I'm interested in making myself a 12V power supply. I need one because I often go out to places with little or no power and need some electricity to power my gear (which is basically a telescope, a camera and some accessories - all run on 12V).

I don't want it to be powered by mains and would instead use a 12V deep cycle battery. For regulation, I'm going to be using 7812. I need about 6 outputs, each capable of 4A. My question is this:

Is it better to regulate each output separately with a 7812 (with a power transistor for increased current capability) or use single 7812 with several power transistors, for a current rating of, say, 25A, and then split this into 6 outputs.

Secondly, since I'm using a 12V battery I understand that regulation is not going to be good as Vout is also 12V. Unfortunately, I can't get a new battery so I was thinking, perhaps, stepping the voltage with a DC-DC Converter to about 14V and then feeding this voltage into the regulator will help with regulation. Is this a good idea?

I require some decent regulation as my telescope is quite sensitive to the power's quality. I apologize if these are boneheaded questions, but searching the web did not return good results.


4 Answers 4


Forget the 7812. If you're going to use a switcher to create a higher voltage first and then use the 7812 to go to 12V, you might as well use the switcher to go to 12V directly.
If the output voltage is close to the input voltage (sometimes below, sometimes above), you want a buck-boost regulator, like National's LM5118. The Webench Designer creates a design to your parameters.


You won't be able to get a good 12V output from a 12V input using a linear regulator like the 7812. Linear regulators need at least a couple of volts more on the input, so they have margin to regulate in, and they aren't very efficient, because that excess voltage, multiplied by whatever current is delivered into your load, is just wasted as heat. Better to use some kind of switching regulator, such as (yep) a DC-DC converter. Don't bother with the follow-on regulation, just select a DC-DC converter that gives you your desired output voltage with suitable regulation, and you're done.

  • \$\begingroup\$ I was hoping to avoid a switching supply because my telescope starts acting up if I connect a switching supply. I'm not sure why this is so, but a linear supply, in my case, has never caused a problem whereas with a switching supply (which had sufficient power) the telescope wouldn't even startup startup sometimes and once it corrupted SRAM inside (I fixed that by removed the internal battery for 30 mins or so). Would you still recommend a switching regulator or should I stick with a linear one just in case? \$\endgroup\$
    – Saad
    Apr 23, 2011 at 20:46
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    \$\begingroup\$ Also, with 6 outputs 4A each, DC-DC converter seems to be much better choice. With 7812, assuming 2V voltage drop, each output would generate 4A*2V = 8W of dissipated power. That's quite a lot! \$\endgroup\$ Apr 23, 2011 at 20:46
  • \$\begingroup\$ @saad - when you're running from a battery, inefficiency means shorter times between charging. instead of following a DC-DC converter with a linear regulator, you could try some passive filtering. \$\endgroup\$
    – JustJeff
    Apr 23, 2011 at 22:46
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    \$\begingroup\$ @saad: "my telescope starts acting up if I connect a switching supply" You need a better-regulated switching supply? \$\endgroup\$
    – endolith
    Apr 25, 2011 at 18:49

You have several issues here, so let me try to address them all:

  1. Most versions of the 7812 are limited to 1 to 2 amps. So you're not going to get 4 amps out of them.

  2. Most versions of the 7812 have a dropout voltage of about 2 volts, meaning that the input needs to be 2 volts higher than the output. Again, for most 7812 type regulators, if the input is not at least 14 volts, then the output will be the Input Voltage - 2. So if the input is 12 volts, then the output will be 10 volts.

  3. If Vin-Vout=2 volts at 4 amps then you're dissipating 8 watts of heat at the regulator. Times 6 outputs and you're spitting out 48 watts of heat. That's a heck of a lot of heat to deal with. You'll need a fan, at least, and it still probably wouldn't work.

  4. A standard 12v lead-acid battery is going to output anything but 12v. Depending on the state of charge, temperature, phase of the moon, and if you've phoned your mother recently it could have up to 15 volts on the output. More typically it is around 13.2 volts, but you must design for the range from 12-15 volts (10-15 would be better). Keep in mind that if it is outputting 15 volts, then your power dissipation goes from 48 watts to 72 watts.

In short, it won't work.

Here's what I recommend:

  1. If you can avoid using a regulator then don't! Many "+12v devices" will work just fine on 10-15 volts. Anything that is designed to operate from a cigarette lighter in a car will work to up 15v (because cars have the same issues). For other devices you might have to read the manual or talk with tech support. But odds are that it'll work just fine. Odds are very good that your telescope will work directly from the battery, as that's what most of their customers are going to do.

  2. For the devices that require a regulator, they are going to need something better than what a simple 7812 is going to do. The 7812 isn't going to handle the current demands, or regulate very well with a 12-14v input. Running a boost DC/DC converter to get 14v and then using a linear regulator like the 7812 doesn't really buy you anything since you're still burning a lot of power needlessly. What you'd want to do then is take a Boost DC/DC converter and follow it with a Buck DC/DC converter-- or use a SEPIC converter (which is like a boost and buck combined into a single converter).

  • \$\begingroup\$ Thanks for the response. The problem is, the scope is very sensitive to power supply fluctuations. If you read the comment above, I've already had the SRAM chip being erased due to a switching supply. Now, weather this was due to a bad supply or not, I don't know. I simply switched to a 12V linear supply and the problems completely went away. Also, do you recommend one regulator with multiple outputs or separate regulators for each output? \$\endgroup\$
    – Saad
    Apr 23, 2011 at 21:07
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    \$\begingroup\$ @saad I seriously doubt that the problems you've experienced are due to a switching supply. Well, let me put that differently... IF your problems are due to a switching supply then it's not the supply that is the problem, it's the scope. Probably the only thing in the scope that actually runs off of +12v would be the motors and maybe the cooling system on the CCD (if you have that). Everything else will run off of 1.0 to 5.0v, so there are regulators inside that will take the 12 and drop it down. If the input power is causing issues, then these regulators are the true problem. \$\endgroup\$
    – user3624
    Apr 23, 2011 at 21:15
  • \$\begingroup\$ Also, for what its worth, the manual for the telescope suggests that a regulated supply is a must (this is written in bold text, in fact!). It can handle a range of 12-18V. I understand that the chips inside the electronics panel will be running at 5.0V, but since switch to a 12V linear supply I've had no problems whatsoever. Its been 7-8 months, so I feel thats a good enough test. Perhaps it was the supply. \$\endgroup\$
    – Saad
    Apr 23, 2011 at 21:17
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    \$\begingroup\$ @saad I think you're confusing a "well regulated supply" with "low-noise supply". If you take the "12v lead-acid" battery output, it will be within the 12-18v range of your scope. It will also be about the most noise free that you can practically get. Try that and I'll bet you it would work. And honestly, if it doesn't then there's a problem with the scope. \$\endgroup\$
    – user3624
    Apr 23, 2011 at 21:45
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    \$\begingroup\$ @saad Look at the specs in the manual, or call their technical support. My guess is that it'll work, but it's worth double checking just to be on the safe side. \$\endgroup\$
    – user3624
    Apr 23, 2011 at 22:48

Simply use you car battery if you are outdoors or something like this http://www.maplin.co.uk/200w-portable-power-pack-225153

  • 1
    \$\begingroup\$ The link comes up "... has been removed". Perhaps a brief explanation of what you wanted to convey, instead of a bare URL, would improve the quality of your answer. \$\endgroup\$ Nov 3, 2012 at 10:21

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