IC that generates +5 V, 0 and -5 V from 16 V single supply input?

Hi,

Does anyone know of an IC that generates linearly regulated +5 V (+50 mA), 0 V and -5 V (-50 mA) from an unregulated 12...18 V single supply input?

That is, 2 input nodes (between which I apply 16 V) and 3 output nodes, and none of the input nodes is at the same potential as an output node.

Noise must be minimum, so I don't want to use switched-mode regulators.

Thank you, Bill

Reply to
Bill
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Well, I was going to suggest you use a dual 5V output DC-DC converter, but now you've just converted that from "easy" to "impossible".

Otherwise, grab six AA's.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Reply to
Tim Williams

Perhaps stack a couple of 5V regs on top of each other. The bottom reg supplies -5, the top one +5 and where they meet is 0V.

Reply to
Polyp

Establish a virtual ground between the two input nodes with a couple of identical resistors and soothing caps, and then reference that for a pair of linear regulator 78L05 & 79L05 on each input node.

--
Adrian C
Reply to
Adrian C

Well, I don't know what I was thinking of (if that was the case), but I meant "the input supply does not provide the 0 V node." There may be coincidence in voltage between the input low node and the output low node, or between the input high and the output high.

I know how to do it with two regulators. I'm just surprised there is no IC with the two in it, being +5V/0/-5V so common.

Thanks,

Reply to
Bill

Why do you say impossible? It is easy with two ICs. I was asking about a single IC solution.

Best,

Reply to
Bill

No, you can't do it with two ICs either. Your vague requirement suggests some level of isolation is required, which is only going to happen with a switching supply (transformer or flying-capacitor). (I suppose you could make a photovoltaic converter, too, but that would be rather expensive, bulky and inefficient.)

If isolation isn't actually required, and one of those float-a-regulator-from-the-middle setups will suffice, then perhaps a four-terminal dual tracking regulator will work. If you can find one.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Reply to
Tim Williams

Soothing capacitors?

Reply to
Spehro Pefhany

Those are used in premium tube amp audio devices. Much more expensive than regular caps.

Reply to
bw

Spehro Pefhany a écrit :

Yup. In anxious audiofools' amps it's bypass caps : the bigger the caps, the bigger the soothing factor.

--
Thanks,
Fred.
Reply to
Fred Bartoli

Just in case someone reads you and believes you...

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It strictly fulfills the requirements I wrote.

My (first) requirement may be strange, but not vague. I did specify something unambiguously.

I didn't mention isolation.

I don't know exactly which configuration you are referring to, but not any one will work. The middle output node must be able to sink and source current.

Best,

Reply to
Bill

Since you only needed 50 mA, why not just use two 5 V 1 W zeners in series and two series resistors to 0 V and +16 V.

Paul

Reply to
Paul Keinanen

Bill pondered

Texas make a "precision rail splitter" called the TLE2426 which sounds kind of like what you want, but it only sources / sinks 20mA, and it doesn't regulate - though you could something like it to split a 10V rail from a linear regulator, or follow it with a dual regulator (if you find one which does both + and -5V in the same package I'd be interested to know what it is).

I've always thought it looked like one of those IC's which *ought* to be useful but would never quite have all the specs you needed for any real-life task!

I realise this doesn't answer the question you asked(!) but I throw it into the debate in the hope that it will spark new lines of thought.

--
Nemo
Reply to
Nemo

I've done this with a 78XXX regulator and an op-amp, but that's two chips. Most regulators will not sink and source current, so you either have to add a dummy load to avoid that requirement, or use one of the regulator chips (for example, some of those called series voltage references) that *can* sink and source current.

Reply to
Spehro Pefhany

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You only need one series resistor

Reply to
David Eather

Bill wrote in news: snipped-for-privacy@4ax.com:

I didn't post before because that wasn't clear, but now that it is, and you want sink and source, try an op-amp as voltage follower. Tie the input to the common point of two 1% tolerance resistors that make a series chain across a

10V zener diode, and a third resistor supplying the zener with enough current to supply it and the small load made by those two resistors. The op-amp can be powered direct from the main supply but the two rails will tap to each end of the zener. You could use two zeners and one resistor as someone already mentioned, if it's not critical that the rails be within 1% of each other's absolute value.

If source and sink is vital, then it must be an op-amp, so you either do the rest with discrete parts (few enough of them), or do something even more elaborate. The op-amp trick does useful things when you want asymmetric rails too.

Reply to
Lostgallifreyan

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The original requirement also included

So you need two resistors.

If the 0 V input and -5 V output can be connected together or the +16 V input and +5 V output are the same pin, one resistor is enough.

Paul

Reply to
Paul Keinanen

Be sure to use nice big decoupling caps on the op-amp output (LOL).

Reply to
Spehro Pefhany

Doubling the price also helps...

Reply to
Robert Baer

Spehro Pefhany wrote in news: snipped-for-privacy@4ax.com:

Good point. I have no idea if that 50 mA is steady, or an average. But if it IS steady it might not matter so much if it's also steady upstream. I'd put a low ESR bypass cap on the op-amp's own supply though, to help with that.

Reply to
Lostgallifreyan

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