Generate -+5V from 12V

You're on the right track, but the 7805 cannot sink current. However, a 7905 can.

The problem you're going to have, with this technique of putting two regulators in series, is that your output "0v" will not be controlled with respect to the input supply, so I don't think it will work the way you've drawn it (even with the bottom regulator being a 7905). You would have to create a (semi) fixed 0v by either tying your 0v node to a beefy resistor divider (between the 12v supply) or a weaker resistor divider with an NPN/PNP follower.

Also, with this technique, you've got to be careful to not short your output

0v with the minus side of the input 12v supply. If you're in a car, for example, you'll have to be careful not to touch your 0v circuit common with the frame of the car.

The best and most efficient method is to use a switching regulator. It's more expensive and complex, however.

Bob

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Spend a few dollars and get something that will work without glitches:

9-18 volts in, 5 volts 200 ma out, isolated grounds so you can use one for +5 and another for -5

John

To keep things simple a combination of the 2 might work - linear in the form of a 7805 and switching in the form of a flying capacitor inverter, such as a 555 driving a diode pump via a coupling capacitor - as long as the OP can accept slightly less than 5V on the negative rail.

If not - run the 555 inverter from 12V and use a 79L05 for the correct negative voltage.

Well, 50 mA times 5V is only a quarter watt; you can use a 7805 with the 'common' pin connected to a 5V zener, with a 50 mA source (100 ohm resistor from 7805 OUT to 7805 COMMON) to bias the zener. As long as you don't draw more than 50 mA from -5, it'll give you the right voltages. All the positive-to-ground current goes through the zener diode, so it'll have to be heatsinked for a watt or so.

The most power-efficient way to do it is a DC/DC converter, but that gets pricey.

Supply + ----+--------[7805]---- +5 V cap | |------+------+------ 0v cap | Supply - ------[7905]------------ -5V

Whats wrong with doing it this way. Use a pair of 1000uf caps. Bearing in mind that if the supply -ve is grounded the 0v will be +5 volts above it.

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Best Regards:
                     Baron.

The problem is that the all the current through the +5V load goes through the load on the -5V. So, unless these currents are exactly equal, the voltage from the node +5V to 0V will not be equal to the voltage from 0V to -5V. That is, you really won't have a +-5V supply.

With this approach, you'll need some method of forcing "0V" to be exactly (or very close to) six volts more positive than the minus side of the input supply. See my earlier post.

Another concern with this approach is the dropout voltage on the 7805 and

7905. Assuming that you're able to keep 0V exactly six volts above the minus side of the input supply (the ideal case), the dropout voltage (i.e., the voltage from input to output on a linear regulator) will only be equal to one volt. I'm not sure the 78xx and 78xx series can work with that low of a dropout voltage at the currents you need. There are things called low dropout regulators (LDO, for short) that allow much lower dropout voltages (at a given load current).

Bob

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For an application powered by a mains transformer you should have a 2V margin (7V) as an absolute minimum, a 3V margin (8V) to allow for peak load dips.

For automotive applications you can recon on a voltage range of 10.8V to

14.4 but the terminal voltage of the battery can drop to 1V/cell during cranking, its clear that even ignoring cranking voltage there is insufficient headroom for a 7805 & 7905 in series unless the battery is maintained at maximum charge at all times.

Doh ! Dropout voltage hadn't occurred to me. Those things need two volts of headroom. Thanks for pointing that out. Its something I should have known. :-(

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Best Regards:
                     Baron.

Thanks for this idea, I'll give it a try.

Yes you are quite right ! I just didn't realise it until it was pointed out to me.

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Best Regards:
                     Baron.

It's not possible to both sink and source current (up to 150 mA) into the node labeled "0v". It might not even be possible to get both regulators properly biased, the 'common' pin has a few milliamps DC from the regulators, even with no l oad.

Giving this a bit more thought... Take the regulator tails to the opposite rail. That would ensure that each regulator had sufficient overhead. Replace the caps with a pair of equal value resistors in series passing say the max expected current. Using the centre point as the 0v. Put the caps across the output.

Would that work ?

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Best Regards:
                     Baron.

There are plenty of DC/DC chips on the market, both flying capacitor inverters and one's that use an inductor but chances are you probably have a

555 more readily to hand in the junk box.

:

The "0v" point isn't low impedance, which makes it unregulated.

It's also a little elaborate, using the (+) and (-) rails as though they were 'common', then not connecting except with power-wasting resistors. Better might be to make (+) the common node, "+5". Attach two negative regulators to make "0V" (that will be a -5V regulator, "A"), and "-5V" (which is a -10V regulator, "B").

Then, knowing that the A regulator can sink any current required at "0V" and current sourced out of the "0V" node can never be more than

50 mA net, a single resistor from "0V" to the (+) rail, 100 ohms, finishes the circuit. You need that, because the A regulator cannot supply positive current, being a negative regulator...

Yes I see how that works. One to stick in the notebook for reference. Thankyou for taking the time to explain things.

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Best Regards:
                     Baron.

See alt.binaries.schematics.electronic for a (linear) circuit that will work up to about 200mA on each supply with a minimum input voltage of 10.6V.

Bob

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== All google group posts are automatically deleted due to spam ==

Sadly, Verizon (and other major service providers) have completely eliminated all alt.* groups from their servers. This affects me, as well. It may yet force me to get an extra-cost service added.

Jon

"Jon Kirwan" wrote in message news: snipped-for-privacy@4ax.com...

I just tried posting the attachments here, too, but it doesn't look like they made it through.

Giganews' cheapest rate is a couple of bucks a month. I haven't had a problem with them.

Here are the LTSpice .asc file guts if you're interested. In fact, even if you're not interested, here they are ->

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Bob

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Thanks, Bob. I got both postings just fine. Apparently, Verizon lets through binaries on non-alt.* groups they support. Perhaps giganews blocks binaries on groups marked as text? If so, that's yet another consideration when I'm out looking for a service to pay for.

Things sure have evolved on internet.

Jon