positve and negative power supply

im building a circuit using the 4052 cmos chip. i need to connect the 4052 Vee to at least as negative as -4V for audio reasons. it cant be at 0 for my purposes. i would like the unit to fit into the 9V supply format (widely used in the guitar fx world) but ive pursuaded myself to go to a 12V supply as i need a regulated +5V supply (not applicable with a +9V trans if i intend to have -4V unless i get a REGULATED 9V trans...hmmm....possible).

now... how do i get a negative voltage without floating the ground (virtual ground).

what i had in mind was to take a 12V in. regulate it to 9V and 4V with a 78L09 and a 317T respectively. assign the +4 to ground (virtual). this would bring the +9V down to

+5V (connected to all the components) and the 0V would then become -4V which i would assign to the Vee pin of the 4052. i may even give the 4052 power pin a +8V feed (from the +12V on the transformer directly) for more audio headroom but the fact that its unregulated may not be a satisfactory solution and +5V is fine.

this would mean that i cannot connect the sleeves of the outside audio connections to internal ground as it really is +4V with respect to the outside ground world but i COULD make sure that they at least all connect together albeit isolated from the internal circuitry.

my trusted friend however does not recommend in the slightest that i have this virtual ground setup.

any ideas?

Reply to
Sean Bartholomew
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This is easy if you have access to the transformer. Connect one side of the transformer secondary through a series capacitor to a clamp diode. Then half wave rectify the negative clamped voltage. Obviously, the voltage will depend on the transformer.

Tam

Reply to
Tam/WB2TT

The idea can work perfectly OK in practice. What you ideally want though is that the incoming 12V supply is 'floating' (like from a wall-wart). Then your (+4V) assigned ground can actually be tied to a true ground point such as the metalwork and incoming signal connectors. Other than that it is sometimes as easy just to take in a 9V supply. Regulate it to give +5V and 0V and then use '7660' chip and 2 capacitors to electronically invert some of the +5V rail to give a -5V rail. Doesn't matter then if the incoming 9V is floating or not. regards john

Reply to
john jardine

ok im liking this. both possibilities intrigue me,

1st, ur saying that BECAUSE its a wall wart transformer, it IS floating. i didnt think of that. there is no direct connection to true ground. nice. so i assign anything i want to ground and connecting that to outside ground from other devices is fine. ok. ill definitely try that.

2nd, this 7660 chip INVERTS power? interesting. i was looking at the MAXIM chip that delivers +-9V from a single 9V supply but all audio experts inform me that its too noisy for audio usage. is this 7660 chip noisy? or would it introduce noise in the 5V reg supply?

Reply to
Sean Bartholomew

also a thought just occured to me. theoretically, if i build TWO regulated 5V supplies from a single 9V transformer supply, i should be able to connect the positive of one to the negative (0) of the other treating THAT as ground ref. then id have +5V, 0 and -5V.

but wouldnt that: A- simply just short out the ground and the +5V of one of the supplies since both their grounds are connected? B- theoretically give me 10V diff total WITHOUT any reclocking from a

9V supply?
Reply to
Sean Bartholomew

hmmm sounds like i should stick to my original plan then what with the half wave of stacking 2 5V and the noise of the 7660. so 9V and 4V here i come. id totally use 10V and 5V but then ill have to have a supply greater than 12V.

Reply to
Sean Bartholomew

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Yes, like the Maxim chip, the 7660 can be a bit noisy. Though a lot depends on how much care is taken with cleaning the supply rails. For simplicity it's easier to go the linear route. regards john

Reply to
john jardine

Yes. They can be stacked like that but you'll then need at least a 12V-14V input as the voltages are now sitting on top of each other.

Unfortunately an awkward problem pops up when stacking positive voltage regulators. They can only source current and not sink it. That opamp powered from across the outer +/-5V rails is fine. The "0V" centre rail is fine. Trouble comes when the opamp drives to drive a load connected to the "0"V rail.

When the ouput swing goes below 0V then it's no problem, as current is taken OUT of the mid rail regulator and passes via the opamp back down the real 0v line. But ... when the opamp tries to swing a positive output voltage to its load, the top regulator is quite happy to feed current via the opamp into the load but this current then can't go back via the pretend 0V rail into the mid voltage regulator output pin. Essentially the circuit can only run as a half wave rectifier. The inefficient work-around is to stick a resistor across the middle rail and the bottom rail which constantly passes a current that's a tad more than the opamp will be liable to ask for when driving its load at maximum. (say

20ma's worth). regards john
Reply to
john jardine

thanks for all the help guys. i love this forum.

Reply to
Sean Bartholomew

You guys are still doing it the hard way. If you want +5 and about -5 V from a 12 V floating supply, connect a 4.7 V zener diode in series with the negative supply lead, and connect the other end to "ground ". Now connect a series regulator between ground and the positive supply lead. The original negative supply lead will be -4.7V. This will work so long as the positive supply current is more than the negative supply current.

Tam

Reply to
Tam/WB2TT

The zener current (plus the negative load current) will be exactly equal to the positive load current. So, if the positive load current is 250 ma, and the negative load is 5 ma, the zener current will be 245 ma.

Tam

Reply to
Tam/WB2TT

The way you describe this, it is impossible for the positive and negative currents to be different, because they are in series with each other. Also, with a voltage regulator from + to ground, and a zener from - to ground, there is nothing to limit the zener current, and it will go pop.

d Pearce Consulting

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Reply to
Don Pearce

"Sean Bartholomew" schreef in bericht news: snipped-for-privacy@posting.google.com...

Perhaps you could use the audio signal, amplify it, ac coupled, schottkey rectifier, and charge a supercap to get your negative supply. It may sound a bit distorted when all is new, but it will get better after a minute or two and stay that way ;) After all, Vee does not draw much current. If it is just for hobby and personal use, I'd might even drop in a lithium button cell or two.

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Reply to
Frank Bemelman

In article , Sean Bartholomew wrote: [...]

The Intersil 7660 blows up too easily. I suggest the Linear equivelent. You can get it from digikey.

With both these parts, there cab be a problem if your circuit pulls the output above ground. Adding a small MOSFET as a common gate stage can protect it from this.

These chips draw their current in spikes. If you RC decouple it from the

+5V, there shouldn't be much trouble with that. You can also add a filter on the -5V produced.
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Reply to
Ken Smith

I think he means this:

---------- -----! LM7805 !-------------- +5V ---------- ! +----------------- GND ! /-/ 4.7V ^ ! --------+----------------- -5V (almost)

The circuit works if the +5V's load is greater than (the -5's load minus

2mA)
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Reply to
Ken Smith

View in a fixed-width font such as Courier.

Reply to
Fred Bloggs

Well- if you have 9V regulated then you only need one standard regulator. If Vee of the 4052 is your only load, then this current is all leakage- you keep the 79L05 in regulation with a small 500uA current through the 10K which guarantees the 5V loading always exceeds the -4V current draw. View in a fixed-width font such as Courier.

Reply to
Fred Bloggs

If your wallwart supplies AC rather that DC you can do it like this:

7805 .-----. -------+---||---+------| |---- -5V | '-----' | | --- === --- GND | 100uF | === GND

created by Andy´s ASCII-Circuit v1.22.310103 Beta

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Of course, you can always open the wallwart and remove any rectifiers and filters inside it. If you don't expect to draw more than 100 mA use 7xL05's.

- YD.

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Reply to
YD

In article , YD wrote: [...]

Never use 7XLXX regulators. If you are going to the bother of wiring up 3 pins use a 7XXX regulator. They don't cost enough more to worry about, they are harder to break and the heatsink hole makes them easier to mount.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

thanks guys. i like the zener diode idea. how can i ensure that the + current remains greater than the negative current? a 10k R? if everything else falls short my plan with the 2 regs can work. incidentally, the diagram shows a 7805. but i need FOUR volts to get a

+5V supply. remember, 9-5=4. thats why im using a 337T cause the 7904 doesnt exist. i wish it did though.
Reply to
Sean Bartholomew

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