Scaling 0-10VDC to 1-5VDC?

Thread Starter

boulderjoe

Joined Nov 21, 2011
5
Hi, I've been lurking for a while in here but can't seem to find an elegant combination for what I'm trying to do.

I have a 0-10VDC output from an instrument I'd like to interface to a 1-5VDC readout. I could use a simple voltage divider if I could use 0-5V, but I'm getting hung up combining the scaling with the offset. Can I use an opamp to add 1V and scale my signal to a 0-4V scale?

Any suggestions?

Thanks, Joe
 

thatoneguy

Joined Feb 19, 2009
6,359
How much current did you need? Is this for Analog to Digital conversion or other purpose?

What is the accuracy required?

One method that comes to mind is a trans-conductance amplifier, so the output current scales with the input voltage.

Put the output current through a precision resistor to produce a 1V-5V voltage across that resistor based on output current, which is directly related to input voltage. Thinking instrumentation 4-20mA through a 250 Ohm resistor will give the 1-5V output.

However, this depends on what you will be using the 1-4V for, if more than a few mA are required, that solution won't work.
 

Thread Starter

boulderjoe

Joined Nov 21, 2011
5
Thanks for the quick reply. I don't think current is an issue. It's for A-D purposes. A pressure gauge with a 0-10V analog signal going to a different manufacturer's pressure readout, with a 1-5V analog input. It then just converts the analog signal to a digital display.

The gauge is 0-1000Torr scaled to 0-10V, so mV accuracy is about all that's needed (0.1 Torr)
 

thatoneguy

Joined Feb 19, 2009
6,359
How much experience do you have with electronics?

There is an IC built specifically for this: XTR110

Not a cheap solution, but a precision one. ($15+ support components and power supply)
 

SgtWookie

Joined Jul 17, 2007
22,230
Here is one way to do it. See the attached.

The LMC6484A is a quad opamp with RRIO (rail-to-rail inputs and outputs) that's reasonably fast with a reasonably low input offset.

U1a is simply used as a voltage follower/buffer, to present a high impedance to your instrument and a low impedance to the voltage divider that follows. Since you have not specified the output impedance of your instrument, I've assumed that it is high, and that a load will affect the accuracy of the output.

The R1/R2 divider gives 4v out for 10v in. U1b's output follows/buffers the output from the R1/R2 divider.

The divider R3/R4 gives 1v out. This depends on the 12v supply and the resistors themselves being accurate. You can use a more accurate voltage reference if you'd like. U1c's output follows/buffers the R3/R4 divider.

R5/R6 are a voltage averaging (summing) network. The voltage at the noninverting (+) input of U1d is (U1b+U1c)/2. Since U1c's output is always 1v, and U1b's output can range from 0v to 4v, the noninverting input of U1d ranges from 0.5v to 2.5v.

U1d multiplies the above average voltage by two, resulting in 1v to 5v output from 0v to 10v input. Three voltage followers/buffers and one multiply by two.
 

Attachments

thatoneguy

Joined Feb 19, 2009
6,359
I could be overthinking this as well.

An op amp with a gain of 2/5 and a +1V bias should work, you would need a rail to rail op am, though.
 

SgtWookie

Joined Jul 17, 2007
22,230
An op amp with a gain of 2/5 and a +1V bias should work, you would need a rail to rail op am, though.
That's just what the circuit I posted does - along with providing buffering for the input.

I didn't show bypass capacitor(s) across the supply pins for the opamp, but they are required.
 

thatoneguy

Joined Feb 19, 2009
6,359
That's just what the circuit I posted does - along with providing buffering for the input.

I didn't show bypass capacitor(s) across the supply pins for the opamp, but they are required.
I didn't refresh the page between my last post and your post. I always forget to do that when I leave a window open and come back to it. :(
 

Thread Starter

boulderjoe

Joined Nov 21, 2011
5
My experience is fairly basic and self taught (enough to get me in trouble?). I'm a ChemE, but never had a full circuits class. I do just fine with relays and most automation stuff. Watchdog ICs I can deal with. I get fuzzy when we move towards complex arrangements.

I'll look through the suggestions and post again later tonight. Thanks!
 

SgtWookie

Joined Jul 17, 2007
22,230
I guess it can look sort of complicated.

It's really just several electronic "building blocks" put together. Three of the opamp channels are simply voltage followers/buffers; that configuration is the least complex of any of them. The last channel has a gain of two, using just two resistors.

Everything else is pairs of resistors used as voltage dividers or summers. It looks like there are lots of parts to deal with, but there are actually just one physical IC, 8 resistors, and one capacitor that isn't shown.

It might help you to have a look at our E-book in the Operational Amplifier section:
http://www.allaboutcircuits.com/vol_3/chpt_8/3.html

I honestly don't know a way to make a circuit that is more simple which will do what you need to do, with the information you have given to us.
 

strantor

Joined Oct 3, 2010
6,782
If this is an industrial application (where amateur-looking homebrew circuit boards in machine panels are frowned upon) and you have plenty of company money to burn, this device will take a 0-10V input and output a 4-20mA signal. Put a 250ohm resistor across the output and you have a 1-5V output.
 

Thread Starter

boulderjoe

Joined Nov 21, 2011
5
Thanks everyone! I've got a handle on it now.

@strantor: you weren't kidding about needing money to burn, I can't believe that guy is $300.
 

SgtWookie

Joined Jul 17, 2007
22,230
Thanks, Ron_H - I figured it could be done with fewer parts, but I don't seem to be firing on all cylinders for the last week or so.

It would be helpful to our OP and others if you provided details how to "fine tune"/calibrate the circuit in order to compensate for component tolerances.
 

SgtWookie

Joined Jul 17, 2007
22,230
@strantor: you weren't kidding about needing money to burn, I can't believe that guy is $300.
Considering how much it would cost to pay an engineer to design a functionally equivalent general-purpose device and then have it built, $300 for an off-the-shelf device is a deal!
 

Ron H

Joined Apr 14, 2005
7,063
Thanks, Ron_H - I figured it could be done with fewer parts, but I don't seem to be firing on all cylinders for the last week or so.

It would be helpful to our OP and others if you provided details how to "fine tune"/calibrate the circuit in order to compensate for component tolerances.
There's always a gotcha.:eek:
As I think you realize, all the resistors interact. That is the advantage of your circuit.
I provided the simpler circuit partly as an intellectual exercise, and partly in case the accuracy of 1% resistors would be adequate. I realize that I did not put 1% resistors in the schematic. The ratios of the 3 resistors need to remain as shown on the schematic when picking values.
 

SgtWookie

Joined Jul 17, 2007
22,230
Yep, I realize that.
But, here are some values our OP could use to either hit or get very close to 129.17 Ohms:
Rich (BB code):
E96 resistance values for series/parallel combinations to get 129.17 Ohms
25 combinations found                                   % of
                   R1             R2          Total R   Error
                 -------        -------      --------   -----
          1)      137.00 ||     2260.00 =      129.17,  0.000%
          2)      150.00 ||      931.00 =      129.19,  0.012%
          3)       38.30  +       90.90 =      129.20,  0.023%
          4)       31.60  +       97.60 =      129.20,  0.023%
          5)       16.20  +      113.00 =      129.20,  0.023%
          6)      200.00 ||      365.00 =      129.20,  0.026%
          7)      133.00 ||     4530.00 =      129.21,  0.028%
          8)      169.00 ||      549.00 =      129.22,  0.040%
Of course, we don't know what kind of accuracy they require - and if they were reading it using a uC, they could just calibrate out errors via the software.
 

thatoneguy

Joined Feb 19, 2009
6,359
Thanks everyone! I've got a handle on it now.

@strantor: you weren't kidding about needing money to burn, I can't believe that guy is $300.
To be fair, that device can operate in current in/voltage out, voltage in/voltage out, or voltage in/current out modes, all with 180V of isolation, so there is quite a bit more design that has gone into it, as well as parts, not to mention $20 worth of dust/grease/moistureproof enclosure for industrial environment (which I assumed you were in when I suggested the 4mA-20mA). Standards and all of that.

If this is in an industrial setting, be sure to put the schematic inside the enclosure and the function of the unit on the outside of the enclosure for anybody that may need to look at it in the future, please! Possibly offer buffered 0-10V and 1-5V outputs of the enclosure as well, in the event that pressure reading is wanted for another use later down the road.
 

strantor

Joined Oct 3, 2010
6,782
If this is in an industrial setting, be sure to put the schematic inside the enclosure and the function of the unit on the outside of the enclosure for anybody that may need to look at it in the future, please!
Yes, nothing more frustrating than coming upon some unknown, unlabeled beast whilst troubleshooting a machine and have to dissect the thing just to understand what it is/what it does/why its there before you can move on. For your application, the function of the board would be fairly obvious (a signal conditioner), but not the specifics of it. Things like this save downtime, and downtime always comes when you least want it.
 

Ron H

Joined Apr 14, 2005
7,063
Yep, I realize that.
But, here are some values our OP could use to either hit or get very close to 129.17 Ohms:
Rich (BB code):
E96 resistance values for series/parallel combinations to get 129.17 Ohms
25 combinations found                                   % of
                   R1             R2          Total R   Error
                 -------        -------      --------   -----
          1)      137.00 ||     2260.00 =      129.17,  0.000%
          2)      150.00 ||      931.00 =      129.19,  0.012%
          3)       38.30  +       90.90 =      129.20,  0.023%
          4)       31.60  +       97.60 =      129.20,  0.023%
          5)       16.20  +      113.00 =      129.20,  0.023%
          6)      200.00 ||      365.00 =      129.20,  0.026%
          7)      133.00 ||     4530.00 =      129.21,  0.028%
          8)      169.00 ||      549.00 =      129.22,  0.040%
Of course, we don't know what kind of accuracy they require - and if they were reading it using a uC, they could just calibrate out errors via the software.
And 620k is not a standard value, although 619k is.
 
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