Generate -+5V from 12V

  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
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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.

 ian field wrote:


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
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