Driving a 12v load from a 5v PIC

Discussion in 'Embedded Systems and Microcontrollers' started by NickWest, Aug 7, 2008.

  1. NickWest

    Thread Starter New Member

    Jul 23, 2008
    I am trying to turn on a small wireless transmitter module that requires 12v at 15mA, from a PICaxe microcontroller that runs on 5v. At this stage I think the wireless module will have about a 10% duty cycle. (probably ON about 2 seconds every 20)

    As I see it I have a few options.

    1. To use a 12v supply for the whole circuit, and a 7805 regulator to drive the PICaxe. Then use a mosfet as described at http://forum.allaboutcircuits.com/showthread.php?t=6290 to switch 12v from the supply to the wireless module. I would prefer to leave the transmitter earthed and switch the high side of its power supply, because I think (I may be incorrect!) that having a direct earth for the transmitter would make the RF output more stable.

    2. To use a 5v supply, and a charge-pump or similar DC-DC converter to power the wireless module. The charge-pump would be turned on by an output of the PICaxe.

    I think option 2 would be more efficient in terms of space and power. I only need a 5v supply (ie 4 NiMH AA cells) and the inefficient DC-DC conversion process is only supplying a fraction of the total power consumption of the whole circuit.

    Oh and the wireless module doesn't need EXACTLY 12v, it works from 3-12v. But apparently I get better range if I can supply 12v. At present when running the module from 5v I get about 100m through trees. I am using this module from Oatley Electronics in Australia : http://secure.oatleyelectronics.com...d=386&osCsid=e626289f6153b3d5edbc6187c99cab3a

    Thanks for your time everybody, this is a fantastic site and the ebooks are a great resource!
  2. SgtWookie


    Jul 17, 2007
    Well, DC-DC converters can be amazingly efficient. Some circuits might even get up as high as 90-95%. However, if you're going to try to boost your 4.8v input to 12v, that'll place quite a load on the batteries.

    Let's say you need 15mA @ 12V to run your transmitter, and you have built a boost converter that's 90% efficient. You'll need about 40.2mA of current from your batteries to get that 15mA, and the current demand will be in a series of pulses at high frequency. Batteries have an internal resistance that increases as the battery discharges. This internal resistance causes power dissipation within the batteries themselves. The higher the load placed on the battery, the greater the wasted power.

    You would likely be much better off to use a 12v battery pack, and a "buck" type DC-DC converter or switching regulator to supply your PIC; your batteries would last a great deal longer.
  3. NickWest

    Thread Starter New Member

    Jul 23, 2008
    Thanks for the quick feedback! The trouble with a 12v supply is that they tend to be bulky and expensive - economies of scale mean that some permutation of AA cells tends to be cheaper and smaller. I'd prefer to use off-the-shelf batteries in this application to minimise difficulties when my transmitter needs replacement batteries (I will not necessarily be responsible for its ongoing maintenance)

    I understand what you're saying about increased current drain when up-converting voltage, but the DC-DC converter would only be working 2 seconds of every 20, as I mentioned.... so roughly a total of 4mA drain on *average*, which should minimise heating and problems of rising internal cell resistance.

    Therefore I guess I should clarify my question to ask, is anyone using a DC-DC converter to convert (roughly) 5v into (roughly) 12v, and if so, what do you recommend?

    Space is more of a concern that cost (within reason!!) so I would prefer a solution that requires minimal components.

    It would be nice if there was something analogous to the 78xx series of linear regulators, but to regulate voltage upwards instead of downwards.

    And yes, I _do_ realise that the principles are entirely different!!
  4. nanovate

    Distinguished Member

    May 7, 2007
    LT1615 or LT1613
    need 2 resistors for setting voltage and a diode and inductor. All can be be very small SMD and the whole thing fits in less than a 1-in square.

    You can also look at the simple switchers from national semiconductor or look at pico electronics for small modules.

    You can also look at low dropout regulators to power your picaxe... less wasted power
  5. NickWest

    Thread Starter New Member

    Jul 23, 2008
    Thanks nanovate, those look like they might do the job... Now I just have to do a lot of reading to make sure:)

    With their tiny footprint maybe I can make a little power-supply "daughterboard" so I don't have to design a whole new PCB... nice.

    I'll try to source a couple here in Australia and get back soon with my progress... or lack thereof!

    Thanks again,

  6. NickWest

    Thread Starter New Member

    Jul 23, 2008
    I ended up using the Intersil ICL7660, mostly because it was cheap and easy to source in an 8pin DIL from futurlec in Taiwan. As a voltage doubler it requires 4 caps and 2 diodes, so the total cost of the supply is very low, and it doesn't take up too much space, and seems to be pretty quiet electrically, although I haven't looked at the output on a scope in any detail... the main thing is that it worked, partly thanks to the extremely informative datasheets, complete with example circuits:)
  7. atferrari

    AAC Fanatic!

    Jan 6, 2004
    I always used the ICL7660 to obtain negative voltages. Surprise!