Newbie Q: Make +/- supply from two wall warts?

[Lance]

I have a need for a quicky ± 12VDC power supply, about 100 mA max
current. I'll be using this supply as inputs to 7805 and 7905
regulators. In turn, the resultant ±5VDC is used to power an instrument
amplifier (AMP04 or AN623) and wheatstone bridge circuit.

I believe if I used two batteries I could connect the + terminal of one
battery to the - terminal of the other. These connected terminals become
my "common". The remaining terminals become my + and - source.

Can I do the same with two wall warts?

Can I connect the common to ground without catastrophe?

What's driving my questions is that we're a bunch of Civil and Mech
Engineers who don't know anything about this stuff. Currently we buy
instrument amplifiers from $300 to $1500 apiece that are very very nice,
but do 10x's what we really need. I've copied and modified some circuits
I found and built one myself that does exactly what we need for around
$25-$50 (not including our labor for assembly). It works great and I'm
proud of myself.

The downside is that my design requires a ± supply. ± supplies certainly
are not hard to find, but I'd like to use little wall warts at Newark
for $5 each that look like they could work (eg, p/n 95B0540).

Thanks for any comments or suggestions,

Lance
*****

 

[Andrew Holme]

I have a need for a quicky ± 12VDC power supply, about 100 mA max
current. I'll be using this supply as inputs to 7805 and 7905
regulators. In turn, the resultant ±5VDC is used to power an
instrument amplifier (AMP04 or AN623) and wheatstone bridge circuit.

I believe if I used two batteries I could connect the + terminal of
one battery to the - terminal of the other. These connected terminals
become my "common". The remaining terminals become my + and - source.

Can I do the same with two wall warts?

Can I connect the common to ground without catastrophe?

Yes, as long as the wall warts don't have an internal connection to ground.
Check them with your multimeter. I've just tested one here (UK) and it
doesn't - so it would be suitable for making a split supply.

You should include 100mA slow blow fuses in series with both inputs to make
catastrophe impossible.

 

[John Fields]

I have a need for a quicky ± 12VDC power supply, about 100 mA max
current. I'll be using this supply as inputs to 7805 and 7905
regulators. In turn, the resultant ±5VDC is used to power an instrument
amplifier (AMP04 or AN623) and wheatstone bridge circuit.

I believe if I used two batteries I could connect the + terminal of one
battery to the - terminal of the other. These connected terminals become
my "common". The remaining terminals become my + and - source.

Can I do the same with two wall warts?

 

[Richard Urwin]

I have a need for a quicky ± 12VDC power supply, about 100 mA max
current. I'll be using this supply as inputs to 7805 and 7905
regulators. In turn, the resultant ±5VDC is used to power an instrument
amplifier (AMP04 or AN623) and wheatstone bridge circuit.

Others have answered directly, but have you considered building your own
PSU? the circuit is simple and, at these currents, cheap, maybe not as
cheap as two wall-warts, but less messy and a lot less annoying when you
find there's only one socket available.

A six volt transformer (at least 1.2VA), pick one output as ground. Diodes
(1N4001 IIRC is well over-specified) from the other in opposite directions
to the +ve and -ve lines. Big electolytic capacitors from each to ground,
select on test (for ripple). That will give you +/- 8.48 volts; IIRC,
enough to drive 7805s.

For bigger current draws you would need a centre tapped transformer and
full-wave rectifiers, but half wave should be enough for 100mA.

 

[John Fields]

Others have answered directly, but have you considered building your own
PSU? the circuit is simple and, at these currents, cheap, maybe not as
cheap as two wall-warts, but less messy and a lot less annoying when you
find there's only one socket available.

A six volt transformer (at least 1.2VA), pick one output as ground. Diodes
(1N4001 IIRC is well over-specified) from the other in opposite directions
to the +ve and -ve lines. Big electolytic capacitors from each to ground,
select on test (for ripple). That will give you +/- 8.48 volts; IIRC,
enough to drive 7805s.

For bigger current draws you would need a centre tapped transformer and
full-wave rectifiers, but half wave should be enough for 100mA.

---
Dropout voltage for a 7805 is 2.5V, and for the 7905 1.1V, so the 7805
would need to see > 7.5V on its input, and the 7905, < -6.5V.

For a transformer capable of putting out 6VRMS with a 100mA load
across it and a half wave rectifier in series with it, the peak output
voltage from the rectifier will be

Vp = (VRMS * sqrt2) - Vf = (6.0 * 1.414) -0.7 = 7.78V.

Since the minimum voltage allowed on the 7805's input is 7.5V, that
means that the ripple on the input can't exceed 7.78V - 7.5V = 0.28V.

The capacitance of the capacitor needed to keep the ripple below 0.28V
will be

Idt 0.1A * 0.0167s
C = ----- = ---------------- = 0.00596F ~ 6000µF
dV 0.28V

Not bad...

Since most electrolytics run from about +/- 20% to +80/-20% the cap
will need to be specified as at least 7500µF to make sure that it
won't be less than 6000µF. Getting 10000µF will assure at least
8000µF, and keep the ripple down to

Idt 0.00167
dV = ----- = --------- = 0.167V
C 0.01

Panasonic ECA-1AM103, Radial Aluminum Electrolytic, 10000µf +/- 20%,
10V, DigiKey, $2.27 ea. qty 1.

 

[James Meyer]

I have a need for a quicky ± 12VDC power supply, about 100 mA max
current. I'll be using this supply as inputs to 7805 and 7905
regulators. In turn, the resultant ±5VDC is used to power an instrument
amplifier (AMP04 or AN623) and wheatstone bridge circuit.

I believe if I used two batteries I could connect the + terminal of one
battery to the - terminal of the other. These connected terminals become
my "common". The remaining terminals become my + and - source.

Can I do the same with two wall warts?

Thanks for any comments or suggestions,

Lance
*****

You *could* use two wall warts. If I were doing the job, I'd use one
wart to power both 5 volt regulators. Put the regulator circuits in series,
power them both from one wart with a voltage output around 15 to 20 volts. The
plus of the wart would go to the input of the plus regulator and the minus to
the input of the minus regulator. The common connection between the two
regulators would then be the "ground" for the amplifier.

You might want to investigate "low dropout" versions of the 7805 and
7905. Then you could probably use a single 12 volt wart. You don't need a
regulated wart either. Unregulated warts tend to run a little on the high side
for voltage out and that would be an advantage too.

If the amplifier requires different currents from the plus and minus
supplies (not very likely) then you should use a dummy load resistor on the low
current side to bring the current flow through that side up to something close
to the other side.

Jim

 

[Jeff]

Use a wall wart that has AC output. Connect one of the output wires to
ground, and the other to 2 diodes, one for the negative voltage, and one for
the positive voltage. Filter the outputs with caps. The wall wart should
have a voltage output of: (5V + ~3V (reg overhead (7905 is less)) + 0.7V
(Diode Drop) + ripple current drop and line regulation (I'll allow ~2V))
/1.414 (sqrt 2) = 7.6V AC Although a bit wasteful, 9V AC wall warts are
commonly available. Since you will be pulling 100 mA from both supplies, you
will need a AC source of at least 200 mA. Add in some head room, especially
since the filter caps will create a poor power factor and since they will be
sucking large chunks of current when the diodes start conducting. 300mA
should be enough. Digikey sells a 600mA 9V AC wall wart (part number:
T702-ND) for $5-6. The diodes can be plain old 1N4002's to 1N4007's. The
filter caps should be rated for at least: ((9V AC * 1.414 (sqrt 2)) - 0.7V
(diode drop)) * 1.20 (to account for poor transformer load regulation
typical in mid size wall warts) = 14.4V DC. Since you likely want the caps
to last a while, use long life, high temp (105 deg C +) capacitors rated for
a least 25V. 35 V or 50 V would be better, especially if you want this thing
to have a really long operational life. Since the rectification is half
wave, the cap value needs to be significantly bigger then with full wave
rectification. Assuming you have a 60Hz power supply, it will need to be at
least: 1/60 Hz = 0.0166666 s (neglection the time at the peak), (9V AC *
1.414) - 0.7V = 12V. The regulator needs about 8V, so we have 4V to play
with. I = 0.1A. I =C (dV/dt), rearranging gives C = I/(dV/dt). C= 0.1 /
(4V/0.0166666) = 416.6uF. Since you want lots of capacitance to carry the
regulator over during a quick brown out, spike/surge, noise, over current
condition on the load, etc, at least triple this value. Digikey's Nichicon
35V 1800uF (part number UHE1V182MHD) cap would be a good choice at a little
over $1 ea. Don't forget to put the anti oscillation ~0.1uF ceramic caps as
close as possible on the inputs and outputs of the regulators. The heat
released in the regulatiors at full load will be about (12V - 5V) * 0.1 A =
0.7W each. About enough to use a small heat sink, such as a clip on unit.

 

[John Fields]

Use a wall wart that has AC output. Connect one of the output wires to
ground, and the other to 2 diodes, one for the negative voltage, and one for
the positive voltage. Filter the outputs with caps. The wall wart should
have a voltage output of: (5V + ~3V (reg overhead (7905 is less)) + 0.7V
(Diode Drop) + ripple current drop and line regulation (I'll allow ~2V))
/1.414 (sqrt 2) = 7.6V AC Although a bit wasteful, 9V AC wall warts are
commonly available. Since you will be pulling 100 mA from both supplies, you
will need a AC source of at least 200 mA.

 

[Fred Bloggs]

I have a need for a quicky ± 12VDC power supply, about 100 mA max
current. I'll be using this supply as inputs to 7805 and 7905
regulators. In turn, the resultant ±5VDC is used to power an instrument
amplifier (AMP04 or AN623) and wheatstone bridge circuit.

Power supply design encompasses much more than just getting the numbers
right. There are other things like safety, conducted and radiated
emissions, electro-magnetic susceptibility, prime power tolerance,
brownout tolerance, isolation, line transient protection, heat
dissipation, flammability, reliability, fault tolerance, OVP and
over-current protection, among others. You have no clue. Look at buying
a +/-5VDC linear supply, fully enclosed- this should be under $50 for
250mA.

 

[Tom Seim]

Power supply design encompasses much more than just getting the numbers
right. There are other things like safety, conducted and radiated
emissions, electro-magnetic susceptibility, prime power tolerance,
brownout tolerance, isolation, line transient protection, heat
dissipation, flammability, reliability, fault tolerance, OVP and
over-current protection, among others. You have no clue. Look at buying
a +/-5VDC linear supply, fully enclosed- this should be under $50 for
250mA.

FREDFRAUD, VIETNAM VET IMPOSTER - Need I say any more?

 

[Lance]

I'd like to thank everyone for their comments and some really intriguing
ideas on producing +/- from + supplies.

I've taken a close look at some battery-powered amplifiers we have
(really great amp with lots of goodies - $1500). I see they use an
ICL7660 chip to make -5 VDC from a positive supply. Since the ICL7660
can't supply a lot of current, the chip is used only for the op-amps
(along with +5 VDC) to get a bipolar output. Regulated +5 VDC from the
batteries is used to drive the wheatstone bridge.

Today, someone at work asked about thermocouple amplifiers because the
commercial ones were too expensive. I found the AD594/5 from Analog
Devices - built-in ice point, linearization and amplification ($10-20).

Once I get a workable design for an inexpensive power supply, it seems
there's no end to the stuff you can do with all these chips floating around.

Thanks again,

Lance
*****

Lance thought carefully and wrote on 10/21/2004 12:11 PM:

 

[Fred Bloggs]

FREDFRAUD, VIETNAM VET IMPOSTER - Need I say any more?

Come clean sissy-how can you not rise to this challenge? You know so
much about what these recent high school graduate kids should do to
defend your greedy, corrupt, and criminally materialistic way of life.
Now please tell us what legacy you left for them- tell us what you did
to secure their world.

 

[Rich Grise]

I'd like to thank everyone for their comments and some really intriguing
ideas on producing +/- from + supplies.

It's been our great pleasure.

I've taken a close look at some battery-powered amplifiers we have (really
great amp with lots of goodies - $1500).

Holy Crap! That's more money than I've ever seen in one place at one time!

I see they use an ICL7660 chip to
make -5 VDC from a positive supply. Since the ICL7660 can't supply a lot
of current, the chip is used only for the op-amps (along with +5 VDC) to
get a bipolar output. Regulated +5 VDC from the batteries is used to drive
the wheatstone bridge.
OK.

Today, someone at work asked about thermocouple amplifiers because the
commercial ones were too expensive. I found the AD594/5 from Analog
Devices - built-in ice point, linearization and amplification ($10-20).

Once I get a workable design for an inexpensive power supply, it seems
there's no end to the stuff you can do with all these chips floating
around.

You have a $1500.00 part, and are worried about power supply budgets? You
could find that kind of power supply lying in the street! - Well, I'm
trying to make a point - a power supply isn't that big a deal, if you
really know what you're looking for.

Good Luck!
Rich

 

[Rich Grise]

Come clean sissy-how can you not rise to this challenge? You know so much
about what these recent high school graduate kids should do to defend your
greedy, corrupt, and criminally materialistic way of life. Now please tell
us what legacy you left for them- tell us what you did to secure their
world.

Freed, you won.

Back off, and let him know that you accept him just the way he is, right
now, and all that hatefulness was never supposed to manifest in the first
place, and he can have his unloving light redeemed, by just letting it go.

What he needs now more than ever is our love.

Heal!
Rich

 

[John Fields]

I'd like to thank everyone for their comments and some really intriguing
ideas on producing +/- from + supplies.

I've taken a close look at some battery-powered amplifiers we have
(really great amp with lots of goodies - $1500). I see they use an
ICL7660 chip to make -5 VDC from a positive supply. Since the ICL7660
can't supply a lot of current, the chip is used only for the op-amps
(along with +5 VDC) to get a bipolar output. Regulated +5 VDC from the
batteries is used to drive the wheatstone bridge.

Today, someone at work asked about thermocouple amplifiers because the
commercial ones were too expensive. I found the AD594/5 from Analog
Devices - built-in ice point, linearization and amplification ($10-20).

Once I get a workable design for an inexpensive power supply, it seems
there's no end to the stuff you can do with all these chips floating around.

 

[Lance]

Rich Grise thought carefully and wrote on 10/22/2004 11:09 PM:

You have a $1500.00 part, and are worried about power supply budgets? You
could find that kind of power supply lying in the street! - Well, I'm
trying to make a point - a power supply isn't that big a deal, if you
really know what you're looking for.

Good Luck!
Rich

Thanks Rich.

The PS budget does sound strange doesn't it?

I actually work in a lab at a university. The cheap power supplies and
amps would be for doctoral engineering students working on their
experiments. I don't want to lend my $1500 amp to an inexperienced
student and they tend to want to keep stuff for 5-7 years until they finish.

Plus they have to justify expenses (go begging) to their professor for
every single penny. This'll make it a lot easier for them.

Lance
*****

 

[Lance]

John Fields thought carefully and wrote on 10/23/2004 8:51 AM:
I dunno. Is it better than sex?

Lance
*****

 

[John Fields]

John Fields thought carefully and wrote on 10/23/2004 8:51 AM:

I dunno. Is it better than sex?

 

[Rich Grise]

I think it's satisfying to the intellect like sex is satisfying to the
inner satyr. ;-)

Cheers!
Rich

 

[Rich Grise]

.

The PS budget does sound strange doesn't it? .....
Plus they have to justify expenses (go begging) to their professor for
every single penny. This'll make it a lot easier for them.

I'm wondering if anything like this would be of any use to you -
maybe you could have them get their own kits and build them as an
assignment. ;-)

Or let the kids come up with any kind of power source they want, that
works and doesn't kill anybody or burn the building down. ;-)

Would something like that be appropriate as a group project? Or maybe
they could chip in on a decent lab supply.

Cheers!
Rich