Generate -+5V from 12V

[vic]

Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply ?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

 

[BobW]

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...

Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply ?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

You're on the right track, but the 7805 cannot sink current. However, a 7905
can.

The problem you're going to have, with this technique of putting two
regulators in series, is that your output "0v" will not be controlled with
respect to the input supply, so I don't think it will work the way you've
drawn it (even with the bottom regulator being a 7905). You would have to
create a (semi) fixed 0v by either tying your 0v node to a beefy resistor
divider (between the 12v supply) or a weaker resistor divider with an
NPN/PNP follower.

Also, with this technique, you've got to be careful to not short your output
0v with the minus side of the input 12v supply. If you're in a car, for
example, you'll have to be careful not to touch your 0v circuit common with
the frame of the car.

The best and most efficient method is to use a switching regulator. It's
more expensive and complex, however.

Bob
--
== All google group posts are automatically deleted due to spam ==

 

Spend a few dollars and get something that will work without glitches:
http://cgi.ebay.com/DC-DC-Converter-Isolated-Power-Supply-In-9V-18V-Out-5V_W0QQitemZ350209167849QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item518a17d1e9&_trksid=p3286.c0.m14&_trkparms=65%3A12%7C66%3A2%7C39%3A1%7C72%3A570%7C240%3A1318%7C301%3A0%7C293%3A1%7C294%3A50

9-18 volts in, 5 volts 200 ma out, isolated grounds so you can use one
for +5 and another for -5

John


On Fri, 05 Jun 2009 15:37:13 +0200, vic <news@bidouille.org> wrote:

Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply ?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

 

[ian field]

"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:2JednZEzesurs7TXnZ2dnUVZ_hydnZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply ?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

You're on the right track, but the 7805 cannot sink current. However, a
7905 can.

The problem you're going to have, with this technique of putting two
regulators in series, is that your output "0v" will not be controlled with
respect to the input supply, so I don't think it will work the way you've
drawn it (even with the bottom regulator being a 7905). You would have to
create a (semi) fixed 0v by either tying your 0v node to a beefy resistor
divider (between the 12v supply) or a weaker resistor divider with an
NPN/PNP follower.

Also, with this technique, you've got to be careful to not short your
output 0v with the minus side of the input 12v supply. If you're in a car,
for example, you'll have to be careful not to touch your 0v circuit common
with the frame of the car.

The best and most efficient method is to use a switching regulator. It's
more expensive and complex, however.

To keep things simple a combination of the 2 might work - linear in the form
of a 7805 and switching in the form of a flying capacitor inverter, such as
a 555 driving a diode pump via a coupling capacitor - as long as the OP can
accept slightly less than 5V on the negative rail.

If not - run the 555 inverter from 12V and use a 79L05 for the correct
negative voltage.

 

[Baron]

ian field wrote:

"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:2JednZEzesurs7TXnZ2dnUVZ_hydnZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V
supply. I need about 150mA from the +5V side, and less than 50mA
from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current
instead of sourcing current, in the case of the 7805 in charge of
the 0V supply ?#

Supply + ----+--------[7805]---- +5 V
cap |
|------+------+------ 0v
cap |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way. Use a pair of 1000uf caps. Bearing
in mind that if the supply -ve is grounded the 0v will be +5 volts
above it.

--
Best Regards:
Baron.

 

[BobW]

"Baron" <baron.nospam@linuxmaniac.nospam.net> wrote in message
news:h0egf2$ae6$1@news.eternal-september.org...

ian field wrote:


"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:2JednZEzesurs7TXnZ2dnUVZ_hydnZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V
supply. I need about 150mA from the +5V side, and less than 50mA
from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current
instead of sourcing current, in the case of the 7805 in charge of
the 0V supply ?#

Supply + ----+--------[7805]---- +5 V
cap |
|------+------+------ 0v
cap |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way. Use a pair of 1000uf caps. Bearing
in mind that if the supply -ve is grounded the 0v will be +5 volts
above it.

--
Best Regards:
Baron.

The problem is that the all the current through the +5V load goes through
the load on the -5V. So, unless these currents are exactly equal, the
voltage from the node +5V to 0V will not be equal to the voltage from 0V
to -5V. That is, you really won't have a +-5V supply.

With this approach, you'll need some method of forcing "0V" to be exactly
(or very close to) six volts more positive than the minus side of the input
supply. See my earlier post.

Another concern with this approach is the dropout voltage on the 7805 and
7905. Assuming that you're able to keep 0V exactly six volts above the minus
side of the input supply (the ideal case), the dropout voltage (i.e., the
voltage from input to output on a linear regulator) will only be equal to
one volt. I'm not sure the 78xx and 78xx series can work with that low of a
dropout voltage at the currents you need. There are things called low
dropout regulators (LDO, for short) that allow much lower dropout voltages
(at a given load current).

Bob
--
== All google group posts are automatically deleted due to spam ==

 

[ian field]

"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:hsWdncmGe8nlVLfXnZ2dnUVZ_vSdnZ2d@giganews.com...

"Baron" <baron.nospam@linuxmaniac.nospam.net> wrote in message
news:h0egf2$ae6$1@news.eternal-september.org...
ian field wrote:


"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:2JednZEzesurs7TXnZ2dnUVZ_hydnZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V
supply. I need about 150mA from the +5V side, and less than 50mA
from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current
instead of sourcing current, in the case of the 7805 in charge of
the 0V supply ?#

Supply + ----+--------[7805]---- +5 V
cap |
|------+------+------ 0v
cap |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way. Use a pair of 1000uf caps. Bearing
in mind that if the supply -ve is grounded the 0v will be +5 volts
above it.

--
Best Regards:
Baron.

The problem is that the all the current through the +5V load goes through
the load on the -5V. So, unless these currents are exactly equal, the
voltage from the node +5V to 0V will not be equal to the voltage from 0V
to -5V. That is, you really won't have a +-5V supply.

With this approach, you'll need some method of forcing "0V" to be exactly
(or very close to) six volts more positive than the minus side of the
input supply. See my earlier post.

Another concern with this approach is the dropout voltage on the 7805 and
7905. Assuming that you're able to keep 0V exactly six volts above the
minus side of the input supply (the ideal case), the dropout voltage
(i.e., the voltage from input to output on a linear regulator) will only
be equal to one volt. I'm not sure the 78xx and 78xx series can work with
that low of a dropout voltage at the currents you need. There are things
called low dropout regulators (LDO, for short) that allow much lower
dropout voltages (at a given load current).

Bob

For an application powered by a mains transformer you should have a 2V
margin (7V) as an absolute minimum, a 3V margin (8V) to allow for peak load
dips.

For automotive applications you can recon on a voltage range of 10.8V to
14.4 but the terminal voltage of the battery can drop to 1V/cell during
cranking, its clear that even ignoring cranking voltage there is
insufficient headroom for a 7805 & 7905 in series unless the battery is
maintained at maximum charge at all times.

 

[whit3rd]

On Jun 5, 6:37 am, vic <n...@bidouille.org> wrote:

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

Well, 50 mA times 5V is only a quarter watt; you can use a 7805 with
the 'common' pin connected to a 5V zener, with a 50 mA source
(100 ohm resistor from 7805 OUT to 7805 COMMON) to bias the zener.
As long as you don't draw more than 50 mA from -5, it'll give you
the right voltages. All the positive-to-ground current goes
through the zener diode, so it'll have to be heatsinked for a watt or
so.

The most power-efficient way to do it is a DC/DC converter, but that
gets pricey.

 

[Baron]

BobW wrote:

"Baron" <baron.nospam@linuxmaniac.nospam.net> wrote in message
news:h0egf2$ae6$1@news.eternal-september.org...
ian field wrote:


"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:2JednZEzesurs7TXnZ2dnUVZ_hydnZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V
supply. I need about 150mA from the +5V side, and less than 50mA
from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current
instead of sourcing current, in the case of the 7805 in charge of
the 0V supply ?#

Supply + ----+--------[7805]---- +5 V
cap |
|------+------+------ 0v
cap |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way. Use a pair of 1000uf caps.
Bearing in mind that if the supply -ve is grounded the 0v will be +5
volts above it.

--
Best Regards:
Baron.

The problem is that the all the current through the +5V load goes
through the load on the -5V. So, unless these currents are exactly
equal, the voltage from the node +5V to 0V will not be equal to the
voltage from 0V to -5V. That is, you really won't have a +-5V supply.

With this approach, you'll need some method of forcing "0V" to be
exactly (or very close to) six volts more positive than the minus side
of the input supply. See my earlier post.

Another concern with this approach is the dropout voltage on the 7805
and 7905. Assuming that you're able to keep 0V exactly six volts above
the minus side of the input supply (the ideal case), the dropout
voltage (i.e., the voltage from input to output on a linear regulator)
will only be equal to one volt. I'm not sure the 78xx and 78xx series
can work with that low of a dropout voltage at the currents you need.
There are things called low dropout regulators (LDO, for short) that
allow much lower dropout voltages (at a given load current).

Bob

Doh ! Dropout voltage hadn't occurred to me. Those things need two
volts of headroom.
Thanks for pointing that out. Its something I should have known. :-(

--
Best Regards:
Baron.

 

[vic]

whit3rd wrote:

On Jun 5, 6:37 am, vic <n...@bidouille.org> wrote:
I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

Well, 50 mA times 5V is only a quarter watt; you can use a 7805 with
the 'common' pin connected to a 5V zener, with a 50 mA source
(100 ohm resistor from 7805 OUT to 7805 COMMON) to bias the zener.
As long as you don't draw more than 50 mA from -5, it'll give you
the right voltages. All the positive-to-ground current goes
through the zener diode, so it'll have to be heatsinked for a watt or
so.

The most power-efficient way to do it is a DC/DC converter, but that
gets pricey.

Thanks for this idea, I'll give it a try.

 

[Baron]

ian field wrote:

"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:hsWdncmGe8nlVLfXnZ2dnUVZ_vSdnZ2d@giganews.com...

"Baron" <baron.nospam@linuxmaniac.nospam.net> wrote in message
news:h0egf2$ae6$1@news.eternal-september.org...
ian field wrote:


"BobW" <nimby_GIMME_SOME_SPAM@roadrunner.com> wrote in message
news:2JednZEzesurs7TXnZ2dnUVZ_hydnZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V
supply. I need about 150mA from the +5V side, and less than 50mA
from the -5V side.

I've thought about using two 7805 in the following configuration
:


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current
instead of sourcing current, in the case of the 7805 in charge of
the 0V supply ?#

Supply + ----+--------[7805]---- +5 V
cap |
|------+------+------ 0v
cap |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way. Use a pair of 1000uf caps.
Bearing in mind that if the supply -ve is grounded the 0v will be +5
volts above it.

--
Best Regards:
Baron.

The problem is that the all the current through the +5V load goes
through the load on the -5V. So, unless these currents are exactly
equal, the voltage from the node +5V to 0V will not be equal to the
voltage from 0V to -5V. That is, you really won't have a +-5V supply.

With this approach, you'll need some method of forcing "0V" to be
exactly (or very close to) six volts more positive than the minus
side of the input supply. See my earlier post.

Another concern with this approach is the dropout voltage on the 7805
and 7905. Assuming that you're able to keep 0V exactly six volts
above the minus side of the input supply (the ideal case), the
dropout voltage (i.e., the voltage from input to output on a linear
regulator) will only be equal to one volt. I'm not sure the 78xx and
78xx series can work with that low of a dropout voltage at the
currents you need. There are things called low dropout regulators
(LDO, for short) that allow much lower dropout voltages (at a given
load current).

Bob

For an application powered by a mains transformer you should have a 2V
margin (7V) as an absolute minimum, a 3V margin (8V) to allow for peak
load dips.

For automotive applications you can recon on a voltage range of 10.8V
to 14.4 but the terminal voltage of the battery can drop to 1V/cell
during cranking, its clear that even ignoring cranking voltage there
is insufficient headroom for a 7805 & 7905 in series unless the
battery is maintained at maximum charge at all times.

Yes you are quite right ! I just didn't realise it until it was pointed
out to me.

--
Best Regards:
Baron.

 

[whit3rd]

On Jun 6, 12:36 pm, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:

Supply + ----+--------[7805]---- +5 V
          cap            |
           |------+------+------ 0v
          cap     |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way.

It's not possible to both sink and source current (up to 150 mA)
into the node labeled "0v". It might not even be possible to
get both regulators properly biased, the 'common' pin has a
few milliamps DC from the regulators, even with no l oad.

 

[Baron]

whit3rd wrote:

On Jun 6, 12:36 pm, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:


Supply + ----+--------[7805]---- +5 V
cap            |
|------+------+------ 0v
cap     |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way.

It's not possible to both sink and source current (up to 150 mA)
into the node labeled "0v". It might not even be possible to
get both regulators properly biased, the 'common' pin has a
few milliamps DC from the regulators, even with no load.

Giving this a bit more thought...
Take the regulator tails to the opposite rail. That would ensure that
each regulator had sufficient overhead.
Replace the caps with a pair of equal value resistors in series passing
say the max expected current. Using the centre point as the 0v.
Put the caps across the output.

Would that work ?

--
Best Regards:
Baron.

 

[ian field]

"vic" <news@bidouille.org> wrote in message
news:4a2aeb3e$0$5636$426a74cc@news.free.fr...

whit3rd wrote:
On Jun 5, 6:37 am, vic <n...@bidouille.org> wrote:
I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V
side.

Well, 50 mA times 5V is only a quarter watt; you can use a 7805 with
the 'common' pin connected to a 5V zener, with a 50 mA source
(100 ohm resistor from 7805 OUT to 7805 COMMON) to bias the zener.
As long as you don't draw more than 50 mA from -5, it'll give you
the right voltages. All the positive-to-ground current goes
through the zener diode, so it'll have to be heatsinked for a watt or
so.

The most power-efficient way to do it is a DC/DC converter, but that
gets pricey.


Thanks for this idea, I'll give it a try.

There are plenty of DC/DC chips on the market, both flying capacitor
inverters and one's that use an inductor but chances are you probably have a
555 more readily to hand in the junk box.

 

[Baron]

whit3rd wrote:

On Jun 7, 2:41 am, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:
whit3rd wrote:
On Jun 6, 12:36 pm, Baron <baron.nos...@linuxmaniac.nospam.net
wrote:

Supply + ----+--------[7805]---- +5 V
cap            |
|------+------+------ 0v
cap     |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way.

It's not possible to both sink and source current (up to 150 mA)
into the node labeled "0v".

Giving this a bit more thought...
Take the regulator tails to the opposite rail.  That would ensure
that each regulator had sufficient overhead.
Replace the caps with a pair of equal value resistors in series
passing say the max expected current.  Using the centre point as the
0v.

The "0v" point isn't low impedance, which makes it unregulated.

It's also a little elaborate, using the (+) and (-) rails as though
they were 'common', then not connecting except with power-wasting
resistors. Better might be to make (+) the common node, "+5".
Attach two negative regulators to make "0V" (that will be a -5V
regulator, "A"), and "-5V" (which is a -10V regulator, "B").

Then, knowing that the A regulator can sink any current required at
"0V" and current sourced out of the "0V" node can never be more than
50 mA net, a single resistor from "0V" to the (+) rail, 100 ohms,
finishes the circuit. You need that, because the A regulator cannot
supply positive current, being a negative regulator...

Yes I see how that works. One to stick in the notebook for reference.
Thankyou for taking the time to explain things.

--
Best Regards:
Baron.

 

[whit3rd]

On Jun 7, 2:41 am, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:

whit3rd wrote:
On Jun 6, 12:36 pm, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:

Supply + ----+--------[7805]---- +5 V
cap            |
|------+------+------ 0v
cap     |
Supply - ------[7905]------------ -5V

Whats wrong with doing it this way.

It's not possible to both sink and source current (up to 150 mA)
into the node labeled "0v".  

Giving this a bit more thought...
Take the regulator tails to the opposite rail.  That would ensure that
each regulator had sufficient overhead.
Replace the caps with a pair of equal value resistors in series passing
say the max expected current.  Using the centre point as the 0v.

The "0v" point isn't low impedance, which makes it unregulated.

It's also a little elaborate, using the (+) and (-) rails as though
they were 'common', then not connecting except with power-wasting
resistors. Better might be to make (+) the common node, "+5".
Attach two negative regulators to make "0V" (that will be a -5V
regulator, "A"),
and "-5V" (which is a -10V regulator, "B").

Then, knowing that the A regulator can sink any current required at
"0V"
and current sourced out of the "0V" node can never be more than
50 mA net, a single resistor from "0V" to the (+) rail, 100 ohms,
finishes the circuit. You need that, because the A regulator cannot
supply positive current, being a negative regulator...

 

[BobW]

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...

Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply ?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

See alt.binaries.schematics.electronic for a (linear) circuit that will work
up to about 200mA on each supply with a minimum input voltage of 10.6V.

Bob
--
== All google group posts are automatically deleted due to spam ==

 

[Jon Kirwan]

On Sun, 7 Jun 2009 17:10:26 -0700, "BobW"
<nimby_GIMME_SOME_SPAM@roadrunner.com> wrote:

"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply ?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

See alt.binaries.schematics.electronic for a (linear) circuit that will work
up to about 200mA on each supply with a minimum input voltage of 10.6V.

Sadly, Verizon (and other major service providers) have completely
eliminated all alt.* groups from their servers. This affects me, as
well. It may yet force me to get an extra-cost service added.

Jon

 

[BobW]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:trro25hqg7j3brjra016rfpi8f2kli1g1v@4ax.com...

On Sun, 7 Jun 2009 17:10:26 -0700, "BobW"
nimby_GIMME_SOME_SPAM@roadrunner.com> wrote:


"vic" <news@bidouille.org> wrote in message
news:4a291f7d$0$696$426a74cc@news.free.fr...
Hi,

I need to generate a symmetric +/-5V supply from a single +12V supply. I
need about 150mA from the +5V side, and less than 50mA from the -5V
side.

I've thought about using two 7805 in the following configuration :


Supply + ----+--------[7805]---- +5 V
| |
`-[7805]----+------ 0v
|
Supply - ---------+------------- -5V


does it make sense ? Would it be OK for the 7805 to sink current instead
of sourcing current, in the case of the 7805 in charge of the 0V supply
?

If this idea is garbage, what would be the recommended solution to this
problem ?

Thanks.

See alt.binaries.schematics.electronic for a (linear) circuit that will
work
up to about 200mA on each supply with a minimum input voltage of 10.6V.

Sadly, Verizon (and other major service providers) have completely
eliminated all alt.* groups from their servers. This affects me, as
well. It may yet force me to get an extra-cost service added.

Jon

I just tried posting the attachments here, too, but it doesn't look like
they made it through.

Giganews' cheapest rate is a couple of bucks a month. I haven't had a
problem with them.

Here are the LTSpice .asc file guts if you're interested. In fact, even if
you're not interested, here they are ->


Version 4
SHEET 1 920 680
WIRE -320 -192 -512 -192
WIRE -144 -192 -320 -192
WIRE 48 -192 -144 -192
WIRE 80 -192 48 -192
WIRE -512 -160 -512 -192
WIRE -320 -128 -320 -192
WIRE 480 -96 464 -96
WIRE 512 -96 480 -96
WIRE 672 -96 512 -96
WIRE 80 -80 80 -192
WIRE 144 -80 80 -80
WIRE 176 -80 144 -80
WIRE 48 -64 48 -192
WIRE -144 -48 -144 -192
WIRE 512 -48 512 -96
WIRE 480 -32 480 -96
WIRE 480 -32 464 -32
WIRE 672 -32 672 -96
WIRE 736 -32 672 -32
WIRE 832 -32 736 -32
WIRE 736 -16 736 -32
WIRE -320 0 -320 -48
WIRE -224 0 -320 0
WIRE -208 0 -224 0
WIRE 672 0 672 -32
WIRE 144 16 144 -80
WIRE 176 16 144 16
WIRE -320 32 -320 0
WIRE 512 32 512 16
WIRE 512 32 464 32
WIRE -80 48 -144 48
WIRE 48 48 48 0
WIRE 48 48 0 48
WIRE -320 96 -400 96
WIRE 48 96 48 48
WIRE 320 96 320 80
WIRE 320 96 48 96
WIRE 672 96 672 80
WIRE 736 96 736 48
WIRE 736 96 672 96
WIRE 752 96 736 96
WIRE 832 96 752 96
WIRE 48 112 48 96
WIRE 672 112 672 96
WIRE 672 112 48 112
WIRE -320 128 -320 96
WIRE 48 128 48 112
WIRE 96 128 48 128
WIRE 672 128 672 112
WIRE 752 144 752 96
WIRE -80 176 -144 176
WIRE 48 176 48 128
WIRE 48 176 0 176
WIRE -320 224 -320 192
WIRE -224 224 -320 224
WIRE -208 224 -224 224
WIRE 176 240 144 240
WIRE 512 240 464 240
WIRE 672 240 672 208
WIRE 672 240 512 240
WIRE 752 240 752 208
WIRE 752 240 672 240
WIRE 832 240 752 240
WIRE 48 256 48 176
WIRE 512 256 512 240
WIRE -320 272 -320 224
WIRE 144 336 144 240
WIRE 176 336 144 336
WIRE 512 336 512 320
WIRE 512 336 464 336
WIRE 96 432 96 128
WIRE 320 432 320 400
WIRE 320 432 96 432
WIRE -512 448 -512 -80
WIRE -320 448 -320 352
WIRE -320 448 -512 448
WIRE -144 448 -144 272
WIRE -144 448 -320 448
WIRE 48 448 48 320
WIRE 48 448 -144 448
WIRE 144 448 144 336
WIRE 144 448 48 448
WIRE -512 464 -512 448
FLAG -512 464 0
FLAG 832 -32 +Vout
FLAG 832 240 -Vout
FLAG 832 96 0Vout
FLAG -224 0 VD+
FLAG -224 224 VD-
FLAG -400 96 VDM
SYMBOL res 688 96 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R3
SYMATTR Value 30
SYMBOL res 688 224 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R4
SYMATTR Value 100
SYMBOL voltage -512 -176 R0
WINDOW 123 0 0 Left 0
WINDOW 39 3 132 Left 0
SYMATTR InstName V1
SYMATTR Value 12
SYMBOL diode -336 32 R0
WINDOW 3 45 31 Left 0
SYMATTR InstName D3
SYMATTR Value 1N914
SYMBOL diode -336 128 R0
WINDOW 3 41 31 Left 0
SYMATTR InstName D4
SYMATTR Value 1N914
SYMBOL pnp -208 272 M180
WINDOW 0 61 55 Left 0
WINDOW 3 62 23 Left 0
SYMATTR InstName Q3
SYMBOL npn -208 -48 R0
SYMATTR InstName Q4
SYMBOL res -336 -144 R0
SYMATTR InstName R5
SYMATTR Value 47
SYMATTR SpiceLine pwr=1
SYMBOL res -336 256 R0
SYMATTR InstName R6
SYMATTR Value 47
SYMATTR SpiceLine pwr=1
SYMBOL res 16 32 R90
WINDOW 0 59 58 VBottom 0
WINDOW 3 57 55 VTop 0
SYMATTR InstName R7
SYMATTR Value 0.1
SYMBOL res 16 160 R90
WINDOW 0 63 53 VBottom 0
WINDOW 3 56 51 VTop 0
SYMATTR InstName R8
SYMATTR Value 0.1
SYMBOL PowerProducts\\LT1964-5 320 288 R0
SYMATTR InstName U1
SYMBOL cap 496 256 R0
SYMATTR InstName C1
SYMATTR Value 0.1ľ
SYMBOL PowerProducts\\LT1962-5 320 -32 R0
SYMATTR InstName U2
SYMBOL cap 496 -48 R0
SYMATTR InstName C2
SYMATTR Value 0.1ľ
SYMBOL cap 32 256 R0
WINDOW 0 -38 9 Left 0
WINDOW 3 -47 59 Left 0
SYMATTR InstName C3
SYMATTR Value 100ľ
SYMBOL cap 32 -64 R0
WINDOW 0 -41 4 Left 0
WINDOW 3 -53 58 Left 0
SYMATTR InstName C4
SYMATTR Value 100ľ
SYMBOL cap 720 -16 R0
SYMATTR InstName C5
SYMATTR Value 100ľ
SYMBOL cap 736 144 R0
SYMATTR InstName C6
SYMATTR Value 100ľ
TEXT 144 -208 Left 0 !.op

Bob
--
== All google group posts are automatically deleted due to spam ==

 

[Jon Kirwan]

On Sun, 7 Jun 2009 19:14:46 -0700, "BobW"
<nimby_GIMME_SOME_SPAM@roadrunner.com> wrote:

snip of ltspice asc and pdf

Thanks, Bob. I got both postings just fine. Apparently, Verizon lets
through binaries on non-alt.* groups they support. Perhaps giganews
blocks binaries on groups marked as text? If so, that's yet another
consideration when I'm out looking for a service to pay for.

Things sure have evolved on internet.

Jon