DC offset dual rail battery supply

David White · Feb 6, 2012

Can someone please have a look at this diagram and advise if it is
workable?

The object is to output two identical signals, one with a +12V shift
above DC/GND, and the other a -12V shift.

IOW there is a 24V differential between the mid-points of the two
waveforms, with respect to each other, centered on the GND reference
of the input signal.

http://www.4shared.com/photo/9Ixd-Ldp/DC-offset-amps.html

Thank you,

David White

Jamie · Feb 6, 2012

David White wrote:

Can someone please have a look at this diagram and advise if it is
workable?

The object is to output two identical signals, one with a +12V shift
above DC/GND, and the other a -12V shift.

IOW there is a 24V differential between the mid-points of the two
waveforms, with respect to each other, centered on the GND reference
of the input signal.

http://www.4shared.com/photo/9Ixd-Ldp/DC-offset-amps.html

Thank you,

David White
Wow, I haven't seen a mess like that in a long time.. I can not say

for sure if that would work, it seems to disconnected to evaluate..

But I think you're looking for a virtual ground?

++--------------+
+ |
| |
+----+---------------+ |
| + | | |
+ .-. | | |
| | | | |\+ |
--- | | +----+|-\ |
24V - '-' | >+-----+
+ +------------+|+/ |
| | |/+ |
| + | ===
| .-. | GND
+---|-|----++--------+
| | +
'-' |
+ |
| |
| |
+------+

Consider this..

The op-amp is your ground and the battery terminals become
your +&- 12Volts.

This only works if you keep the battery or isolated power source
lifted from common. Thus, the op-amp output becomes the common.
THe R's at the (+) input of the op-amp are a pair of equals to
give you a divider.

Use the proper Op-amp that can handle the current requirements.

Jamie

Phil Allison · Feb 6, 2012

"David White"

Can someone please have a look at this diagram and advise if it is
workable?

** No way.

http://www.4shared.com/photo/9Ixd-Ldp/DC-offset-amps.html

** What dual op-amp has outputs on pins 1 and 2 ??

..... Phil

Kaz Kylheku · Feb 6, 2012

On 2012-02-06, David White <davidwhite@optonet.com> wrote:

Can someone please have a look at this diagram and advise if it is
workable?

I've not seen a diagram that isn't workable. Not while it is just a diagram.

The object is to output two identical signals, one with a +12V shift
above DC/GND, and the other a -12V shift.

IOW there is a 24V differential between the mid-points of the two
waveforms, with respect to each other, centered on the GND reference
of the input signal.

http://www.4shared.com/photo/9Ixd-Ldp/DC-offset-amps.html

Your diagram is somewhat confusing. The two op-amps are on one IC, U1.

What is the U2 box? and why isn't its mating pair labeled?

Are those separate boxes just the power connections to the op-amp IC?

It seems like the circuit you're trying to design actually has two separate
op-amp IC's, U1 and U2. You're trying to give each of these a single voltage
supply: one from 0 to 12V and the other from -12 to 0.

Firstly, a dual supply based on batteries is risky. The reason is that the
batteries may not discharge evenly over time, leading to unequal voltage rails.
The same comment applies to the +12/-12 midpoint on each side between
BT4 and BT3, and BT2 and BT1.

Secondly, batteries are not ideal voltage sources and have an internal
resistance. As AC signals leak back into the power supply circuit, the signal
voltage will then appear in the power rails, thanks to that resistance. You
need some bypass capacitors to combat this problem.

Third, your + connections to the op-amps are left floating. You cannot do this;
the + connections must be connected to a reference voltage, otherwise the
op-amp circuit is meaningless. If you have a chip with multiple op-amps and
some of them are not used, even the unused op-amps have to be properly
connected, never mind used ones.

You have no feedback to control the gains. Basically your circuit is quite
incomplete. (Please post what you think is a complete circuit.) The reference
voltage is different for the different op-amps because you want a DC offset
between them. Thinks about what you want the output to be when the input is at
0V with respect to ground. How are you going to make the top op-amp put out 12V
(the midpoint between 0 and 12) when the signal at the - input is at 0V?

One obvious way to make the implied circuit topology to do the required job is
to generate a +12V voltage reference at the + input of the top op-amp, and a -12V reference at the + input of the other op amp, with some crappy voltage
dividers (that are well bypassed with capacitors). (You have the battery
dividers for this already, but see the note of caution above: if BT4 discharges
faster or slower than BT3, that reference level will shift!)
Add negative feedback to stabilize the op-amps (are you looking for gain, or
just unity buffering?). Then, use capacitors to couple the input signal to
the - inputs (with separate capacitors, one in front of each op-amp). Your
input will be centered on 0V, and the coupling capacitors will take care of
bridging the AC component of the input signal into the 12V and -12V DC levels
present at the op-amp inputs.

Jasen Betts · Feb 7, 2012

On 2012-02-06, David White <davidwhite@optonet.com> wrote:

Can someone please have a look at this diagram and advise if it is
workable?

no, you need some negative feedback, and for each op-amp you need to
keep the inputs between the supply voltages.
easiest way is probably this.

The object is to output two identical signals, one with a +12V shift
above DC/GND, and the other a -12V shift.

maybe this (view with a fixed-pitch font)

+24
| .--R-----.
| | |
| | +24 |
| | |\| |
R +12-R-+--|-\ |
| | >--+--
+----------|+/
| |/|
R 0V
|
| .--R-----.
----+ | |
| | 0V |
| | |\| |
R -12-R-+--|-\ |
| | >--+--
+----------|+/
| |/|
R -24
|
-24

where R are all resistors of the same value (eg: 10K)

--
ââ 100% natural

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David White · Feb 7, 2012

On 7 Feb 2012 08:04:58 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

no, you need some negative feedback, and for each op-amp you need to
keep the inputs between the supply voltages.
easiest way is probably this.

Sorry. I can see I have confused the issue by using op amp symbols to
represent the amps.

In practice, I intend to use two commercially supplied kits, based
upon a 10W audio amp IC (mono's) in push/pull configuration. IOW the
only ground reference is between the battery common and input signal.

Rather than the amp details, I really just wanted to know if the
battery arrangement was OK.

Once again, sorry for the inaccuracy of my OP.

David White

Jasen Betts · Feb 7, 2012

On 2012-02-07, David White <davidwhite@optonet.com> wrote:

On 7 Feb 2012 08:04:58 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

no, you need some negative feedback, and for each op-amp you need to
keep the inputs between the supply voltages.
easiest way is probably this.

Sorry. I can see I have confused the issue by using op amp symbols to
represent the amps.

In practice, I intend to use two commercially supplied kits, based
upon a 10W audio amp IC (mono's) in push/pull configuration. IOW the
only ground reference is between the battery common and input signal.

Rather than the amp details, I really just wanted to know if the
battery arrangement was OK.

yeah, that should work, if you're expecting DC performance you'll
need some sort of level shifting like I drew
else you can probably use capacitors instead,

most amplifiers don't like seeing a voltage that's not between the
supply voltages.

Once again, sorry for the inaccuracy of my OP.

but if you're using batteries another way to get what you want is this

---------------
+12 +
|\| 12V battery
| \ -
amp >------|
| / +
|/| 12V battery
-12 -
---------------

with the advantage of possibly much less energy lost in the amplifier.

--
ââ 100% natural

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Jim Thompson · Feb 7, 2012

On 7 Feb 2012 08:04:58 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2012-02-06, David White <davidwhite@optonet.com> wrote:

Can someone please have a look at this diagram and advise if it is
workable?

no, you need some negative feedback, and for each op-amp you need to
keep the inputs between the supply voltages.
easiest way is probably this.

The object is to output two identical signals, one with a +12V shift
above DC/GND, and the other a -12V shift.

maybe this (view with a fixed-pitch font)

+24
| .--R-----.
| | |
| | +24 |
| | |\| |
R +12-R-+--|-\ |
| | >--+--
+----------|+/
| |/|
R 0V
|
| .--R-----.
----+ | |
| | 0V |
| | |\| |
R -12-R-+--|-\ |
| | >--+--
+----------|+/
| |/|
R -24
|
-24

where R are all resistors of the same value (eg: 10K)

Tim Wescott · Feb 7, 2012

On Tue, 07 Feb 2012 08:45:58 +0000, David White wrote:

On 7 Feb 2012 08:04:58 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

no, you need some negative feedback, and for each op-amp you need to
keep the inputs between the supply voltages. easiest way is probably
this.

Sorry. I can see I have confused the issue by using op amp symbols to
represent the amps.

In practice, I intend to use two commercially supplied kits, based upon
a 10W audio amp IC (mono's) in push/pull configuration. IOW the only
ground reference is between the battery common and input signal.

Rather than the amp details, I really just wanted to know if the battery
arrangement was OK.

You don't show where the amplifier power is coming in. If the amplifiers
have capacitively coupled inputs and DC coupled outputs, if those DC
outputs are centered between the supply rails, if they can sustain
whatever standing current there's going to be between their outputs
without either failing to work correctly or outright frying, if U1.1 is
powered from +24V to ground, and if U1.2 is powered from ground to -24V,
then, well, maybe it'll work.

But you really can't depend on any precision to this, and you certainly
can't depend on DC coupling. (Nor, IMHO, is it safe to assume that the
amplifiers will work as desired, or even survive, if you have standing
current from one to another).

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

David White · Feb 7, 2012

On Tue, 07 Feb 2012 12:06:05 -0600, Tim Wescott <tim@seemywebsite.com>
wrote:

You don't show where the amplifier power is coming in. If the amplifiers
have capacitively coupled inputs and DC coupled outputs, if those DC
outputs are centered between the supply rails, if they can sustain
whatever standing current there's going to be between their outputs
without either failing to work correctly or outright frying, if U1.1 is
powered from +24V to ground, and if U1.2 is powered from ground to -24V,
then, well, maybe it'll work.

The diagram was meant to show that one amp is powered from +24 to +12
and the other -24 to -12.

I assume you noted this and had an objection. If so, can you please
advise further? I don't understand why this would be a problem if the
output signal is not referenced to the rails.

(Nor, IMHO, is it safe to assume that the
amplifiers will work as desired, or even survive, if you have standing
current from one to another).

Yes, this is what I am interested in finding out.

While on the subject. What about eliminating one battery per rail, and
offseting one leg with a spower diode?

That would reduce the standing current you refer to, and provide +12
and +0.5, and -12 and -0.5 rails respectively.

David

Tim Wescott · Feb 8, 2012

On Tue, 07 Feb 2012 20:15:21 +0000, David White wrote:

On Tue, 07 Feb 2012 12:06:05 -0600, Tim Wescott <tim@seemywebsite.com
wrote:

You don't show where the amplifier power is coming in. If the
amplifiers have capacitively coupled inputs and DC coupled outputs, if
those DC outputs are centered between the supply rails, if they can
sustain whatever standing current there's going to be between their
outputs without either failing to work correctly or outright frying, if
U1.1 is powered from +24V to ground, and if U1.2 is powered from ground
to -24V, then, well, maybe it'll work.

The diagram was meant to show that one amp is powered from +24 to +12
and the other -24 to -12.

I assume you noted this and had an objection.

No, actually -- I noted that the amplifier power inputs seemed to be
completely disconnected.

But now that you specify, note that an audio amplifier is probably going
to hold the output at about half way between it's positive and negative
supplies, so your standing voltages will be (very) approximately +18 and
-18, giving you 36V between the amplifier outputs.

If so, can you please
advise further? I don't understand why this would be a problem if the
output signal is not referenced to the rails.

Well, how is the output referenced in your particular amplifier? If it
just comes out a honkin' big electrolytic (which, come to think of it, is
probably the case) then all the messing around with power supplies isn't
going to do you any good at all.

(Nor, IMHO, is it safe to assume that the amplifiers will work as
desired, or even survive, if you have standing current from one to
another).

Yes, this is what I am interested in finding out.

While on the subject. What about eliminating one battery per rail, and
offseting one leg with a spower diode?

That would reduce the standing current you refer to, and provide +12 and
+0.5, and -12 and -0.5 rails respectively.

I think that, if you don't have schematics for the amplifiers in question
and know what you're doing, a better way to go would be to just do the
whole shebang with op-amps, possibly with transistor followers to provide
power.

If you are always going to be having current flowing from the + output to
the - output, and your frequencies aren't too high (f < 5kHz) you should
be able to do this easily with fairly wimpy opamps and source or emitter
followers.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

George Herold · Feb 8, 2012

On Feb 7, 3:15 pm, davidwh...@optonet.com (David White) wrote:

On Tue, 07 Feb 2012 12:06:05 -0600, Tim Wescott <t...@seemywebsite.com
wrote:

You don't show where the amplifier power is coming in. If the amplifiers
have capacitively coupled inputs and DC coupled outputs, if those DC
outputs are centered between the supply rails, if they can sustain
whatever standing current there's going to be between their outputs
without either failing to work correctly or outright frying, if U1.1 is
powered from +24V to ground, and if U1.2 is powered from ground to -24V,
then, well, maybe it'll work.

The diagram was meant to show that one amp is powered from +24 to +12
and the other -24 to -12.

Why do you want to do that?

It'd be pretty simple to capacitiviely couple a signal into two
different amps operated from different power supplies.... but why?

George H.

I assume you noted this and had an objection. If so, can you please
advise further? I don't understand why this would be a problem if the
output signal is not referenced to the rails.

(Nor, IMHO, is it safe to assume that the
amplifiers will work as desired, or even survive, if you have standing
current from one to another).

Yes, this is what I am interested in finding out.

While on the subject. What about eliminating one battery per rail, and
offseting one leg with a spower diode?

That would reduce the standing current you refer to, and provide +12
and +0.5, and -12 and -0.5 rails respectively.

David

David White · Feb 8, 2012

On Tue, 07 Feb 2012 20:54:28 -0600, Tim Wescott <tim@seemywebsite.com>
wrote:

While on the subject. What about eliminating one battery per rail, and
offseting one leg with a spower diode?

That would reduce the standing current you refer to, and provide +12 and
+0.5, and -12 and -0.5 rails respectively.

I think that, if you don't have schematics for the amplifiers in question
and know what you're doing, a better way to go would be to just do the
whole shebang with op-amps, possibly with transistor followers to provide
power.

You may feel this is a frustrated practical app rather than the
off-beat experiment it is ... but what about the diode idea described
above?

As far as the way the audio amps are wired, abeit strangely, no one
here seems to be saying it wouldn't work.

There is no ground reference at the output as such. After setting this
up, I just want to see what happens when the outputs of one amp are
connected to the other through a resistive load. And how this differs
between single ended and push pull.

Of course, if it is less than instructive, I would rather not buy the
parts.

David

Jasen Betts · Feb 10, 2012

On 2012-02-07, Jim Thompson <To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

OK. Students: Which parts in Jasen's drawing are redundant?

Most of them, here's a better way:.

all resistors 10K

.--------+-------------+----------
R | |
| |\| _|_ +
|------|+\ _
| | >---. ___ 24V
---R-+------------|-/ _|_ _
| | |/| /// |
| R | |
| `--------+-------------+--------
=== 10n
|
_|_
///

--
ââ 100% natural

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DC offset dual rail battery supply

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