I have a large board which asks for three power supplies, +3V, +3.3V and
+-5V. First two can be easily satisfied with two HP power supplies (each has its own GND pin also). Regarding +-5V, can I replace with an HP power supply set to 10V, -5V pin connect to HP's ground and +5V to power?
If I read you correctly, that's four supplies: +3V, +3.3V, +5V, and -5V.
No. Doing so would provide zero volts to the board's -5V rail and +10V to its +5 rail.
You need four supplies.
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will an ATX supply not do? rw
Maybe not. Compound switchers need a load on the main supply in order to come up, and the auxiliary supplies aren't always well regulated.
Jerry
-- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
They also make more noise than a neo-natal nursery. :-)
Regards, Steve
If he knows what he's doing, ha can load the 10V supply with a beefy op-amp connected as a follower to a divider across the rails and ground its output. If something goes wrong, it can blow the board unless he uses Zener-cum-fuse protection. There are DC-DC power-supply bricks and chips that can probably supply all the -5 needed from a +5 supply.
Jerry
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I don't know what you people are talking about. Back to my question, how do I make a +-5V?
You might with some caveats. First, the intended +-5v supply needs to be floating with respect to the other supplies. Then the 10v voltage difference can be reference wherever you want in theory and often in practice.
The challenge is: now that you've floated the supply, how will you reference it to the ground or 0v point on the board?
Think of the +-5v supply as a 10v battery. A battery "floats" with no problem. Unless you do more, the result looks like this:
+------------------------------>+5v | | +----+ +--------------->+3.3v | 10v| +----+ +----+ |3.3v| +----------+3v | +-+--++----+ | | |3.3v| | | +-+--+ | | | | | | | +-----+---------> 0v: the reference for +3.3v, +3v | | | +------------------------------>-5vWith the 10v battery floating, there is no reference to the other batteries. Current flowing through the circuit board will cause the +/-5v terminals to go almost anywhere relative to 0v. Depending on what's on the board, the +5v terminal could end up at -6v and the =5v terminal at -15v (both relative to 0v of course).
+------+----------------------->+5v | | | | +--------------->+3.3v +----+ | +----+ | 10v| | |3.3v| +----------+3v +----+ ++-+ +-+--++----+ | |R1| | |3.3v| | ++-+ | +-+--+ | | | | | | | | | +-------+-----+---------> 0v: the reference for +3.3v, +3v | +--+ | |R1| | ++-+ | | +------+----------------------->-5vA resistor divider with current much higher than the +/-5v loads and connected to the 0v reference will refer the +/-5v to the rest of the batteries. It's not a very elegant or even practical solution but it makes the point to address your question.
Fred
...
If you don't know what we're referring to, you're better off buying what you need. Power supplies are cheap compares to your board. If you post your current requirements, we can suggest specific hardware.
Using supplies with adjustable voltage but without knob locks is a fairly common way to smoke parts. Be careful.
Jerry
-- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
You can make Fred's solution more practical by adding an operational amplifier that can deliver the difference between +5 and -5 currents.
+------+-------------------------------------------->+5v | | | | +--------------->+3.3v +----+ | +----+ | 10v| | |3.3v| +---------+3v +----+ ++-+ +-+--+ +----+ | |R1| +----------+ | |3.0v| | ++-+ | |\ | | +-+--+ | | +---|-\ | | | | | | +---+----------+------+---------> 0v | +--------|+/ | +--+ |/ | |R1| | ++-+ | | +------+----------------------->-5vIs it worth it?
Jerry
-- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Here's a cheap and quick, potentially noisy solution:
Make sure the wall worts are rated to deliver enough current otherwise you'll get too much voltage drop for your app. Note that this supply is UNREGULATED, that is, increasing load will decrease your voltage.
If you need a regulated supply, use 7-12VDC wall worts (to compensate for the voltage drop across the regulators), buy a 7805 positive and a 7905 negative 5V regulator, two 3300uF electrolytics and two 1uF ceramic caps to filter the output and a small PCB. There are tons of diagrams how to arrange things to get what you want. Google is your friend. Total cost of parts should be about $5 to build the regulator, plus some sort of enclosure if you require.
If you happen to have two matching wall worts of >6VDC you can probably use them if you regulate, the regulator chips can accept up to ~35VDC. note that with larger input voltages you'll need to use heatsinks on the regulators to dissipate the extra power/heat.
Dave
We need answers to these basic questions:
1)You need -5 volts, +5 volts, +3.3 volts, and +3.0 volts - correct? (can you use just +3.3 volts instead of both +3.0volts _and_ +3.3volts?) 2)How much current do you need for each voltage? 3)What voltages and currents do your 2 HP power supplies produce?Assuming that the HP's can produce negative voltage (with respect to the HP's GND), then set one HP for -5 volts and set the other one for
+5 volts. Then use Low-DropOut (LDO) voltage regulators to produce the +3.0 volts and +3.3 volts.Dave Pollum
Swing and a miss
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If those hp supplies are like mine, either the positive or negative terminals may be grounded by a link.
Jerry
-- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Only if you buy unregulated wall warts. You can get regulated
5V wall-warts and cable-lumps. They usually cost more of course.
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The main problem with this scheme is that if either of +5 or -5 volt rails tries to draw more than a few mA, the opamp won't be able to source or sink that much current, so that rail will droop. (The other rails will just follow whatever they've been grounded to.) This isn't too bad, but it'll cause the _other_ rail to suddenly spike by the same amount in the opposite direction. So, a 5A spike on the +5 rail might cause the -5 rail to go to -10V, possibly smoking something funny (like components.)
I'd say you need some kind of beefy totem pole output to drive it, meaning a couple of power transistors in addition to the circuit above. Now, is it worth it? Particularly when you can buy a +-5V regulated supply that will source 20A for $50.
-- Regards, Bob Monsen Music is the pleasure the human soul experiences from counting without being aware that it is counting. - Gottfried Leibniz
It's not hard to get op amps that handle a few tens of milliamps, and that represents the difference between the drains on the two rails. The quiescent drains can be approximately balanced by a single resistor.
I don't recommend this except for emergencies. It seems penny wise and pound foolish.
Jerry
-- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
You can get opamps that handle far more current than that, but the op said a 'big card', so I was thinking maybe a PC motherboard or something like that. A 300W ATX power supply can supply 30A through the +5 rail, but only
300mA through the -5V rail. Hook up a load that is designed for one of those, and you'll be in for quite a surprise.-- Regards, Bob Monsen What good your beautiful proof on (the transcendence of) PI. Why investigate such problems, given that irrational numbers do not even exist? - Leopold Kronecker
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