Hey guys, thanks for all you help before with helping me choose which booster pack to buy. I bought the motomaster eliminator 700ccr one with a 22ah battery in it. I came home let it charge for 72 hours, i go into my garage to hook up the + and - Clamps to my amplifier and the thing wouldn't work. I did a bit of research and it turns out it only outputs power through the clamps when it senses there is at least 3v present in the battery it is suppost to help boost. How can i over ride this 3v so i can run my amplifier off of it?

 

You might fool it with a 9v transistor battery connected with a diode so that current can't flow back into the transistor battery.

 

Brandon,
Sgt has given you a VERY informative and important bit of advice.
Do not attempt to do this without the diode!


so here are some attachments (not to scale) that might help.
as I had nothing much else going on.


I don't know your level of expertise so here are a couple attachments to explain the procedure.

we could also ad a large elctrolytic capacitor in parallel to this little circuit and a switch between the
diode and + jumper end. After the amp was turned on, the switch could be shut off and the capacitor
would maintain the stay-alive signal.
This is a method that is used often in battery charger systems on industrial and construction equipment.





stuck in a hotel in Seattle (actually Redmond) tonight, obviously nothing better to do.

I have nothing against anyone from Seattle, and 50 degree weather in Seattle is better than the 20 degrees in Baltimore, Seattle is just not my favorite place. I do not know why, maybe so far away from reality.
Of course being back at corporate headquarters is what sux...
but hey, we all have our druthers....
but I guess if this is where Jimi came from, it can't be all bad..

 

Hey thank you all that responded. I like the diagrams that you have attached. This seems very simple to do up, i wouldn't say i know a lot about this stuff but ive been playing around with this low voltage stuff since i was about 8, im 17 now...... I have a couple of quick questions relating to you diagrams. #1 i do not own a 9v battery, they are all 6v 7ah lead acid batteries. Can i use a 9v smoke detector battery or is this dangerous because it might explode if hooked upto the booster because it is not rechargable? My second question is, do i specifically need a 6a02? Ive got a few other ones kicking around but none are the 6a02, can i use just any diode? once again thank you all that have helped me out with this little project.

 

Brandon,
Sgt has given you a VERY informative and important bit of advice.
Do not attempt to do this without the diode!


so here are some attachments (not to scale) that might help.
as I had nothing much else going on.


I don't know your level of expertise so here are a couple attachments to explain the procedure.

we could also ad a large elctrolytic capacitor in parallel to this little circuit and a switch between the
diode and + jumper end. After the amp was turned on, the switch could be shut off and the capacitor
would maintain the stay-alive signal.
This is a method that is used often in battery charger systems on industrial and construction equipment.





stuck in a hotel in Seattle (actually Redmond) tonight, obviously nothing better to do.

I have nothing against anyone from Seattle, and 50 degree weather in Seattle is better than the 20 degrees in Baltimore, Seattle is just not my favorite place. I do not know why, maybe so far away from reality.
Of course being back at corporate headquarters is what sux...
but hey, we all have our druthers....
but I guess if this is where Jimi came from, it can't be all bad..

voodoomojo can you sketch me up a diagram with the large elctrolytic capacitor, i have one from a computer power supply its 330uf 200v mk.... not sure if it will work but its worth a shot.

 

The 6A02 represents a 6 amp diode, 100v peak inverse voltage diode.
you do not need this exact diode. In the scheme of things, a 6 amp diode is a pretty large diode and can handle the current used. i use it often because I have at least 100 of them at any give time.
A 9 volt battery can give out quite a bit of current and the amp will be asking for a lot of it. Until the booster pack kicks on, the 9 volt battery will be trying to power up the amp. This condition only lasts for a minute (as in "my-noot") amount of time.
The diode stops the 12-13 (or so) volts from the booster pack reaching the 9 volt battery. When the switch is turned on, the 9 volt battery will satisfy the 3 volt demand of the booster pack and turn it on. The capacitor will charge up to near booster pack voltage almost instanteously and the amplifier will turn on. Once the amplifier starts up, the switch can be turned off. The capacitor will then just maintain a stay-on condition for the booster pack.Actually the booster pack at this time is providing the stay-alive signal to itself. The capacitor is used to steady that signal and permit the battery circuit to be disconnected by using the switch.

 

Thanks again voodoomojo, its amazing how a diagram can say a 1000 words.... I appreciate all the time you have taken to help me out. I just got home so now its time to build the circuit. Ive got 1 more question then ill leave you along haha... the 9v battery can it be an alcaline or does it have to be rechargable? also the booster pack needs to sence 3.5v present so can i just use 1 of my 6v 7ah batteries instead of a 9v? Thanks alot buddy

 

The 9v battery could be either. However, it may not have enough power to charge any output capacitance your booster might have; I didn't think about that before.

You'd be better off using one of your 6v 7ah batteries - however, use a heavy-duty rectifier diode and a current limiting resistor - just in case there are large caps in the output of your booster.

Radio Shack carries both 1n500x series 3A rectifier diodes, and 10 Ohm power resistors. Pick up at least one of each, and wire them in series. Make sure that if you wire the diode to the positive terminal of your 6v battery, that the cathode (banded end) of the diode is away from the battery terminal. If you wire the diode to the negative terminal of the battery, the cathode (banded end) must be towards the negative terminal.

The 10 Ohm resistor will limit maximum current flow from your 6v battery to less than 1A. This is for safety.