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.

[MegaWatt]

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.

Yes, I have noticed that!
Is there any good replacement chip with some same functionallity?

[Chasm]

I would not call it a problem. I already know how to generate magic smoke in large quantities.  
MegaWatt wants to power a +/-5V device out of his 0-15V supply. The virtual ground circuit will enable him do that.
Now I would like to make it a bit more operator proof. (In a different way than a pair of zener diodes.)

Maybe I have a brain fart, but I still think that a pair of 7805/7905 will work on the output side of the virtual ground generator.
The real problem is the combined dropout voltage which will increase the minimal input voltage to 15V, and I've ignored it.

Using one 10V regulator on the input side, and then the opamp to generate the virtual ground will definitely work. But so will zener diodes as long as you are not using low power types.


"Just" using a pair of 78xx/79xx regulators without a (virtual) ground in between them will not work.
That is actually a nice thing to try and proof if you have a pair of them.

[Zero999]

Is there any good replacement chip with some same functionallity?

There's no pin out compatible replacement.

Function generator ICs don't seem to be very popular nowadays. If you want to make a square/triangle and sine wave generator, your best bet is to build a triangle wave generator and use a wave shaper circuit to get the sine.

[Bored@Work]

Is there any good replacement chip with some same functionallity?

I am not aware of any. The MAX038 was really special, and I don't know any chip with similar performance. The old XR2206 is still available, but generates rather mediocre signals. The ICL8038 (notice the similar number?) is also obsolete.

The state of the art for some time now are DDS chips, e.g. from Analog Devices. To use them in a functiong enerator you usually need to have an MCU to control and configure them.

[MegaWatt]

Hmm, too bad! For me it doesn't really matter, I'm making a function generator for home use, and I got a MAX038 from ebay, but too bad they don't make it anymore, it's a nice chip!

[Bored@Work]

If I remember the story (don't quote me on this), they had to stop production after the fab burned down, they couldn't find another fab with the same process and didn't invest in rebuilding the fab or building a new one.

[Simon]

I must say it seemed odd that they are no longer available, last i looked they were not too cheap (10 years ago) but I've never seen such a versatile chip. maybe they lost the blueprints ? or it's cost meant it was not popular enough

[alm]

I believe they outsourced the production of the MAX038 to TSMC, and TSMC discontinued the process it used due to lack of demand. Selling a few MAX038s for building function generators isn't enough to keep a semiconductor process profitable I guess .

This product was manufactured for Maxim by an outside wafer foundry using a process that is no longer available.

[Zero999]

What amazes me is that there are still lots of ICs which are obsolete for commercial devices and only useful to hobbyists still in production such as the NE555 timer and the µA741 op-amp.

[Simon]

there must be a lot of us out there using them 

[Chasm]

There are a zillion designs that use the parts. It is easier to keep the part alive than to redesign them.

As for the µA741...
I'm currently ramping up to build a linear power supply that uses two of them (or the various replacement types).
The design is from 1973 - and it was and is still the underlying principle in many brand name linear power supplies from Gossen, HP and Agilent and the like.

The design may be old design but it is still very applicable, still easily changed for different voltages and currend. Especially with a couple of mods that fix bugs and make life easier. Not that I came up with any of the modifications, but I'm not going to reject them just because of that.

And on that note, back to the layout.

[Wim_L]

I think the real question was why you'd put a 741 in there when you can get another opamp that's better and cheaper these days. The only reason I can think of is that they're the typical textbook opamp and people just naturally reach for one when they want a noncritical opamp and don't want to bother searching for devices.

[Zero999]

Most modern op-amps follow the same pin out as the old 741 so it can be replaced without changing the PCB most of the time.

The NE555 can be directly replaced with the ICM7555 or the TS555 in old designs or anything from CMOS logic gates to a comparator or microcontroller for new designs.

[Chasm]

One part of the answer is that the circuit does does not really need a that much faster opamp.
Here is a link to the lastest, modified version. The author has recently measured his power supply. Switching a (FET) load from 0.3 to 1 Amp at 300Hz takes 20 micro seconds to regulate. Others have confirmed these values.
Ok...  admittedly using a TL071 but still not to shabby for a not much newer old op-amp.


A less obvious part of the answer are the involved voltages.
This is old(er than me) analog tech. Most of todays op-amps do not survive Voltages in the 30V region.
(This modified version has been a bit defanged, it it uses different voltages in the voltage regulation than the original.)


Back to topic:
The  important "trick" and reason for the fast response time in this form of circuit is that the ground of the op-amp rails and the and reference voltage float on the positive output voltage. This way you always have the full regulation output at the base of the power transistor(s). That is why the op-amps and reference voltage have their own transformer.

Several Gossen Konstanter, Several Anatek power supplies, HP/Agilent E361xA, E3630A, E6236, Leader LPS 151&152 - all use the same principle.
There is also a HP Appnote, No.90 from 1967 on this topic, it was base for the development and publishing of the circuit in 1973.

There are some later clones that use the negative output as reference. They are much worse because you now need a much higher voltage difference - even worse, depended on the output voltage- to regulate the system. In comparison a quite obviously stupid idea.

The mentioned dual power supplies the ground for the op-amps floats on the (middle?) ground of the output. A contradiction? No, look where the power transistors are in that case.


If I'll manage to get this thing build I'll post a bit more about it.

[MegaWatt]

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.

Could you tell me what the influence is of the 10k resistor at the op amp output?
I'm struggling a bit with understanding what is happening with the resistor and the two transistors

[Zad]

Texas Instruments TLE2426 Precision Rail Splitter / Virtual Ground

http://focus.ti.com/docs/prod/folders/print/tle2426.html

Used in quite a lot of low power audio circuits (<20mA) for simulating a dual power supply.

[Psi]

When building just one of them yourself the component prices add up quickly.

Sometimes its just easier (and often cheaper for hobby stuff) to buy a little module to do it for you.

USD $4.22

10-13V input / +5V and -5V output at up to 100mA each

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=102-1342-ND

Of course it depends if you have enough other stuff you want to buy so you get free shipping

[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.

Could you tell me what the influence is of the 10k resistor at the op amp output?
I'm struggling a bit with understanding what is happening with the resistor and the two transistors

That resistor is not essential. It will work without it. The resistor just improves the transient response. I hope you can see that the transistor buffer the output of the op-amp and that negative feedback overcomes the 0.7V voltage loss caused by the base-emitter junctions. The 10k resistor bypasses the base emitter junctions so the op-amp's doesn't have to jump around quite so much when the load changes.

[MegaWatt]

So if I understand correctly, the current my circuit can draw to/from ground depends on the collector currents of my transistors?