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'[EE] virtual ground or regulator'
2007\10\17@135325 by Martin

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Hello,
I'm designing an analog circuit which will run from +18v. I need a
bipolar supply of which +-9v will suffice. I am considering two options:

1. Run an inexpensive opamp supplied from +18v pos and 0v neg. This will
simply buffer a resistor divider between 18v and 0v to create a virtual
ground at 9v. - Possibly using an NPN emitter follower for higher current.

2. Use an analog regulator to output the 9v virtual ground. The problem
with this is that the bipolar supply will not be symmetric when the 18v
battery sags.

The issue with either circuit is that the virtual ground will start out
at 0v if turned on abruptly.

Thank you for advice.

Martin K

2007\10\17@141342 by Dr Skip

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How 'bout a pair of zener diodes in series? Center is gnd.

Martin wrote:
> Hello,
> I'm designing an analog circuit which will run from +18v. I need a
> bipolar supply of which +-9v will suffice. I am considering two options:
>
> 1. Run an inexpensive opamp supplied from +18v pos and 0v neg. This will
> simply buffer a resistor divider between 18v and 0v to create a virtual
> ground at 9v. - Possibly using an NPN emitter follower for higher current.
>
> 2. Use an analog regulator to output the 9v virtual ground. The problem
> with this is that the bipolar supply will not be symmetric when the 18v
> battery sags.
>
> The issue with either circuit is that the virtual ground will start out
> at 0v if turned on abruptly.
>
> Thank you for advice.
>
> Martin K
>

2007\10\17@142321 by Marcel Duchamp

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Martin wrote:
> Hello,
> I'm designing an analog circuit which will run from +18v. I need a
> bipolar supply of which +-9v will suffice. I am considering two options:
>
> 1. Run an inexpensive opamp supplied from +18v pos and 0v neg. This will
> simply buffer a resistor divider between 18v and 0v to create a virtual
> ground at 9v. - Possibly using an NPN emitter follower for higher current.
>
> 2. Use an analog regulator to output the 9v virtual ground. The problem
> with this is that the bipolar supply will not be symmetric when the 18v
> battery sags.
>
> The issue with either circuit is that the virtual ground will start out
> at 0v if turned on abruptly.
>
> Thank you for advice.
>
> Martin K
>
Number 2 will be able to source to your "negative" rail but probably not
sink from the "positive" rail.

Number 1 with the NPN will act similarly.

One solution uses a class B output; NPN/PNP with base to base, emitter
to emitter and a resistor from the base to emitter junction to help
crossover current.

If you go with an opamp solution, check it with a scope to verify it is
not oscillating from capacitive loading.

Depending on the current you require, a charge pump may work for the
negative rail.  Watch out for pump frequency noise.

2007\10\17@142930 by Bob Blick

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--- Martin <spam_OUTmartinTakeThisOuTspamnnytech.net> wrote:

> Hello,
> I'm designing an analog circuit which will run from
> +18v. I need a
> bipolar supply of which +-9v will suffice. I am
> considering two options:
>
> 1. Run an inexpensive opamp supplied from +18v pos
> and 0v neg. This will
> simply buffer a resistor divider between 18v and 0v
> to create a virtual
> ground at 9v. - Possibly using an NPN emitter
> follower for higher current.
>
> 2. Use an analog regulator to output the 9v virtual
> ground. The problem
> with this is that the bipolar supply will not be
> symmetric when the 18v
> battery sags.

This method is usually my favorite since you will
likely have other parts of your circuit which
reference true ground and it'll be nice not to have
your virtual ground bouncing around.

Also opamps quite often have better supply rejection
on the positive rail.

Make sure if you use a voltage regulator to create the
virtual ground it has enough load on it so it is
solid.

Symmetry is rarely a requirement, although you may
choose to have a lower voltage than vcc/2 for your
virtual ground.

Cheerful regards,

Bob

2007\10\17@153555 by Mark Scoville

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{Quote hidden}

Hi Martin,

I used a TI TLE2426 railsplitter 10 or so years ago. Might be worth taking a
look at.

http://focus.ti.com/docs/prod/folders/print/tle2426.html

-- Mark



2007\10\17@155824 by Harold Hallikainen

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I'd use the voltage divider and op-amp, or the
http://focus.ti.com/lit/ds/symlink/tle2426-q1.pdf "rail splitter." Do you
really need to source or sink much current on this virtual ground? If so,
an NPN follower would only source, not sink. If you do indeed need to
source and sink a fair amount of current, you could have a voltage divider
drive an audio power amp to supply half supply.

Harold


{Quote hidden}

> -

2007\10\17@163534 by M. Adam Davis

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On 10/17/07, Martin <.....martinKILLspamspam@spam@nnytech.net> wrote:
> 2. Use an analog regulator to output the 9v virtual ground. The problem
> with this is that the bipolar supply will not be symmetric when the 18v
> battery sags.

Perhaps there's a way to make vcc/2 using an adjustable regulator.
How close do you need to be?

-Adam

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2007\10\17@180524 by David VanHorn

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What's wrong with an op-amp, fed on the + input with two equal
resistors across the supply, and configured as a zero gain buffer?

2007\10\17@180855 by David VanHorn

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On 10/17/07, Dr Skip <drskipspamKILLspamgmail.com> wrote:
> How 'bout a pair of zener diodes in series? Center is gnd.

You're mean..  :)

2007\10\17@182015 by Dario Greggio

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David VanHorn wrote:

> What's wrong with an op-amp, fed on the + input with two equal
> resistors across the supply, and configured as a zero gain buffer?

I always go like this, though the TI parts looks nice :)

Only thing that has been pointed out is to beware of self-oscillating
(in the op-amp)

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Ciao, Dario

2007\10\17@193242 by peter green

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David VanHorn wrote:
> What's wrong with an op-amp, fed on the + input with two equal
> resistors across the supply, and configured as a zero gain buffer?
>  
nothing wrong per-se but a single rail incoming supply suggests that
some inputs and outputs will likely be related to the supply negative.
In such a situation it is far more important to keep the midpoint-ground
voltage stable that to keep the ratio of the voltages stable.

also the output impedance of an op-amp can be fairly high compared to a
chip intended as a power regulator.

also a power rail style point may be a highly capactive and this can
send some op-amps into oscilation


2007\10\17@195834 by Jinx

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> also the output impedance of an op-amp can be fairly high compared
> to a chip intended as a power regulator

But you could use an op-amp simply as a voltage tracker or buffer

Just Google for split supply virtual

eg

http://tangentsoft.net/elec/vgrounds.html

2007\10\18@090239 by David VanHorn

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> also the output impedance of an op-amp can be fairly high compared to a
> chip intended as a power regulator.
>
> also a power rail style point may be a highly capactive and this can
> send some op-amps into oscilation

So use the right op-amp?  :)

There are plenty that have low output impedance (check the LM-12) and
are unconditionally stable.

2007\10\18@090249 by David VanHorn

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> also the output impedance of an op-amp can be fairly high compared to a
> chip intended as a power regulator.
>
> also a power rail style point may be a highly capactive and this can
> send some op-amps into oscilation

So use the right op-amp?  :)

There are plenty that have low output impedance (check the LM-12) and
are unconditionally stable.

2007\10\18@100027 by Martin

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David VanHorn wrote:
>> also the output impedance of an op-amp can be fairly high compared to a
>> chip intended as a power regulator.
>>
>> also a power rail style point may be a highly capactive and this can
>> send some op-amps into oscilation
>>    
>
> So use the right op-amp?  :)
>
> There are plenty that have low output impedance (check the LM-12) and
> are unconditionally stable.
>  


Priced an LM12CLK lately? I'm glad I don't need that much current.

Thanks for the advice, everyone it was very helpful.

-
Martin K

2007\10\18@104407 by Harold Hallikainen

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{Quote hidden}

I favor the use of a voltage divider driving an op amp (with appropriate
stability precautions) over the use of a voltage regulator since a
regulator can normally "pull up" (for a positive regulator) and not
"pull-down." Since you're simulating a bipolar supply, current could go
either direction on the regulator output pin.

However, speaking of regulators, I thought Linear Technology's new series
of regulators
(http://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1040,C1055,P38503)
was very clever. Instead of a reference voltage, they use a reference
current. The regulator output is then compared to the voltage a the top of
the "voltage set resistor," which is driven by that reference current.
This reduces the number of resistors required from two, with a normal
adjustable regulator, down to one (the ideal design has zero parts). Also,
the output is now adjustable all the way down to zero volts instead of
down to the reference voltage. Also, the set pin can be driven directly
with a DAC to give an adjustable output at the same voltage as the DAC.

Pretty clever!

Harold




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2007\10\18@111113 by Spehro Pefhany

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Quoting David VanHorn <.....microbrixKILLspamspam.....gmail.com>:

>> also the output impedance of an op-amp can be fairly high compared to a
>> chip intended as a power regulator.
>>
>> also a power rail style point may be a highly capactive and this can
>> send some op-amps into oscilation
>
> So use the right op-amp?  :)
>
> There are plenty that have low output impedance (check the LM-12) and
> are unconditionally stable.

You can stabilize any op-amp circuit with a capacitive load by
providing a separate AC feedback path. The LM12 typical circuit
uses an inductor, paralleled with a small resistor, in series with the load,
but it's also common to use a 50-100R resistor and a feedback cap if there
isn't much load current.

(sorry about the private reply, David, I'll remember to hit 'reply to
list' one of these times).

Best regards,
Spehro Pefhany
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2007\10\18@195746 by David VanHorn

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> Priced an LM12CLK lately? I'm glad I don't need that much current.
>
> Thanks for the advice, everyone it was very helpful.

Yeah, but it's plenty stiff!  :)

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