Virtual Ground on Power Supply?

Thread Starter

KLillie

Joined May 31, 2014
129
I am building a simple power supply using a wall-wart. It supplies a little over 16 volts output with a max 650ma draw. I am entertaining the idea of adding this voltage divider (virtual ground) in the power supply. I bought the wall-wart for a dollar from goodwill. This is all being built on the cheap side. I'm not buying a lm337 or a virtual ground IC. Thoughts?
Blurb.jpg

1. Should I add a diode?
2. Are 100KΩ resistors too much/too little?
3. Is this going to be a problem? :eek:
 

ronv

Joined Nov 12, 2008
3,770
I am building a simple power supply using a wall-wart. It supplies a little over 16 volts output with a max 650ma draw. I am entertaining the idea of adding this voltage divider (virtual ground) in the power supply. I bought the wall-wart for a dollar from goodwill. This is all being built on the cheap side. I'm not buying a lm337 or a virtual ground IC. Thoughts?
View attachment 86647

1. Should I add a diode?
2. Are 100KΩ resistors too much/too little?
3. Is this going to be a problem? :eek:
I think you have a couple of problems.
I think you really want the positive output to come from the output of the regulator.
With such high value resistors you wont be able to draw much current without the voltage changing. Add a load to one supply in tour simulation and you should see the problems.
 

crutschow

Joined Mar 14, 2008
23,506
You have a large problem.
You need an active virtual ground circuit to carry the ground current for any imbalance in the currents between the plus and minus load.
 

Thread Starter

KLillie

Joined May 31, 2014
129
I have not given this too much thought. I was hoping it could just be elegant and simple. ronv - I added a a stepping resistor in parallel with the whole divider and I didn't understand my results. Negative current!? Gotta get to work. I see I once again have lots to think about. crutschow-:confused:
 

Thread Starter

KLillie

Joined May 31, 2014
129
Here is what I don't understand. If I build just the power supply and make a virtual ground on a breadboard "outside" the power supply (just for op amps) ...what's the difference? Any ideas on how this could work?
 

Thread Starter

KLillie

Joined May 31, 2014
129
You are wasting your time. Cheap yes, but time spent?

Go for something small, thus cheap, but useful and with future.
Thanks for your 2 cents. Any venture that causes me to think is valuable. I probably will not go this direction in building "my" virtual ground PS. I still don't understand what is going on.
 

atferrari

Joined Jan 6, 2004
3,486
That common must sink / source current from/to any loads you decide to connect, offering definitely low impedance.

The suggestion of a simulation is really good for you to understand.

Even if beating a horse dead already, have you considered that your "change" of connection (red trace) is an illusion?

Some ideas like yours here, I had too, but a that time there was not forum nor Internet to post about them. :p

Buena suerte.
 

SgtWookie

Joined Jul 17, 2007
22,210
Try it like this:

R5 and R6 are used here for example loads.
D1 doesn't have to be a Schottky; it was simply a convenient 1A diode I had a model for.
You can use an L272 power opamp for the LT1886; available from vendors like Digikey and Mouser; two for a buck if you buy 10 or more.
http://www.digikey.com/product-detail/en/L272AM/L272AM-ND/1053997
http://www.mouser.com/ProductDetail/Fairchild-Semiconductor/L272AM/?qs=sGAEpiMZZMtOXy69nW9rM4JypmRX5pBdV0VzagItYbc=
You'll need to provide ways for both the regulator and the opamp to dissipate the heat generated.
R2 can be a 2k potentiometer (pot) wired as a rheostat/variable resistor.
 

Attachments

Thread Starter

KLillie

Joined May 31, 2014
129
That common must sink / source current from/to any loads you decide to connect, offering definitely low impedance.

The suggestion of a simulation is really good for you to understand.

Even if beating a horse dead already, have you considered that your "change" of connection (red trace) is an illusion?

Some ideas like yours here, I had too, but a that time there was not forum nor Internet to post about them. :p

Buena suerte.
I do notice my red trace is not what I expected since you pointed it out. Thanks
 

Thread Starter

KLillie

Joined May 31, 2014
129
SgtWookie! I hope that you had that floating around or did you go to the trouble of putting that together just for me? I think I am going to stick to my first plans. I am always amazed at how helping this site is.
Try it like this:

R5 and R6 are used here for example loads.
D1 doesn't have to be a Schottky; it was simply a convenient 1A diode I had a model for.
You can use an L272 power opamp for the LT1886; available from vendors like Digikey and Mouser; two for a buck if you buy 10 or more.
http://www.digikey.com/product-detail/en/L272AM/L272AM-ND/1053997
http://www.mouser.com/ProductDetail/Fairchild-Semiconductor/L272AM/?qs=sGAEpiMZZMtOXy69nW9rM4JypmRX5pBdV0VzagItYbc=
You'll need to provide ways for both the regulator and the opamp to dissipate the heat generated.
R2 can be a 2k potentiometer (pot) wired as a rheostat/variable resistor.
 

WBahn

Joined Mar 31, 2012
24,853
SgtWookie! I hope that you had that floating around or did you go to the trouble of putting that together just for me? I think I am going to stick to my first plans. I am always amazed at how helping this site is.
If you stick to your first plans, be prepared to learn some valuable lessons from its failure (which is a success in its own right).

While you say that the walwart powering this thing is 16V/650mA, you never say what the desired output from your power supply is. Let's say that it's ±5V, meaning that you need PosOutput to be 10V. How much current can the regulator deliver (assuming adequate heat sinking)? You need about 2V of dropout for the regulator, so the regulator output can't go above about 14V. That means that you have 4V across that 220Ω resistor, so you will be limited to no more than about 18 mA.

Now let's say that you connect a 1 kΩ resistor between the positive output of your supply and your ground. Ideally it should have 5 V across it and should have 5 mA through it. Even with the 220Ω resistor this is well within the supply's capabilities. But what do you actually get? Well, analyze the circuit:

split.png

What's the voltage at VGND going to be? It is anywhere close to 0V? What is the voltage across the 1kΩ resistor? What is the current through it?
 

crutschow

Joined Mar 14, 2008
23,506
................... I think I am going to stick to my first plans. I am always amazed at how helping this site is.
Well I'm amazed that you (apparently) read all the reasons we gave that your plan won't work and you are building it anyway??
Truly an example of belief trumping logic. :confused:
 

atferrari

Joined Jan 6, 2004
3,486
Well I'm amazed that you (apparently) read all the reasons we gave that your plan won't work and you are building it anyway??
Truly an example of belief trumping logic. :confused:
In that case, the best the OP can do, is to build the circuit in a properly designed PCB that couuld cost maybe 15 USD and take 15 to 25 days (YMMV) so when he sees it failing will be certainly sure it was a wrong design properly implemented.

Buena suerte again. (Why not?)

Edit to add

Enough said.
 

SgtWookie

Joined Jul 17, 2007
22,210
SgtWookie! I hope that you had that floating around or did you go to the trouble of putting that together just for me?
I actually threw it together just for you - and anyone else that might need a power supply with a virtual ground.
One major item I omitted is the fact that many power opamps are not stable at unity gain (when used as a buffer/voltage follower), but the L272 is. So, if one wants to substitute another power opamp, they need to carefully peruse the datasheet to ensure that it is stable at unity gain.

Another item that I didn't show is that the unused half of the L272 should be wired as a voltage follower (output connected to - in) and the + in connected to the junction of R3/R4. Failure to do so may result in the unused channel oscillating unpredictably, possibly causing difficult-to-diagnose problems.

The buffer/voltage follower in combination with the R3/R4 voltage divider makes the virtual ground actively track halfway between +OUT and GND(NEG). R3/R4 could actually be replaced with a single 10k to 50k pot, to enable you to have, say, 7v on positive and -3v on the negative side. However, build it as shown first, and get THAT working - THEN experiment if you wish.

I think I am going to stick to my first plans.
You are, of course, free to do so if you wish. However, the way you have wired R2 in your original schematic causes the LM317 to act as a constant current supply rather than a voltage regulator supply, and the regulator will try to keep the voltage across R2 (from OUT to ADJ) to be ~1.25v to ~1.3v by supplying current from the OUT terminal. Your output voltage will be at least 1.7v lower than the input voltage, but will depend upon your load current and R3.

Since you have your load current flowing through R2, the output impedance of your regulator will be very high, and voltage regulation will be practically nonexistent. Your voltage divider for the virtual ground has a similar problem in that it has a high impedance, and won't actively seek the "middle ground" between the regulator output voltage and the "real" GND.

You really don't have to believe me if you don't wish to, as I've only been doing this for, oh, 45 years? And there are some replying in this thread that have been at it longer than I have.

We would really like for you to have a power supply that works right after you put it together, as failures can be very discouraging, and may cause neophyte hobbyists to abandon the field. On the other hand, success inspires confidence and enthusiasm, and you feel accomplished.

I am always amazed at how helping this site is.
Many of us actually enjoy it. ;)
 

WBahn

Joined Mar 31, 2012
24,853
Duh! I didn't even recognize the original circuit as a constant current source. :oops:

I should have, but was too focused on a different viewpoint.

The PosOutput voltage will rail until there is about 5.7mA of current being drawn at which point it will drop to whatever voltage is needed to keep the current at about 5.7mA.
 

Thread Starter

KLillie

Joined May 31, 2014
129
Whoa whoa whoa guys! My original plans of just making a regular power supply! No voltage divider. I know there is a right way to do a dual -/+ power supply, just with the parts I have right now I may be doing that later. Also I haven't read all your posts yet. I just wanted to stop the hate mail as soon as possible. When I said this site was helpful, I meant it. I value your opinions. Sometimes (often) I don't understand the responses. But it seems if you get riled you do start spitting out numbers and I like that. It does help. Thanks again and NOW I am going to finish reading your posts.
 

Thread Starter

KLillie

Joined May 31, 2014
129
OK, I'm back.
atferrari - I do notice my red trace is not what I expected since you pointed it out. That it is an illusion. Thanks
WBahn - When you asked me to analyze the circuit you should have seen the deer in the headlights look that came over me! I don't know! I'm guessing the voltage @ VG will not be zero!
SgtWookie - Along with WBahn, I had not noticed I hooked up that resistor that way. But it is cool to know that sets it up as a constant current source! I'll have to check the way I wrote out the circuit so I don't do that again. I think I will invest in some of those L272's and try your circuit. If I actually made a power supply like this, would I only be able to use it as a virtual ground (split supply)?
 

WBahn

Joined Mar 31, 2012
24,853
You could use it as a single supply with no problems.

You still haven't indicated what kind of voltage and current you are hoping to achieve.
 
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