Question about the Diodes used in Ricky's battery based tank circuit

**nobb** · 08-15-2006, 05:09 PM

Hi.

I was wondering if someone could please clarify this whole tank circuit issue. I have not setup my carputer yet and I am on my planning stages. I am concerned over the computer shutting down during crank (in winter, I've had cranks which lasted ~5 sec) so I would like to build a tank circuit. Before I had even read any of this on the forums, I was thinking maybe just get a big diode and put it on the positive wire connecting to my backup 4Ah battery. This would be nice and simple and stop current going from my backup battery to the starter motor during a crank. However, after reading through the site, people have said that diodes drop voltage, therefore my backup battery would not be charged fully. Does there even exist a cheap high current diode which would allow me to do this with minimal voltage loss (I dont really need the lead acid to be fully charged, just enough to last for a couple of seconds)?

After some searching, I came across Ricky's tank circuit schematics.

Battery based tank circuit (tested)

It makes sense and seems like it would probably work. However, it seems that this circuit is designed for a laptop, whereas I plan on using an inverter and full sized PC. Later on, I will be plugging in an old home theatre sub into the inverter. Total current draw at 12 volts may be at around 45-50 amps (overestimating, just in case). I am not that familiar about diodes, but the diodes being used in Ricky's schematics seem to only be able to handle a few amps. So if I were to replace the diodes in that schematic with some higher current ones, this would work right? I am kind of weary about putting so much current through a diode. If someone has any diodes they could recommend, I would really appreciate it. Prefereably, I want a diode that can be bought from here:
http://active-tech.ca/

Also, I have a question about the resistor used. Ricky mentioned that the resistor should limit charging current to 1/4 the Ah rating of the battery. So for my 4Ah battery, I should use a resistor at ~13.8 ohms? Since the resistor is put in series between the main battery and the backup battery, wouldnt the resistor lower the voltage which appears to the backup battery? Eg, in a series circuit involving light bulbs, putting another bulb in series with an existing bulb causes that bulb to be dimmer since it is not recieving the full voltage.

Input regarding this tank circuit issue would really be appreciated, since I am a bit confused. Thanks.

**v8 scimitar** · 08-15-2006, 06:43 PM

cant help you out with the circuit but resistors don't limit voltage they limit current. Voltage drops in a circuit due to resistance because you are trying to take too much power from the circuit.
To visualise this think of water in a pipe with a tap.
Voltage is the pressure you feel when you put your finger over the pipe.
Current is the amount of water coming out the pipe.
The resistor is the tap.
Now imagine you turn the tap so you just get a trickle out the pipe. There is hardly any current there I.E not much water coming out the pipe but if you put your finger over the end there is just as much pressure.

**Scouse Monkey** · 08-15-2006, 06:46 PM

try PMing a guy called Chris31 as he is really good with this kind of stuff and tell him Scouse sent you.

**nobb** · 08-15-2006, 08:43 PM

Thanks for your replies (I liked your analogy, schimitar). Anyways, I guess I could just trust Ricky's schematics. I found a 18.5 ohm resistor lying around and I was just wondering how much power would be dissipated through it. To calculate this, what voltage would I use? Do I use the 13.8 volts that come from the alternator, or would the backup battery act as a resistor too?

Anyways, doing some searching I found that the MBR6045 and MBR3045 high current diodes keep getting mentioned. My local electronics part store only caries the MBR3045, which only carries 30 amps. It was also mentioned that the MBR3045 has three pins, with the outer pins being anodes and the outer pins being a common cathode. I am wondering, with the MBR3045, could I just connect the two anodes together, therefore doubling the maximum current that this diode could carry? This would be ideal, since the only place I can get the MBR6045 is online, and they charge enormous shipping fees!

**ming** · 08-15-2006, 09:20 PM

with the resistor, the current won't even reach 1A, thus no need to have a 30 A diode...

I just use a backup battery with a relay, no diode or resistor. don't know why people bother with them