±5V from up to +12v

I'm trying to design a remote controlled volume controller, and the chip I've chosen to control the volume needs ±5v for the analog side of the circuit, and +5v for the digital.

I have several options for input voltage, mostly based on old AC adapters I have laying around. In particular I have a "selectable" adapter which can select up to 12v, but I may have others with higher voltages (I haven't looked through them recently).

One idea I had was to chain 7805s together:

+12v connected to VinA & VinB

GndIn connected to GndA

VoutA connected to GndB

If I understand correctly, I could then use VoutA as my reference point, making GndA -5v and VoutB +5v.

If I do this, am I going to let the magic smoke out? Is there an easier way to do what I'm trying to do? It seems like most ±5v chips I've found require a bit more complicated external components, and are far more expensive.

Thanks, Daniel.

Reply to
Daniel Pitts
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It won't work like that. The 7805 needs the ground connected to the minus with respect to the positive input. You may be able to rig up something with a 7805 and a 7905.

Reply to
Tom Biasi

The VoutA would be -7 with respect to the +12v, so the VinB would be at a at the right place respective the +12v.

Reply to
Daniel Pitts

Hook it up and get back to us.

Reply to
Tom Biasi

You need to create a "splitter" which will force a ground node and absorb the current imbalance between the +5V and -5V outputs.

Do you have knowledge of the operating currents for each portion? ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Reply to
Jim Thompson

WW1 +-----+ |~ +5|------>+5 | | |~ 0V|--+ +-----+ | +--->GND +-----+ | |~ +5|--+ | | |~ 0V|------>-5 +-----+ WW2

Reply to
John Fields

Huh? Didn't Daniel imply he had a single +12VDC adapter as input? Currents have to balance. ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

I could probably calculate it, but I don't yet have that, no.

If I don't have the appropriate "splitter", what will happen? Will I be drawing more current across the regulators, or will they just not regulate reliably?

Is there a better approach than this? One or more DC to DC converters perhaps? I'm open to alternative suggestions as well.

Reply to
Daniel Pitts

I was planning on using only one +12VDC adapter.

Here's the diagram of what I was thinking:

7805(A) +------+ | +5|------>+5 +--|Vin | WW | | 0v|--+ +-----+ | +------+ | |~ +12|--+ 7805(B) +--->GND | | | +------+ | |~ 0v| | | +5|--+ +---|-+ +--|Vin | | | 0v|-+---->-5 | +------+ | | | +----------------+

So, looking at this, I think I saw your earlier point about needing the splitter...

Well, I think I do, but every time I try to write it out, I can't figure out how to phrase it.

Reply to
Daniel Pitts

Example of problem... load on upper +5V is greater than load on lower

+5V, so current tries to flow _into_ lower +5V, but can't :-( ...Jim Thompson
--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

What's going to upset the "bottom" 7805, if it happens, is the "top" 7805 dumping more current onto its output than is being consumed. If that happens then it'll fall out of regulation (it only supplies current, it does not sink it).

The 7805/7905 should sorta-kinda work, maybe -- I'd give it a higher chance of success than the stacked 7805 chips.

You could load the "bottom" 7805 with a resistor, to make sure that it's always delivering current.

Or you could womp up a regulator from op-amps, particularly if the current draw is light.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

Turns out, +12V input is insufficient, headroom-wise, to power a back-to-back combination of 7805/7905.

Rolling your own would still have headroom issues unless you use MOSFET's as the pass devices. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

Yes, I thought of that but the OP said he had higher voltages available.

Reply to
Tom Biasi

b +12 in -+--------[7805]-- +5 | | | a | `---[7805]-+--- 0V | | gnd in -------+-------- -5 that will work, but only if the -5 half has a larger load than the +5 half, you might need to add something there (a resistor?) to make sure.

if you have at-least one of the DC-DC convertsers with an isolated output you can stack their outputs

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Reply to
Jasen Betts

A big enough load resistor from GND to -5V would fix that, at the expense of always burning up some power.

--
Tim Wescott 
Control system and signal processing consulting 
www.wescottdesign.com
Reply to
Tim Wescott

What advantages would that give over the stacked 7805 topology, It seems to me you'd have twice the hassle keeping the 0V where you want it than you have with the stacked 7805 topology,

With two 7805s where you can do something like this.

---------+-----+----------[7805]---------- +5 ^ | | | | |e [100] | | _\| | | Q1 |---+-[7805]---+-+---+----+---- 0V + /| | | | | 12V |c | [1K] | | - | | | |/ | | | +---| | | | | | |\| |/ | | | |/ e+--| | `--[1K]--------| | |\| | | |\| [1K] e| v | e| | | -------------------+------+-----+----+----- -5 Q2 Q3 Q4 Q1 eg: TIP42 Q2,Q3 eg: BC547, PN2222, or 2N2903 etc Q4 eg: TIP41

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Reply to
Jasen Betts

Possibly. I'd probably opt for a TL431 to provide the load balance... your approach smacks of instability. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

d

As others have said you've got issues with the current flow. I did something similar, splitting 48V to +30, GND, -5. But I still needed to know the direction of current flow into the 'Gnd' ter minal... in my case it was always a current sink.

Can you use LM317's and LM337's? Then I used the power supply negative lea d as -5V, used the 337 to make a 'ground' 5 volts above that, and the 317 f or the +30. You may be able to do something similar.. or if bi-directional current in the 'ground' then some sort of load resistor to the appropriate rail... (as others suggested.)

George H.

Reply to
George Herold

Or make the 12 V into 10 Volts, and then a power opamp as rail splitter. (I've got these TCA0372's in my parts box waiting for a project.) I guess it depends on what sorts of currents are involved.

George H.

Reply to
George Herold

On Mon, 04 Nov 2013 13:56:39 -0700, Jim Thompson wrote:

--- He said he had a bunch of wall-warts lying around, so my suggestion was to use two wall-warts to get plus 5 and minus 5V.

Current balance doesn't seem to be important:

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Reply to
John Fields

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