# ±5V from up to +12v

D

#### Daniel Pitts

Jan 1, 1970
0
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.

T

#### Tom Biasi

Jan 1, 1970
0
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.
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.

D

#### Daniel Pitts

Jan 1, 1970
0
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.

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

T

#### Tom Biasi

Jan 1, 1970
0
The VoutA would be -7 with respect to the +12v, so the VinB would be at
a at the right place respective the +12v.
Hook it up and get back to us.

D

#### Daniel Pitts

Jan 1, 1970
0
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?
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.

D

#### Daniel Pitts

Jan 1, 1970
0
Huh? Didn't Daniel imply he had a single +12VDC adapter as input?
Currents have to balance.

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.

T

#### Tom Biasi

Jan 1, 1970
0
Turns out, +12V input is insufficient, headroom-wise, to power a
back-to-back combination of 7805/7905.

MOSFET's as the pass devices.

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

J

#### Jasen Betts

Jan 1, 1970
0
On 11/4/13 12:16 PM, Jim Thompson wrote:
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.
Is there a better approach than this? One or more DC to DC converters
perhaps? I'm open to alternative suggestions as well.

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

J

#### Jasen Betts

Jan 1, 1970
0
Turns out, +12V input is insufficient, headroom-wise, to power a
back-to-back combination of 7805/7905.

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

G

#### George Herold

Jan 1, 1970
0
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.

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' terminal... in my case it was always a current sink.

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

George H.

G

#### George Herold

Jan 1, 1970
0
Possibly. I'd probably opt for a TL431 to provide the load balance...

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.

P

#### Phil Hobbs

Jan 1, 1970
0
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.

Careful with those--they work great, except that the heat sinking is
quite a bit less great. The best package runs 22 K/W junction to
"case", whatever that means in a plastic-encapsulated package with no
power tab. That's quite a bit considering that the chip can dissipate
over 40W before thermal shutdown kicks in. (Steady-state thermal
transfer calculations would put the die at about 900C with an infinite
room-temperature heat sink--a nice bright red-orange glow.)

They run warm to the touch just with their quiescent current.

As a rail splitter, you have to watch out for the chip hitting thermal
shutdown, which will probably make one rail collapse. A couple of beefy
Zeners on the output would help prevent Joergish noises.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

G

#### George Herold

Jan 1, 1970
0
Careful with those--they work great, except that the heat sinking is
quite a bit less great. The best package runs 22 K/W junction to
"case", whatever that means in a plastic-encapsulated package with no
power tab. That's quite a bit considering that the chip can dissipate
over 40W before thermal shutdown kicks in. (Steady-state thermal
transfer calculations would put the die at about 900C with an infinite
room-temperature heat sink--a nice bright red-orange glow.)

They run warm to the touch just with their quiescent current.

Grin, Hey, the TCA0372's are burning a hole in my parts bin :^)
(maybe the OP only needs ~100mA)
I must admit when you (or someone else?) pointed out these 1 amp IC's in a dip8 package I was a bit 'concerned'.
(I'd really like a nice to-220 style metal tab.. or a surface mount with 'power pads' on the bottom.)

George H.

I

#### Ian Malcolm

Jan 1, 1970
0
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.

The other replies have covered in some detail the disadvantages of
starting with +12V DC in.

Do you have any adapters with AC output between 8V and 12V (RMS)?

If so, a much better approach would be to half-wave rectify the AC with
a diode and electrolytic reservoir capacitor for each supply to get
positive and negative unregulated supplies with a common 0V, then using
7805 and 7905 regulators respectively to get your regulated +/-5V
output.

A 8 or 9V AC supply is ideal for +/-5V output, but higer voltages can be
used although the regulators will run hotter. Component choices depend
on input voltage, load current on each output and mains supply
frequency, so tell us what AC output supplies you have available
(nominal voltage and current + measured output voltage unloaded) and the
required output currents so we can help you pick the best one.

If you dont have any AC output adapters, the use of a DC-DC converter
module to get a negative rail has already been mentioned. There are a
few issues with ripple on the output, and potentially with power supply
sequencing so you had better tell us which volume control chip you
intend to use.

D

#### Daniel Pitts

Jan 1, 1970
0
---
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:

Version 4
SHEET 1 880 936
WIRE -2000 -16 -2144 -16
[snip]

Is there a program to interpret that script? Hopefully one that runs on
Mac, DOSBOX, or the web?

Thanks,
Daniel.

P

#### petrus bitbyter

Jan 1, 1970
0
Daniel Pitts said:
---
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:

Version 4
SHEET 1 880 936
WIRE -2000 -16 -2144 -16
[snip]

Is there a program to interpret that script? Hopefully one that runs on
Mac, DOSBOX, or the web?

Thanks,
Daniel.

petrus bitbyter

D

#### Daniel Pitts

Jan 1, 1970
0
Thanks! I'll see if this is in my capability to understand and build,
but it looks nice at first glance.

D

#### Daniel Pitts

Jan 1, 1970
0
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.
Thanks for all the suggestions everyone. I see how my original thoughts
were flawed.

I also realized I should probably get a better idea of what my amperage
will be. I'm guessing <100ma, but I'm not entirely sure.

I'm also considering using just a pair of digital POTs instead of this
fancy schmancy volume control IC. That way I only need +5v, which is a
much easier problem for me to solve ;-)

Thanks again,
Daniel.

K

#### Kaz Kylheku

Jan 1, 1970
0
I'm trying to design a remote controlled volume controller, and the chip
I've chosen to control the volume needs Â±5v for the analog side of the
circuit, and +5v for the digital.

Idea:

* Use a 5V (unregulated) adapter. Regulate to 5V with a 7805.

* Use a charge pump IC to generate a -5V rail from the regulated +5V rail.
I have several options for input voltage, mostly based on old AC

You can get a center-tapped transformer for a couple of dollars, or rip it out
of something. You can build your own dual-voltage AC adapter easily with a
center tapped transformer and bridge rectifier or discrete diodes. Your device
then just has to provide the reservoir capacitors and regulators.

Why not find some discarded old device (e.g. piece of audio gear) which has
a dual supply in it already and re-use its power entry and perhaps more.
Throw out the circuit board and build your own device in its place.

You can also use two identical AC adapters to get the equivalent of a
center-tapped transformer, if you can live with the clunkiness.

S
Replies
4
Views
2K
Shahid Sheikh
S
Replies
16
Views
166
Replies
17
Views
274
Replies
10
Views
182
Replies
10
Views
309