# Need a negative voltage

Discussion in 'Electronic Basics' started by [email protected], Oct 4, 2005.

1. ### Guest

In an attempt to repair a presumably faulty preamp I realised the
construction deals with two voltages; +14 and -14 (as well as ground).
I don't have access to the transformer unit, so thought I could
construct my own.

Am I correct in my assumption that what I need is a 28 volt power
source, and to connect two resistors in series across the two leads,
and I'll end up with +14 volts on the positive lead, -14 on the
negative, and a 0 volt "ground" where the two resistors join?
Kinda like this (hoping the ASCII-schematic looks like it's meant to)..

--------- +14
|
__|__
| |
| R1 |
|____|
|
|---- 0/"ground"
__|__
| |
| R2 |
|____|
|
|
--------- -14

So how do I calculate the resistors? I need less than 1 A (probably
much less as 1 A is what the fuses in the amplifier are rated at).

Cheers!

2. ### Jonathan KirwanGuest

Probably not a good idea.

To get the zero volts, you'd need R1=R2. The thevenin equivalent
would then be a 0V source through an R of (R1)/2. To keep the
variation of your "ground" to, let's say, 1/2V about ground at up to
one amp you'd need to have (R1)/2*(1A) = .5V. This means R1 of 1 ohm.

So, R1 and R2 would each need to be at least able to handle 200 watts.
Just in case, make them 1 kW resistors. Your 28V supply should need
to support about 15 amps. About 500 watts worth. This would then
supply your up-to-1A pre-amplifier, assuming it could tolerate the .5V
variations in the ground.

You will probably want to avoid that solution, though.

Jon

3. ### PeteSGuest

Using a resistor pair is not a particularly good idea for any
significant current (significant being more than a few milliamps)

A better solution might be to use about 36V in total across a centre
tapped transformer. i.e. each side relative to the centre tap has 18V
above the ripple after rectification and *under load*. Use the centre
tap as your local common (ground).

Then use Linear regulators (if you are less than 1A and you stay close
to the drop out - about 2.5V each - then the heat sinking won't be too

I know the drop out is a little highly spec'd here, but that's standard
for the old style 78xx beasts (and various others). A low dropout
regulator has it's own very special problems that will bite you in the
behind.

Cheers

PeteS

4. ### Bob MonsenGuest

Using your idea, you'll end up with something that looks like this:

----o-----------o------ +14
| |
__|__ __|__
| | | |
| R1 | | L1 |
|____| |____|
| |
o-----------o------ 'gnd'
__|__ __|__
| | | |
| R2 | | L2 |
|____| |____|
| |
| |
----o-----------o---- -14

Where L1 and L2 are indeterminate impedances. GND will only be at 0V
in your diagram when L1=L2, which you probably can't guarantee.

If you use very low value resistors for R1 and R2, then you can be
sure that those values will cause L1 and L2 to be large in comparison.

However, a better way to go would be to use a couple of transistors,
like this:

----o---------o-----o------ +14
| | |
__|__ | __|__
| | |/ | |
| R1 | .-| NPN | L1 |
|____| | |> |____|
| | | |
o-----o o-----o------ 'gnd'
__|__ | | __|__
| | | | | |
| R2 | | |> | L2 |
|____| '-| PNP |____|
| |\ |
| | |
----o---------o-----o---- -14

If you do this, it'll keep GND pretty much at 0V, even if the
impedances of L1 and L2 are not equal.

Note that the current through R1 and R2 should be about 1/10 of what
the maximum current through either transistor is. This will depend on
how different L1 and L2 are. If L1 is not there, for example (an
infinite impedance) and L2 is 14 ohms, then it'll draw 1A through
the NPN transistor, and no current through the PNP transistor. So,
you'll need about 100mA through R1/R2, meaning R1 and R2 should be
140 ohm 2W resistors. The Transistors should be beefy, since they may be
dissipating 14W. Big TO-3 cases and heatsinks are probably a good idea.

If you use an opamp to drive the transistors, with feedback from the
virtual ground, it'll be far more precise. Something like this:

----o------------------------------o------o------ +14
| | |
__|__ | __|__
| | |/ | |
| R1 | .-----| NPN | L1 |
|____| | | |> |____|
| |\ | | |
o--------+| \ | o------o-----------.
__|__ | \ | | __|__ |
| | | >--------o | | | |
| R2 | .-o| / | |> | L2 | |
|____| | | / '-----| PNP |____| |
| | |/ |\ | |
| | | | |
----o------)-----------------------o------o---- -14 |
| Rf |
'---------/\/\/\/--------------------------'

With this circuit, your virtual ground would be far more accurate, and
the resistors R1 and R2 can be large, like 100k. Make R1 = R2, and Rf
= R1/2. That way, your opamp inputs will see similar impedance.

An even better way would be to just generate the right voltages to
begin with. You can get 'center tapped' transformers, which provide AC
voltages which are mirror images of each other, and a ground. If you
use one of these, you can generate +-14V fairly easily.

For example, assuming a 1A current draw on either or both, you'll need
a 28VAC transformer that is rated for something like 35 VA. Using
this, use the following circuit:

.---------.
| o-----|>|----o--------- 19 max
| | |
| | --- + C
| | ---
| | |
| | |
---o + | |
| | |
| o------------o--------- gnd
---o - | |
| | |
| | --- + C
| | ---
| | |
| | |
| | |
| o----|<|-----o--------- -19 max
'---------'

The unregulated output goes into linear regulators, maybe LM317 and
LM337 linear regulators, suitably configured. (see the datasheet at
www.national.com)

These regulators require the supply to be about 2V above or below,
respectively. Thus, the supply must stay above 16V. That means that
the droop across the smoothing capacitor must be less than 3V in 1/60
of a second. So, using the famous formula:

1A = C * 3*60

we compute that each C needs to be at least 5600uF, and rated for at
least 25V. Using a transformer that puts out more volts will allow a
smaller C, at the cost of more dissipation in the regulator. You are
already looking at an average power of 3.5W, which is pushing an LM317
unless it is heatsinked.

This supply will waste far less energy than the one above, because it
will not have to dissipate energy in the case where the positive load
is more than the negative load.

You can also buy cheap +- 15V supplies. I've seen them for \$10
mailorder from www.mpja.com. Here is one for \$15 that might work for
you:

http://www.mpja.com/productview.asp?product=15695+PS