How do I convert ac voltage to +, - and GND differential

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I have a controller circuit that outputs 10V peak to peak AC on a
single wire. I need to convert this to 3 lines - GND, +5v and -5v for
input to a laser galvanometer amp.

(This is all low current stuff).

I was told that either a differential chip will do the job, or an op-
amp wired as an invertor. Can someone suggest either a suitable chip,
or show me how I wire an op-amp up to do this?

Analog electronics is a complete mystery to me...

Thanks,
Dave

 

[John Fields]

I have a controller circuit that outputs 10V peak to peak AC on a
single wire. I need to convert this to 3 lines - GND, +5v and -5v for
input to a laser galvanometer amp.

(This is all low current stuff).

I was told that either a differential chip will do the job, or an op-
amp wired as an invertor. Can someone suggest either a suitable chip,
or show me how I wire an op-amp up to do this?

Analog electronics is a complete mystery to me...

---
If you can stand the diode drop, use Schottkys and view in Courier:


ACIN>--+--[DIODE>]---------+----> +OUT
| |
+--[<DIODE]--+------|----> -OUT
| |+
[BFC] [BFC]
|+ |
ACIN>---------------+------+----> GND

 

[John Popelish]

I have a controller circuit that outputs 10V peak to peak AC on a
single wire. I need to convert this to 3 lines - GND, +5v and -5v for
input to a laser galvanometer amp.

(This is all low current stuff).

I was told that either a differential chip will do the job, or an op-
amp wired as an invertor. Can someone suggest either a suitable chip,
or show me how I wire an op-amp up to do this?

Analog electronics is a complete mystery to me...

No kidding. ;-) Well, that is why this group exists.

First of all, you cannot have a signal on a single wire
dangling out there in space. All signals are the current
through a path (circuit) or a potential difference between
two nodes. So your 10 volt peak-to-peak signal implies some
zero volt reference node with respect to that it is
measured. So that is two wires.

It sounds to me that you need to rescale this signal to some
other voltage range and also send it (and its ground
reference) to a galvanometer along with + and - 5 volt
supply lines, to power the galvanometer. So that is 4 wires.

Please confirm that I have not lost the gist of what you are
asking about, before I go into the means to rescale the signal.

 

[Jamie]

I have a controller circuit that outputs 10V peak to peak AC on a
single wire. I need to convert this to 3 lines - GND, +5v and -5v for
input to a laser galvanometer amp.

(This is all low current stuff).

I was told that either a differential chip will do the job, or an op-
amp wired as an invertor. Can someone suggest either a suitable chip,
or show me how I wire an op-amp up to do this?

Analog electronics is a complete mystery to me...

Thanks,
Dave

You can use a pair of diodes, ( half wave)/
cathode and anode of 2 diodes will connect to the
single line. the common being the other side/
you would need a couple of caps to filter it how
ever.
there will be a drop in voltage how ever of
~ 0.6 volts.

 

[BobW]

You can use a pair of diodes, ( half wave)/
cathode and anode of 2 diodes will connect to the
single line. the common being the other side/
you would need a couple of caps to filter it how
ever.
there will be a drop in voltage how ever of
~ 0.6 volts.

Please explain how this will convert a single-ended signal to a differential
pair?

Bob

 

[default]

I have a controller circuit that outputs 10V peak to peak AC on a
single wire. I need to convert this to 3 lines - GND, +5v and -5v for
input to a laser galvanometer amp.

(This is all low current stuff).

I was told that either a differential chip will do the job, or an op-
amp wired as an invertor. Can someone suggest either a suitable chip,
or show me how I wire an op-amp up to do this?

Analog electronics is a complete mystery to me...

Thanks,
Dave

A precision rectifier (op amp circuit that converts AC to DC and
automatically eliminates the diode drop to the signal) (Op Amp
cookbook has the circuits and explanations)

The op amp output needs to feed a differential output op amp if the
signal needs to be isolated from ground. (Cookbook again)

Finally, you may want to add a gain stage in there somewhere to adjust
zero (offset) and gain (span) and to get the original signal back to
the correct polarity - if the other stages invert the signal)

"Laser galvanometer" You'd want to use an efficient low current galvo
or select an op amp capable of driving the galvanometer you have.- not
all galvanometers are low current devices.

Op Amp Cookbook

 

[John Fields]

Please explain how this will convert a single-ended signal to a differential
pair?

---
The OP didn't ask for a differential pair, he asked for something
like this: (View in Courier)

ACIN>--+--[DIODE>]---------+----> +OUT
| |
+--[<DIODE]--+------|----> -OUT
| |+
[BFC] [BFC]
|+ |
ACIN>---------------+------+----> GND

 

[John Popelish]

---
The OP didn't ask for a differential pair, he asked for something
like this: (View in Courier)

ACIN>--+--[DIODE>]---------+----> +OUT
| |
+--[<DIODE]--+------|----> -OUT
| |+
[BFC] [BFC]
|+ |
ACIN>---------------+------+----> GND

I am not so sure. I am beginning to think he is asking for
a bridged amplifier output, to drive a galvanometer coil,
directly. But that would not need a ground. I haven't
gotten any clarification back from him.

 

[Jamie]

Please explain how this will convert a single-ended signal to a differential
pair?

Bob

rather simple.
You specified AC, this assumes you have a + & - signal with respect to
an axes point, the base line. So in turn, you have a + and - with a
common leg. The use of 2 diodes are polarized to pass the 2 different
polarities to generate a half wave voltage pairs. The caps are there
to help maintain the voltage..

or did you make an error and actually mean you have a 0..10 volts Peak?