Symmetric power supply from 0-15V Power supply

[MegaWatt]

Hello everybody,

I am making a project, and I have a chip that has a GND pin, a +5V and a -5V pin. Now I only have a 0-15V DC power supply. This is probably an easy to solve question, but how can I get these voltages from a single 0-15V DC power supply?

Thank you! 

[Simon]

well actually it's not so easy to solve.

What chip is it, what does it do and how much current does it draw, what is the total current draw of the circuit.

You could use two regulator circuits, one at 5V and one at 10V but the one at 5V much also sink current so would be tricky, maybe use a power opamp.

[MegaWatt]

Hello,

the chip is use is the MAX038 function generator. What would be the easiest solution to solve this? Do transformators exist who can give me these voltages?

Thank you

[Zero999]

It's actually pretty easy to solve.

Set the voltage to 10V and use a virtual earth circuit consisting of an op-amp and a couple of transistors.

[Simon]

well a transformer needs drive circuitry, how much of your circuit needs 5-0-5 volts ? what current do you need at these voltages ?
you might get away with a potential divider but that is a very wasteful method and you would need to somehow level shift the output for the rest of the circuit if you required any amount of power

[Simon]

It's actually pretty easy to solve.

Set the voltage to 10V and use a virtual earth circuit consisting of an op-amp and a couple of transistors.

yea you hit the nail on the head, of course silly me forgot you can buffer an opamp with transistors although essentially its a power opamp and of course the OP have 0-15 available to can fix at 10V

[MegaWatt]

Ah yes thank you Hero and Simon! 
I will definatelly look more on the net about this circuit and use it.
BTW Simon: what is a power op amp 

Greetings!

[Simon]

a power opamp is an opamp that can output a fair amount of power, usualy they are used to drive speakers directly, they are generally not as refined as the small opamps used for signal conditioning but that's not a problem as they are meant to supply a number of amps to a speaker so really are power amplifiers that are setup the same way as opamps (feedback etc), typical opamps can only output a few hundred mA if that

[MegaWatt]

So do you think I need a power op amp for my virtual ground circuit? Or does it depend on how much current i draw to ground, and I should calculate it first?

EDIT: I think I already know the anwser, no I do not need a power op amp 

[Simon]

I'd use the opamp with buffer transistors, they will probably supply a smoother current as they will react faster than a power opamp made for audio frequencies. the cost of the two solutions will be about the same or the power opamp will cost more anyhow

[MegaWatt]

Ok, thank you for all the help, it was really helpfull.
I'm going to see if I can build this thing!
Greetings and a happy newyear

[Simon]

happy new year and make sure (although not essential) you select a complementary (similar) pair of transistors, mosfets can also be used

[Zero999]

Your circuit hardly uses any power so the circuit I posted a previously will be fine. You don't need a high powered op-amp, a crappy old LM741, a LM358 or a TL082 will be fine. Most low powered transistors will do but I'd recommend the BC328 and BC338 because they have a high gain and are cheap. They have maximum power dissipation of 625mW so the maximum continuous current into 0V circuit is 0.625/5 = 125mA and your circuit uses less than that and don't forget that it's the current into the 0V that counts so if your circuit takes 200mA from both the +5V and -5V no current will flow in or out of 0V.

http://www.datasheetcatalog.org/datasheet/GeneralSemiconductor/mXqxsxy.pdf
http://www.datasheetcatalog.org/datasheet/GeneralSemiconductor/mXtwuqt.pdf

[Chasm]

One quick reminder:

This kind of circuit provides a virtual ground at half the input voltage.
It does not limit the output voltage to 5V.  


Here is a similar circuit, this time using the LM386 power opamp.

The potentiometer is used to change the symmetry of the output. If you do not need this you could use two Resistors, say 10k, instead.
Some 100nF caps like in the circuit Hero999 posted are missing and would help to kepp the voltages stable.
If you take a look at the datasheet of the LM368 you'll notice that the circuits are the same.

[MegaWatt]

Thank you Hero, I was just looking for some transistors, to buy a couple so I could use them in some future projects. Just some basic transistor (i'm just a beginner in electronics, and am just prototyping) so I guess the BC328 and BC338 are a good choice for overall doing projects with? And I have a bunch of LM324 op-amps, are they any good?

Hello Chasm, thank you for the circuit, what do you exactly mean with "It does not limit the output voltage to 5V." ?

[Simon]

if you look at the diagram, assuming the 12V supply showed there is a 50/50 potential divider supplying the voltage reference, so 6V will be the output.

if your using a stable 10V supply you will get 5V output, if you want the circuit to be immune to supply voltage then replace the lower resistor with a 5V zenner diode and set the top resistor value so that 5-10 mA flows through the zenner (470-1000R)

[Zero999]

Thank you Hero, I was just looking for some transistors, to buy a couple so I could use them in some future projects. Just some basic transistor (i'm just a beginner in electronics, and am just prototyping) so I guess the BC328 and BC338 are a good choice for overall doing projects with? And I have a bunch of LM324 op-amps, are they any good?

Yes, the LM324 is fine and those transistors are good al-rounders

Helle Chasm, thank you for the circuit, what do you exactly mean with "It does not limit the output voltage to 5V." ?

I think he means to say that it's not a voltage regulator, the output voltage will be equal to the input divided by two.

Anyway, he read my mind, I was thinking about suggesting the LM386 which biases its output voltage to half the supply.

If you want the circuit to be immune to supply voltage then replace the lower resistor with a 5V zenner diode and set the top resistor value so that 5-10 mA flows through the zenner (470-1000R)

That won't work because it'll only regulate the negative voltage. With a 5V zener and a 15V supply, the negative will be 5V and the positive will be 10V. The best way to make it regulate the voltage is to power it from a regulator such as the LM317.

[Simon]

I don't understand, if you supply 5V to the opamp input it will maintain the output at 5V regardless of how you obtain the reference ?

[Zero999]

Feeding the input with 5V will hold the output (0V) 5V relative to the negative rail so the positive will be equal to the power supply voltage minus 5V. In order to regulate the circuit the power supply voltage needs to be regulated.

[Simon]

oh I see your point yes well we are assuming the OP is using a 0-15V supply set at 10V, but yes for the complete setup to be supply independent this stage needs feeding with an already stable supply

[Chasm]

Helle Chasm, thank you for the circuit, what do you exactly mean with "It does not limit the output voltage to 5V." ?

This kind of circuit provides a virtual ground at half the input voltage.
(Or the point you selected, if you use other resistor values or a potentiometer.)

There are some limits:
The Input voltage must stay within the operational limits of the opamp.
The current is limited by the output stage of the the power opamp, or the transistors if you use a normal opamp.
Say up to 250mA for the examples posted, extending this much more usually does not make too much sense.


So the caveat is that you have to watch the input voltage, or you can easily damage the device you are powering.
Because you need a fixed voltage next step is to add some form of voltage regulation.

Two zener diodes (at least 1W!) in the output in the have been mentioned. But getting this right is actually a bit more complex.
The simplest solution is to use a pair of linear voltage regulators, 7805 and 7905. They are cheap, about 50ct at the local electronics store for the TO-220 version.
There are of course also many other (and better) linear regulators. But they are not used in abundant quantities and thus usually more expensive and harder to get.

[Simon]

The simplest solution is to use a pair of linear voltage regulators, 7805 and 7905. They are cheap, about 50ct at the local electronics store for the TO-220 version.
There are of course also many other (and better) linear regulators. But they are not used in abundant quantities and thus usually more expensive and harder to get.

WRONG! the 7805 will not sink current, this is not to create a 5V output but a virtual ground in a dual voltage supp,y system so it must source AND sink current a linear reg will only source, it has one output transistor the opamp circuit has two transistors, one to source from VCC and one to sink to the supply's GND in order to act as a new GND

[Zero999]

Simon,
No, he's talking about adding the LM7805 and LM7905 to the output of the circuit to regulate the voltage. The problem with this is the higher drop-out voltage required at 2V to 3V per regulator (4V to 6V in total), using a single regulator will mean it'll have a drop-out voltage of only 2V to 3V.

I think he's worried that if some devices are only rated to 5V and are connected between 0V and either rail, there's a risk that the start-up transients could damage the components. This won't be a problem here because the IC used is connected from +V to -V. If there are components connected from 0V to either rail which could be damaged by higher voltages, they can be protected from transients using zener diodes.

[Simon]

oh i see, yes if used as references and general power supply fine, I suppose a beefy pair of zenners across the output would do the trick, make them a volt or two higher so that during normal operation they are not part of the circuit

[Bored@Work]

the chip is use is the MAX038 function generator.

The MAX038 is no longer produced. If you don't have access to old stock it is not a good choice for a new development.