I'm just doing bench work for a large PLC job.
The servo module exports -10 to + 10 VDC to control speed and direction of the drive. But my drive on the bench only supports 0 to +10VDC (the final will use the -10 to +10) It will work as 0-5 being reverse and 5-10 being forward speeds, and 5dvc being stopped.

I feel I have two options:
1 - Build an amp that will drive a small hobby DC motor to spin the encoder. (the output of the servo module won't have enough current to drive the motor outright)

2 - Build an amp that converts the -10 to +10 where Zero volts in the input side would equal 5volts on the output side.

I'd prefer option 2, that way the safeties and drive enables, etc can also be tested.
I'm *rather* analog challenged. help?
Thanks,
Dave

 

The simplest circuit is an analog voltage divider. Basically 2 resistors. You need a 10V supply.

The left plot shows the output, the right the input.

This won't protect you against exceeding the limits, and could apply a negative voltage to the servo if for example you apply ±12V. If this is a problem, let me know.

 

Thanks!

I'm a bit confused.
The output signal is a DC voltage; neg for reverse and positive for forward.
And amplitude defines the speed it will run.

I don't see how a simple voltage divider is going to invert the neg. (but then like I said, I'm analog challenged)
What I need is:
Input of -10 needs to equal zero on output
Input of 0 to equal +5 on output
Input of +10 to equal +10 on output

But I'll pay on a breadboard.

 

Thanks!

I'm a bit confused.
The output signal is a DC voltage; neg for reverse and positive for forward.
And amplitude defines the speed it will run.

I don't see how a simple voltage divider is going to invert the neg. (but then like I said, I'm analog challenged)
What I need is:
Input of -10 needs to equal zero on output
Input of 0 to equal +5 on output
Input of +10 to equal +10 on output

But I'll pay on a breadboard.

It doesn't do any inverting - it just modulates an already existing +10V supply. If you're clever, you could even generate the 10V off the input signal, though it would be tricky.

Simulate it on a breadboard and it should work. It's just a voltage divider with an offset that can move. It divides by two, to get -5V to +5V. The 10k upper resistor and power source then shift it up by +5V to get you a 0V to +10V signal.

 

From what I see, if you overlap the two signals you are getting exactly what you want. Think of the sign wave as the variable pot.
When Input = -10V, Output = 0V
When Input = 0V, Output = +2.5V
When Input = +10V, Output = +5V

iONic

 

From what I see, if you overlap the two signals you are getting exactly what you want. Think of the sign wave as the variable pot.
When Input = -10V, Output = 0V
When Input = 0V, Output = +2.5V
When Input = +10V, Output = +5V

iONic

The output is actually 10V for 10V in, 5V for 0V in and 0V for -10V in - you divided the signals by two.

 

From what I see, if you overlap the two signals you are getting exactly what you want. Think of the sign wave as the variable pot.
When Input = -10V, Output = 0V
When Input = 0V, Output = +2.5V
When Input = +10V, Output = +5V

iONic

Thanks, but there is only one signal, and it needs to be changed.
Device A gives -10 to +10
Device B only accepts 0 to +10

the 'second' signal is the adapter I'm trying to make.

I'm gonna play with the divider method and see what happens.

 

Thanks, but there is only one signal, and it needs to be changed.
Device A gives -10 to +10
Device B only accepts 0 to +10

the 'second' signal is the adapter I'm trying to make.

I'm gonna play with the divider method and see what happens.

This is exactly what my circuit does. It converts -10V into 0V, 0V into +5V, and +10V in +10V.

You could also try an inverting op-amp and a DC offset, but it's far more complicated.

 

The output is actually 10V for 10V in, 5V for 0V in and 0V for -10V in - you divided the signals by two.

I think I calculated correct, when you take into consideration I was only "half" awake!

Yes, you are right, A simple voltage devider is the ideal testing solution.

i

 

Tom, thanks! If I'm ever over that side of the pond I owe you a pint!

Took me a bit to figure out what was going on. It hit while gathering up the parts.
I was looking, incorrectly, to convert/invert the signal. Yours needs a steady 10V, not a prob for this project (since its just bench work with correct levels in the field.)

I was thinking of a single supply solution, once again over complicating things.
(and sorry for the massive delay)

 

Tom, thanks! If I'm ever over that side of the pond I owe you a pint!

Took me a bit to figure out what was going on. It hit while gathering up the parts.
I was looking, incorrectly, to convert/invert the signal. Yours needs a steady 10V, not a prob for this project (since its just bench work with correct levels in the field.)

I was thinking of a single supply solution, once again over complicating things.
(and sorry for the massive delay)

No problem, glad to hear!