Battery based tank circuit (tested) - Page 11

Arathranar · 10-17-2004, 04:02 PM

If you have the chance measure the charging current flowing into the battery when connected through a diode...see if its much higher than what the graph says. I havnt done this so I wouldnt know.

Me neither. I'll let you know when I've assembled all the bits. I'm assuming that the current will be acceptable because the battery will find it's own charging current that's within acceptable bounds (assuming it hasn't been fully discharged - which should always be the case for a tank battery only present to ensure the pc survives the crank):
Battery will seek its own current level and float fully charged

Ricky327 · 10-17-2004, 04:06 PM

Many people on this forum have used the diode only solution with great success. So clearly having a fully charge battery is not a requirement.

This is where my problem is, Im using a small battery and fully charging it is very important. Although I can probably have the PC running off it for a few mins. If the diode can only charge up the 0.8Ah battery to 40% what chance has it got to survive a crank?

Ricky327 · 10-17-2004, 04:15 PM

Posted months ago :

I have done a test on the backup battery being fully charged, it did last about 3mins at 2.7A drain.

By using the diode, the battery lasted seconds...can you see the difference now?

Try to fully discharge the battery if you gonna do the test, so then we know whats the maximum it can ever need to charge up.

Arathranar · 10-17-2004, 06:53 PM

Quote: Originally Posted by Ricky327

I have done a test on the backup battery being fully charged, it did last about 3mins at 2.7A drain.

That's only 0.135Ah. Are you sure your battery is ok? It should have lasted a fair bit longer with a full change of 0.8Ah.

Ricky327 · 10-17-2004, 11:15 PM

Quote: Originally Posted by Arathranar

That's only 0.135Ah. Are you sure your battery is ok? It should have lasted a fair bit longer with a full change of 0.8Ah.

The battery is new...

Thats because the DC-DC adaptor I was using cuts out at around 10.5-11.0V cant remember the exact value now. If you read the whole thread Im sure you find it there somewhere.

Now if the PSU cuts out at 10.5V, remember we also have a diode. The PC will now fail to survive the crank if the battery voltage goes below 10.5+0.6 = 11.1V

Even though the battery is fully charge but if the diode is droping the voltage then the PSU will cut out sooner. You can see why Im not getting the full running time.

The reason for finding a diode with the lowest drop out.

I dont know if you seen the circuit I have done months ago, its a ZERO voltage drop tank circuit. It still consist of 2 diodes but theres a relay to short the diode so theres no voltage drop at all and therefore putting the full voltage into the PC and so making the most out of the battery capacity.

The only time theres a voltage drop is when the relay swithes over which happen in a split second anyway.

Ricky327 · 10-17-2004, 11:30 PM

If your inverter have a high cutout voltage (11.00V lets say) and you got a standard diode voltage drop (0.7V lets say). Then I wont be too surpised if the tank is not able to deliver the full expected running time.

11.7V from the tank and the inverter will cut out...something to think about.

Heres the discharge curve :

Arathranar · 10-18-2004, 12:07 AM

Now if the PSU cuts out at 10.5V, remember we also have a diode. The PC will now fail to survive the crank if the battery voltage goes below 10.5+0.6 = 11.1V

The diode is between the main battery and the tank, so will not cause the above problem. However, I'm considering adding a number of diodes to drop the tank battery voltage to nearer 12V as all the DC-DC psu's do not regulate this (bar the Opus). The diodes should result in a less variable voltage drop than a resistor.

Thats because the DC-DC adaptor I was using cuts out at around 10.5-11.0V cant remember the exact value now. If you read the whole thread Im sure you find it there somewhere.

From what I've read I would have expected a lead acid battery voltage to still be adequate well beyond the 16% discharged point. But I suppose batteries vary. I imagine I'll have fewer issues withs the bigger 5.1Ah I intend to employ.

The only time theres a voltage drop is when the relay swithes over which happen in a split second anyway.

I don't believe this circuit will work, as I've read that the split second of zero voltage as the relay switches will reset a PC.

Ricky327 · 10-18-2004, 12:25 AM

The diode is between the main battery and the tank, so will not cause the above problem.

No, Im talking about the circuit on the beginning of this thread. Since you ask why Im only getting short running time. Even on the circuit you propose you will still have a shorter running time if it was being powered by the main battery...no mather what you do you aint gonna get away with that voltage drop.

I don't believe this circuit will work, as I've read that the split second of zero voltage as the relay switches will reset a PC.

How? when theres 2 diodes there to backup the supply as the relay swithes over. That relay can be removed and the circuit can still work. It was only added there to remove the voltage drop

Beside where did you read that? If those diode never existed then Ill would have agreed with you, but this time no. Im not sure if you even understand it at all.

I dont claim it gonna work 100% as I have never tested it, but please study it first before you ask a question or make a comment.

Arathranar · 10-18-2004, 12:29 AM

I did study the circuit. As the relay switches, the voltage to the PC will be briefly zero.

Ricky327 · 10-18-2004, 12:30 AM

However, I'm considering adding a number of diodes to drop the tank battery voltage to nearer 12V as all the DC-DC psu's do not regulate this (bar the Opus).

Are you kidding?

Ricky327 · 10-18-2004, 12:38 AM

Look at the top diode, its connected to the main battery and its supplying to the PC...is that correct?

Look at the bottom diode, its connected to the tank battery and also supplying to the PC...is that correct also?

So now tell me when is the time there will be ZERO volt into the PC? The answer is when both battery are dead.

Havent you noticed its exactly the same circuit as post #1 only with a relay added in...

Argg Im going around in the circle, this remind me of the CAP vs BAT

Arathranar · 10-18-2004, 12:38 AM

no mather what you do you aint gonna get away with that voltage drop.
True. But I'm still far from convinced the resistor is better than just a diode.

Here's the circuit I'm going to try:

Arathranar · 10-18-2004, 12:40 AM

Quote: Originally Posted by Ricky327

Look at the top diode, its connected to the main battery and its supplying to the PC...is that correct?

you are right. I repeatedly viewed the "box" around the relay as just that. It's too similar to standard notation.

Ricky327 · 10-18-2004, 12:44 AM

Quote: Originally Posted by Arathranar

no mather what you do you aint gonna get away with that voltage drop.
True. But I'm still far from convinced the resistor is better than just a diode.

Here's the circuit I'm going to try:

By all means try it and let us know how it goes