how can i turn a 3 volt current into a 12 volt continuous current to keep a relay switch open?

i am building an r/c tank as some may already know from previous questions, anyway my batteries came in and now i have gone to my local jarcar shop only to find out that the relays i require to run a 12 volt 9 amp current need 12 volts at 150ma to stay open!
but the reciever from a toy rc truck can only supply 6 volts for which the motors run off and another 3 volts for which the servo runs off .
so how can i amplifie this current into 12 volts continuous to keep the relay open?
i know whats needed, the voltage will be doubled but the amphours will be cut in halfe or quarters or whatever the opposit of what is needed to make a 12 volt current. i need to know what it is i need and how to assemble it , i am good with most instructions and im presuming that this should be quite simple and similar to that of a solar powerd sla battery charger.

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Better plan - get a relay with a 5V coil, and run it from the 6V motor battery.
oldmanbeefjerky (author)  steveastrouk7 years ago
i may not have explained my situation . right now i have 2 motors that run off 12 volts at 9 amps, this 12volt 9 amps are brought on from 4 relays (each relay is for forwards and backwards for either motor). anyway the reciever i am using gives out a 6 volt current at about 700ma or so which is not enough to set off the relays which were the only ones aquirable that could handle a current above 3 amps and i was not about to buy multiple relays and set them parralel when i could just amp up the power! anyway another thing i should point out is that im using this motor in my tank from jaycar and the power is from a large battery pack not from the reciever, the reciever and battery pack are not linked in any way other than through the relay,ok!
so anyway, how can i get a power amp? in jarycar there a kits fro making a 9volt batterie from a pair of aa's wich seems simple enough to make, but how am i to double the voltage from the receiver, i need instructions people a website or instructable that shows what parts i need and how to assemble them, or just answer here,please, i feel as if i have just wasted $230 on what i thought to be all that i needed i need to know that i can before the 15 day return policy expires
Use the 6V output to switch the relays with some transistors ? Costs a few cents in bits to do it.

Do the 700mA outputs sink or source current ?
oldmanbeefjerky (author)  steveastrouk7 years ago
the 700 ma current is driven directly off the battery thorugh the reciever as it is used to drive the motor as for the transistors, how exactly should i do this, what do the transistors do exactly anyway? im still learning the basics of electronic components but i do very well understand motors and batteries or at least enough to get up to this point anyway. i can read schematics sort of , as long as the components are labeld out on schematics
Here is a drawing with two different ways of doing what I think you want. Fewest parts, use the top solution. Don't forget you can see it full size if you click the i button at the top left.
oldmanbeefjerky (author)  steveastrouk6 years ago
i canot read a word of that nor possibly understand how that works, could you possibly direct me to the site that this image is from, or a higher quality image of it anyway? i dont see any "i" buttons on this image by the way, now since i cannot see what it is thats going on in the picture ill asume its a voltage multiplier using capasitors and diodes correct?, or is there something else, i cant realy read electronic schematics properly , not un less the components are labeld with what they are, then i would understand, perhaps i should borrow a book from the library on read electronic schematics? anyway , that does seem to lack large amounts of components which is good but i must ask, what is that box with the numbers and alot of slots (ill asume) on it anyway? or is that just a pcb board? i realy am clueless when it comes to these things, not unless i draw them
Here's a bigger picture. You need to learn how to read schematics. The "box" is a single chip solution for you. There is no voltage multiplier, it is not needed.
The top circuit, above the line, uses a single chip, the ULN2003 or the ULN2803 to use your 6V signals to switch 12V relays. Connect your 6V signals to the left hand side, and connect your relays to the right hand side.

The lower circuit shows how to do it with a transistor for each relay. You need a 150 Ohm resistor, a transistor (TIP121) and a diode (1n4007) for each relay.
mtdesigns4 years ago
Easy way would be to use a step-up voltage regulator. On ebay they have small ones for $5 or $6 bucks, wires in series, output voltage adj from 2v to 30v, and at 12v handles 1.3a max. You'll just need a cheap volt meter to adjust the output to 12 volts.
oldmanbeefjerky (author) 6 years ago
can somebody pleeeaaaase ! post a real picture of a real relay conected to a uln2003an , or at least a similar setup with a fake relay but still using the chip as i dont know were to put the 6v into the chip to produce the 12 v output on the other side, its all too confusing, i thought i understood how it was supposed to work but everyone keeps telling me what it is and does not how to actually use one, same with tech drawings, all it says in the schematics or the uln2003an is that its a bunch of diodes all lines (like this ||||||| ) and techinacaly it should be able to turn 6 v into 50v 500ma
oldmanbeefjerky (author) 6 years ago
theres one more thing i must ask, can this change any voltage into 12 volt?, because my recievers internal relay just burnt out (it was quite old), and now i am using a receiver that uses a 8 volt motor and a 4 volt servo, i need to know if these currents will still change into 12 volts using the uln2003
aslo whats the dirrerence between the two transitors anyway
oldmanbeefjerky (author) 6 years ago
i have just found the chip , i think, could you please take a look at this so i can be sure that it is the correct one, if it is i wont need to produce 4 voltage multipliers for the 4 relay cards click here to see the chip

oh and i should probably point out that this here is the relay card im using
you are getting a best answer from me!! XD
i wish you guys had told me sooner about this, but thanks anyway
lofty7 years ago
What voltage batteries are you using? If you've got 12V batteries, you could just use a transistor to switch the current through the relay straight from the battery pack.

If your battery pack isn't 12V then you could do one of the following:
- Add more batteries just for the relay, give them both a common ground
- Use a different relay with a different primary voltage
- Use power transistors or MOSFETs with a low on-resistance (make sure to heatsink them) instead of the relay
- Use a buck-boost or boost converter (google it) to make a 12v rail from a lower voltage
oldmanbeefjerky (author)  lofty7 years ago
do you mean like placing a battery between the reciever and the relay to give it the extra voltage when the receiveris putting power to the relay? e.g it has 6 volts and putting a 6 volt batterie on the relay means it constantly has 6 volts exept when the reciever is turned on then it has 12 volts
oldmanbeefjerky (author) 7 years ago
i cannot get another relay that can handle a 30 amp current which has a 5v coil or any coil that is below 12 volts anyway, i am using the lowest avalible coil i have that will not be fryed by the potential voltages and amps produced by the battery and motor, i need to know how to multiply the voltage
orksecurity7 years ago
Assuming you mean DC: You'd need a DC-to-DC converter, which conceptually uses the DC to produce AC which can be stepped up through a transformer and regulated back down to to DC. There are some inefficiencies in this process -- and don't forget that, like any transformer, you'd be losing current to at least the same ratio that you gain voltage. (There's no free lunch; power is volts times amps and there's only so much power available.) It might probably be cheaper and easier to find a relay designed to operate on 3V...
... or, as Steve points out, 6V. I missed that the 6V source was available.