Connect with us

+/- 12 V power supply acting weirdly(?)

Discussion in 'Power Electronics' started by Ch33f, Jun 19, 2016.

Scroll to continue with content
  1. Ch33f


    May 18, 2015
    First of all: Yes I know, you should always measure the performance of a power supply under load. BUT:

    I just conjured up a little +/- 12 V supply with a 7812 and a 7912. After soldering the thing together I just hooked it up to my 30V lab suplly (which has only one channel, hence the need for this) and checked the output with my multimeter. I thought that is fine because I wasn't interested in any noise measurement or the exact output voltage. I just wanted to see if we were in the ballpark.

    Oddly enough the positiv output was perfectly fine but ne negative one showed -3 V. So I quickly turned it off and looked intensely for faulty wiring. With nothing to be found in that respect I replaced the 7912 with a spare one I had orderd along with it. But that gave me exactly the same behaviour.

    When testing the 7912 parts in an isolated environment (pure -12V supplied with -15V) it showed up fine, so no defective parts. After several hours of frantic resoldering and looking for bad or false connections I came up with the idea of hooking up a little load (just two 4.7k resistor because I had it handy) and all of a sudden the ba***rd gave me +12 V and -12 V.

    So I guess there is a lesson to be learned here: ALWAYS always always put a load on if you want to measure output voltage.

    Here is my question however: Since I am relatively new to electronics I wanted to ask if this behaviour is normal or have I found a slightly strange version of the 7912?
    Like I said I would have expected maybe something from -11 V to -13 V or so since there was no load (apart from the 1Meg in the multimeter) but I really didn't expect a discrepancy as high as almost 10V!

    Here is the datasheet for the exact 7912 I used:

    Thanks in advance!
  2. davenn

    davenn Moderator

    Sep 5, 2009
    you do know that the neg reg has different pinouts compared to the pos reg ??

    show us YOUR circuit and show us a sharp and well lit pic or 2 of your construction

  3. Ch33f


    May 18, 2015

    Input on the bottom: Red +30V; black 0V
    Output on top: Red +12; black GND; yellow -12V

    Each column is connected all the way from top to bottom and in green I added the invisible connections on the backside. I added the 1N4007 after I experienced the weird behaviour. It didn't change the behaiviour. The scematic suggested 1N4001 but I only had the 07 type lying around and as far as I can tell that shouldn't be a problem.

    Didn't have a 0.33uF so I hooked up 4 0.1uF in parallel.

    Yes, it is far from pretty because of all the resoldering. I quadruple checked each and every connection: No connections where non should be and all connections that are supposed to be there are.

    I used the typical application circuit on page 21 in here:

    And to stress it again: It does work under load!

    I also quickly tried the circuit on a breadboard without any capacitors and it showed the exact same behaviour.
  4. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    You did notice that in the application circuit, Figure 16, that a bi-polar +20 V and -20 V with respect to circuit common or ground was used? The circuit you built does not do this.

    If your 30 V supply has its negative side connected to the input of the 7912 negative regulator and its positive side connected to the input of the 7812 positive regulator there is no common for the 30 V supply. By adding load resistors to the outputs of the regulators, you have created a floating common. This pretty much shoots down the possibility of driving a significant load with the voltage regulators. Back to the (circuit) drawing board. Pay closer attention to makey-makey when copying application notes.
    ModemHead likes this.
  5. Ch33f


    May 18, 2015
    Huh... I guess I missed a core concept there. :oops:
    I thought that might work. Well one followup then:
    Would it work if I hook up two 1Megs two create the floating common?
    That way I should only waste/lose 12 μA and thus should still be able to supply a fair amount of current right?

    Edit1: I found some reading to do:
    I will work my way through this these days and maybe come up with some more followup questions.

    Edit2: Nevermind the 1Meg question. That would mean my ground has a high impedance and thus any connections between + or - and ground could only be supplied with a max of 12 μA.
    Or am I wrong? I am currently confusing myself greatly. :confused:
    Last edited: Jun 19, 2016
  6. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    It would be better, IMO, to create the "floating" ground at the input, between the terminals of the (hopefully) floating 30 V supply. A pair of low-valued resistors will do it at the expense of a lot of wasted power. Another approach is to use a power op-amp output as the "floating" ground, which then becomes a "driven" virtual ground. You connect the op-amp as a unity-gain non-inverting buffer and provide a voltage divider at the non-inverting input. The voltage divider is two equal-valued resistors connected across the 30 V supply, and the op amp is powered by connecting its +V and -V power inputs across the the 30 V supply. The current you can draw from the regulators will depend on how much current the op-amp can provide.

    You aren't wrong, but it would be a whole lot simpler to just use a bi-polar power source as shown in the app note. So-called floating grounds and driven virtual grounds are mainly useful when you need to do low-level, low-current, signal processing in battery-operated equipment that is otherwise powered from a single polarity power source.

    The link you posted is an excellent resource and explanation of "floating" grounds.
    Last edited: Jun 19, 2016
  7. Ch33f


    May 18, 2015
    Ok, so I have looked into this a bit further and here are some of the possible aproaches:
    • Create a virtual ground on the input side using resistors. Problems: It's a balancing act between wasing power and still having a fairly low impedance on the ground rail. Especially if you want +/-12V you really can't go below 200Ω or you will have trouble finding resistors that can handle the heat.
    • Create a buffered virtual ground using an opamp. Problems: You can't supply a lot of current anymore because most opamps are made for low current. I found some that can go up to 1 or 2 A but they are extremely expensive (about 13€/~15$ where I found them).
    So it sound pretty bad all in all. But on the bottom of the page I posted earlier is a link to another page which shows an interesting aproach.

    Take a look at this:

    For a more complete circuit (including capacitors and so on) and some notes please visit the site!

    In my eyes this compares surprisingly well to the approaches I mentioned above. The regulators now don't actually regulate the input voltage but are used to create the virtual ground.
    The out terminals of U1 and U2 now have (theoretically) the same potential. Of course we don't live in a perfect world and the difference here will depend on how precisely you set up the big input regulator U3. But you can easily get away with 1 Ω resistors on the devider here (even less if you get it really precisely set up).

    More importantly you don't depend on an expensive opamp in order to supply ample current. The only limiting factor here (apart from U1 and U2) is U3. Since I have a lab supply I don't need a third regulator, so I havent looked into how expensive a 3 A regulator is (my U1 and U2 can each supply 1.5 A hence the 3 A on U3) but I assume it's less expensive than a 1 A opamp.

    I tested it and using a lab-supply instead of U3 I got the difference as low as 4 mV.
    In the end the circuit drew 8.5 mA at 24.55 V without a load, which equates to 20.8 mW of loss. However I think this power was lost inside the regulators because the divider only wasted about .3 mA.

    I haven't tested any other behaiviour of this (noise, stability etc.) but as far as I am able to do so I might do that in future and if so I will post my results here.

    I quite like this approach because it seems to vastly outweigh it's drawbacks (i.e. only a third regulator). But I am still a newbie in the field of electronics, so I would like some oppinions of you guys.
    What do you think of this circuit?
  8. Ch33f


    May 18, 2015
    Ok, first problem found:
    Looking at asymetrical loads I found the following problem.
    I put a 470 Ω resistor between the + rail and the VGND => everything ist fine.
    I put a 470 Ω resistor between the - rail and the VGND => The VGND suddenly shifts massively and the + rail shows 21 V and the - rail shows -3.5 V (both relative to VGND).

    470 Ω should only draw about 25 mA which should be a piece of cake for a 1.5 A regulator.
    Why does this happen?
  9. Ch33f


    May 18, 2015
    To elaborate a bit:
    If this was a simple resistor based voltage divider I would not have been surprised, because by drawing current asymetrically you basically change the impedance of half of the divider, thus shifting the VGND.

    BUT: To my knowledge the two resistors play only a minor part in creating the VGND in this circuit. They basically just even out the 4 mV difference that occures due to imprecisions. I would have expected the regulators to regulate any shifting in the VGND because thats basically theire job: Supllying the same voltage no matter how much current (to a limit of course) is drawn. Since I only draw about 25 mA (in theory; in practise it's much less due to the shift) a 1.5 A regulator should handle that easily, right?

    And odly enough it only happens when I place the resistor across - and VGND, not if I place it across + and VGND.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day