positve and negative power supply

[Sean Bartholomew]

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?

 

[Tam/WB2TT]

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?

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

 

[john jardine]

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?

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

 

[Sean Bartholomew]

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?

 

[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?

[clip]

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

 

[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?

 

[john jardine]

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?

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

 

[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.

 

[Tam/WB2TT]

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.

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

 

[Don Pearce]

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

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
http://www.pearce.uk.com

 

[Frank Bemelman]

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.

<silly thinking>
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.
</silly thinking>

 

[Sean Bartholomew]

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

 

[Ken Smith]

2nd, this 7660 chip INVERTS power?

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.

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?

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.

 

[Ken Smith]

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.

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)

 

[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
http://www.pearce.uk.com

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

 

[Fred Bloggs]

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?

View in a fixed-width font such as Courier.

 

[Rich Grise]

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.

No it won't. The Zener diode will pass the quiescent current of the
regulator when there's no load. You just stick it in the '-' lead and pick
up 0V at their junction.

But it is true, that for the circuit to work, the current through the
"positive half" of the load has to be greater than the "negative half" by
a value which is inside the operating range of the Zener current.

A possibly more effective/efficient way to make a virtual ground is with
an opamp follower driven from that voltage divider someone has mentioned.

Cheers!
Rich

 

[Fred Bloggs]

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).

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.

 

[YD]

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?

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



7805
.-----.
-------+---|<---+------| ----- +5V
| | '-----'
From | | |
Xfmr | --- ===
| --- GND
--+ | | 100uF
| | |
=== | ===
GND | GND
|
|
| 7905
| .-----.
---->|---+------| |---- -5V
| '-----'
| |
--- ===
--- GND
| 100uF
|
===
GND

created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

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.

 

[Ken Smith]

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.

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.