikeMl]

A 15mA draw to power a LED will kill an Alkaline 9V battery (typical capacity of 565mAh) in just 37 hours. You want to light three LEDs.

btw- read this

 

[More Coffee]

I dont have anything to test it right now ..
if there were two 1n4007 in series in the place of D2
and led two with a 220ohm resitor was hooked in between these diodes ..to ground ..wouldnt that operated the led when the 9 volt source failed

Im thinking along the lines of blown fuse indicator circuit for this ..

 

[Renardak]

 

[Renardak]

So I ended up getting the backup battery circuit working by itself and been really struggling on getting the Green, Yellow and Red lights to switch on at their proper times (mainly the yellow to come on without manually using a push button reset). I'm probably going to use a 12 V pack of batteries in series but I'm running the back up battery circuit I posted above without the LED (that was just to ensure it was working) and I replaced the one diode with two which somehow make sure that when the main supply was on it was reversed bias (I don't know why the single one didn't do the job). We also put a LM3809 regulator to make sure we have 9V coming out to the load and the indicator circuit is in parallel with the battery.

Now this circuit I'm posting isn't the whole indicator circuit cause I've been fooling around with it for the past two days moving things in and out such as control relays and other things but I can't seem to get it working. The pots are there for variation so that the lights come on representing the battery levels on the load from the back up battery. I haven't even thought about adding the buzzer or speaker for when the red lights comes on I figure I'll use a 555 timer for that but thats another story. I'm just trying to get these lights figured out first I'm not sure really where I'm going wrong or we making this circuit much harder then it needs to be.

 

[MikeMl]

You also need to know something about the circuitry you are powering, and the voltage regulator. Presumably, you will use a Low-Drop-out regulator. The minimum battery voltage Vbat has to be > (Vout +Vdropout), otherwise the circuit following the regulator may not operate properly. Obviously, the Red LED (Battery too low) should come on as Vbat approaches (Vout+Vdropout).

The yellow LED should indicate an impending problem. Since the output voltage of most batteries drops rapidly after discharging below a specified voltage published by the battery maker, that might be a good point for the Green Led to go off and for the Amber Led to go on.

If your circuitry requires 9V, and you are using a voltage regulator with a Vropout of 1V, your trip points should be something like:

Vbat >12V shows green,
10V<=Vbat<=12V shows amber,
Vbat<10V shows red.

 

[Renardak]

Yeah those values that we have are ones given by medical standards and what the team wants that is why I have them written down. If I can use just a 9V battery that would be ideal but the setup I have now seems to draw more then that, not by much. The Vout at the load which will go back into the motherboard of the ECG has about 8.8V after the Schottkey Diode and the 7805. I just can't get these LED's working how they should.

 

[Renardak]

Hey guys,

So I've tried many different things to get this indicator circuit running along with the battery, the backup battery part is solved but I still can't get this indicator to work. I'm trying to get this working now with a more simple circuit using comparitors. I can't get the Vreferance levels to trigger at the voltages I need those LED's to come on they just stay on the entire time. The values on the resistors are not right but I was just messing around with numbers cause I'm not sure if the circuit is hooked up entirely right either.
I'm trying to get the:
Green light on at 9V - 8.5ish V
Yellow from 8.4V to 7.9V
Red on for < 7.9 V.

Any suggestions would be greatly apprecaited because the deadline is slowly creeping up and I'm running out of idea's on how to get this working properly. We are actually just starting to use comparitors so it's a fairly new subject to me so trying to pick the knowledge up as I go along.

Thank you folks.

 

[cowboybob]

For starters, you're "-IN"s are floating.

You need to bias each one, individually, to set the lower edge of the comparator window you wish each to trigger within.