Battery based tank circuit (tested) - Page 14

justICE · 10-19-2004, 05:06 PM

Quote: Originally Posted by Arathranar

If I'm trying to imitate an upto 200W load (may be as low as 100W) at ~12V, then I need a resistor of only ~0.8 Ohm don't I? But it needs to be capable of getting rid of up to the 200W. I suspect I'll burn myself if I try and do that a resistor - it'll make a nice 0.2kW bar heater inside my car.

My idea was to test it before you get your PS and other components. You do not want to be running tests on your spiffy new stuff=). This is just a resistance in between the terminals of your 7.2ah battery to test the voltage drop under a load. This set up will show you the effect--on your batter--of a very conservative current (only ~6ish amps). This way you can tell what value the potential will be at and for how long=).

I agree, putting this resistance inline with your ps would not be a good idea LOL. It is merely a test bench with a replaceable 80 cent item rather then your brand new gear.

I also just noticed the resistor I was looking at is only 10W. For some reason I was thinking 100W. Better find a higher rated resistor=). Or put eight 16-ohm @ 10W in parallel.

davesaudio2 · 10-19-2004, 07:31 PM

just use some old headlights as load.
doesn't everybody have some old sealed beams w the low beam burned out?

Arathranar · 10-19-2004, 07:46 PM

Hmm - that's a cunning idea.

I have a new reversing bulb from a 2-pack I bought and the old one I was replacing where the reversing filament had gone but the other filament worked. Does anyone know what current these take at ~12V? Is it safe just to hook it up and see using a ammeter? (assuming I hold it in something that can take the heat and don't look directly at it...

)

Arathranar · 10-20-2004, 12:28 AM

And the answer is:- about 0.55A.

Arathranar · 10-20-2004, 01:12 AM

That was for the low beam in the bust bulb.
It's about 2.15A for the high beam in the working bulb.

Dennis5587 · 11-16-2004, 09:31 PM

Hey, sorry to bring up a pretty much dead thread, but which circuit was considered to be the best anyway. There's so many in here and I'm not sure which one was finally decided to be the best... help...

Maukio · 11-17-2004, 12:53 AM

Quote: Originally Posted by Dennis5587

Hey, sorry to bring up a pretty much dead thread, but which circuit was considered to be the best anyway. There's so many in here and I'm not sure which one was finally decided to be the best... help...

What he said.

Ricky327 · 11-18-2004, 09:36 AM

Quote: Originally Posted by Dennis5587

Hey, sorry to bring up a pretty much dead thread, but which circuit was considered to be the best anyway. There's so many in here and I'm not sure which one was finally decided to be the best... help...

I think its best to build the zero voltage drop circuit I posted before. I have never built it but theres a member here called "curiosity" who have and claimed it to work.

Hes also modifying it such that it will autoswitch when the main battery voltage goes below a certain level.

The circuit on this thread and the zero voltage drop work in a very similiar way except that theres a relay to short out the diodes therefore eliminating the voltage drop...putting full power into the inverter/PSU.

Ricky327 · 11-18-2004, 09:42 AM

Heres what he propose post #23 :

http://www.mp3car.com/vbulletin/show...1&page=2&pp=15

Based on this circuit :

Arathranar · 01-15-2005, 02:37 AM

Finally have some real details from my setup. The current draw for my system is ~5.3A when idle and 7A running Prime 95. Way lower than I was guessing it would be. My 7.2Ah battery can supply that for many minutes (I got bored waiting

). The zero load voltage is 12.85V and it drops down to 11.8V and then stabilizes and drops way slower (mabe 0.05V a minute - possibly less). I left it running for something like 5 minutes running Prime 95 (the extra current draw takes about 0.1V off the battery voltage) and it got down to 11.55V.

So my 7.2Ah battery is easily capable of keeping the PC running during crank and after the ignition is switched off. Huzzah!

dtomgr · 04-11-2005, 05:39 AM

finaly i have read the last post in this article , but , i have some questions...

in the circuit i've attached , what kind of backup battery i sould use, and what about a diode circuit ?

my inverter stop his output whene his input falls down 10.5 volts....

i want to stop rebouting my pc , and keeping alive for some minutes.

thanks

kungmpton · 04-21-2005, 12:37 PM

Quote: Originally Posted by Ricky327

Someone might have already done this before but anyway I havent seen it on this board so I thought Ill post it

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Alot of poeple here seems to have problem fully charging the tank battery. The standard tank circuit uses a diode typically the MBR6045. Because of the voltage drop of the diode, the charging voltage is always 0.6-0.7V lower than it should be...13.8V. With this standard setup the backup battery never get fully charge and may not function properly during crank.

On the diagram I have added a limiting resistor and an extra diode. The limiting resistor is directly connected to the main battery, this ensure that the backup battery get the full voltage from the main battery.

The typical charging current for a sealed lead acid should be 1/10th of the Ah rating. It is not neccessary to have this limiting resistor in a normal setup since the battery will find its own charging current. Having read the datasheet from Yuasa it says that the charging current can be limited to 1/4 of the Ah rating. Using a 0.8Ah battery a 68 Ohms 3 Watt is found to be suitable.

This resistor is very important in this circuit :

-To limit the charging current to 1/4 of the Ah rating.
-To unsure that the back battery get the full voltage of the main battery.
-To limit the current being drawn from the back up battery during cranking.

The additional diode is there so that the PC PSU cannot drain the charging current going into the backup battery therefore maintaining the full charging voltage.

The diode on the top ensure that the backup battery never get drained during cranking...but only through the limiting resistor. Since the current is limited a heavy drain cannot occur on the backup battery during cranking.

I have tested the circuit and it does work as expected

Compaq Evo N400C
Targus DC-DC PSU

Engine off = 12.1V
Engine cranking 10.1V
Engine running 13.5V

PSU cut off = 11.0V
PSU/Laptop curent drain = 2.7A

The circuit was tested with a 1N5822, this diode have a drop of 0.525 at 3A. It is better to use 1N5820 as labelled on the diagram and this only drop by 0.475V. So an even better results can be expected with the 1N5820.

With the test, hardly any heat is generated on any components. The only time alot of heat can be expected is when the backup battery is fully drained but then the 3W resistor should cope with it.

The backup battery was delibrately drained down such as the PC die completely. This cut out happened when the backup is supplying 11.2V. The backup battery is then charged for 3 mins by running the engine. A recrank is made, and yup again it survive a crank after only being recharged up for 3 mins

I havent tested how long the backup battery will last if it was charged up for 1 hour of driving as I didnt have the time. Im expecting like 3-4 mins if the battery was fully charged up. But 15 seconds of crank survival should be more than enough.

The maximum current can be drain with this circuit is only 3A as limited by the diode. I dont have a more powerful diode to test it out

A bigger backup battery can also be used together with a recalculated limiting resistor value and suitable wattage.

For some reason the standard tank circuit uses MBR6045, its better to use MBR3045 or something with a lower voltage drop. MBR6045 drop by 0.7V while the MBR3045 drop by 0.5V

Even with this improved setup I dont think its good enough for ITPS. The only way around the ITPS is to add a relay that short out the diode so no voltage is ever lost. Even with that zero voltage drop circuit, the ITPS minimum operating voltage of 13.3V is still pushing it to the limit.

Ill use a higher rating diode next time on a bigger PC...If anyone wanna try first please do so and report back

Sorry to bother you ricky, but as being n00b on electrics, can you tell me what would be correct resistor for 1,2ah accu, and about 5a drain ?

And if possible, even explain how to calculate those =)

Thnx :P

KyleYankan · 06-30-2005, 02:25 AM

I'm sorry but I don't have time to read this thread all the way through. What part numbers should we use for the diodes and resistors? Also - waht size battery do you recoment for a 200 watt system? 400 watts? (theroetically)

KyleYankan · 07-24-2005, 06:36 PM

I'm sorry: I searched and searched:

What power rating should the resistor be? I'm imagining about 100 watts? I'm not sure - any help? Anyone selling one?

KyleYankan · 07-25-2005, 03:39 AM

What's wrong with a simple circuit like this one:

The diode can be an 1N5822. This is only a 1/2 volt drop, I beleive.

When the car is on, at 13.8 volts, the backup battery gets 13.3 volts, more than enough to charge it, and to keep the 12 volt inverter/PSU happy. Best of all, only 1 component. Simple.

Here's my only problem: the 1N5822 can only handle 3 amps. I need like oh, call it 30 amps to be safe. Can I throw 10 of them in parallel to acheive this? I think this would make for a neater install. I mean 10 of those suckers all sodderent together and in shrink-wrap looks like a thick wire, that gos to the battery terminal to being with, Or is there somnething here I'm not seeing?