Voltage follower with variable offset?

[ws6transam]

I'm stumped tonight as I thought this would be a simple circuit. I'm trying to find the simplest method that involves the least amount of parts. If anyone has the time, and the solution looks easy to you (which to me it does not), would you take a crack at it?<p>I've got a temperature input signal that varies from zero to ten volts, which covers a thousand degrees C. This oven application only requires me to measure temperatures for the last five hundred degrees, which translates into an output voltage of five to ten volts. I want to send this five volt signal span to an A/D converter, which has an input requirement of zero to five volts.<p>So, my voltage follower only needs a unity gain, but needs to be able to offset the signal by -5 volts. These are all single-ended voltages, and both the temp sensor and the A/D converter are referenced to ground. I have +12 volt power available, but I do not, as of yet, have a negative voltage source.<p>Which seems to be the simpler solution?<p>a) Somehow create the non-inverting voltage follower with a 'virtual' ground on the negative input pin that sits at five volts, such that the output is 'zero' when the signal input is five volts ? <p>b) Introduce a negative voltage by using a LT1026 to create a -5V voltage potential. Bring in the signal and the -5V into an interting summing amp configuration, then invert with a second op-amp. <p>Option 'B' seems to make the most sense to me, but it sure does use a lot of expensive parts by doubling my op-amp requirement, and adding that $4 inverter into the mix.<p>Forgive my ignorance, I'm way out of practice on op-amp design. I hope these options dont look *too* dumb, but I'm stuck!<p>[ January 05, 2005: Message edited by: D Burk ]</p>

[Chris Smith]

If your equipment doesn’t share a common ground, you can make a voltage divider from two or more resistors and center tap them for a Zero reference point or ground, and below the tap will be less potential than this point, or a negative voltage. Above the center point is the positive.

[Ron H]

Your function is Vout=Vin-5.
You can do this with an op amp, a 5 volt reference (the A/D's reference?), and two pairs of equal-valued, well-matched resistors. One pair forms a voltage divider of your temperature signal, with the center tap connecting to the non-inverting input. The other matched pair connects in series from the 5V reference to the op amp output, with the center tap connecting to the inverting input. You may need a voltage follower between the temperature signal and the voltage divider, depending on the source impedance of the temperature signal.

[grant fair]

Steve Ciarcia (in Circuit Cellar Ink) wrote an article on temperature monitoring in Feb 1993, issue 31. He shows a signal conditioning schematic. Its simply a unity gain buffer, followed by a gain stage if you want it, followed by an offset stage. It uses a total of 3 op amps (two LM358's would do the job), or just two op amps if you omit the gain stage. It requires positive and negative supplies. I added an ICL7660 to provide the negative rail. I have a pcb pattern for the 3 opamp version with a 7660 which I would share. While this may not have the lowest parts count it's an easy solution.<p>Incidentally the schematic mistakenly has the gain and offset stages in reversed order, as a later issue points out.<p>Grant

[rocket scientist]

Ron H's solution reminds me of the pre-digital days when op-amps were amplifiers used in analog computers for arithmetic operations -- in this case subtracting 5. Hence the name op-amp, of course.

[cato]

If you want simple, how about 2 100K resisters in the form a voltage divider. Knock your 10 volt temperature signal down to 0 to 5 and take it directly (or buffer it if you like) into your A/D converter. If its a 12 bit converter, that should still give you 2048 counts over the 500 degree range for 1/4 of a degree resolution. <p>If its only an 8 bit conveter, then even if you condition the signal for 500-1000 over the 0-5 input range, thats only a resolution of 2 degrees, which probably isn't good enough for stable control (if thats what your doing)<p>By the way. It sounds like there is already a signal conditioner in this set up, probably converting a signal from a thermocouple to your 0 to 10V signal. Are you sure it can't be reconfigured to give you 0-5 or even rescaled so the 0-5 represents 500-1000?

[MrAl]

Hi there D Burk,<p>Im not sure if Chris was talking about this or
not, so...<p>The simplest way for sure is to use two resistors
to divide the +12 power supply down to 5 volts.
You would then connect the ground of the ad
converter to the center tap of the two resistors
(+5 volts). Doing this would make the signal
that spans 5 to 10v look like 0 to 5v to the
ad converter.
A 10uf cap in parallel with a 0.1uf cap across
the bottom resistor would reduce noise and
power supply variations.<p>Pitfalls are if the +12v supply isnt well
regulated over temperature there will be some
inaccuracy with changing ambient, so depending
on your accuracy requirements you may need to
use a reference diode as the bottom resistor
adjusted to +5 volts exactly.<p>Please let us know what you end up doing and
how well it works out if you can...<p>Take care,
Al

[jimandy]

What if you invert the 0-10v input, clip the resulting 10 - 0 v with a diode, then re-invert that? - This makes so much sense to me that I know there must be a hole in it wide enough for flaming arrows.

[rshayes]

As RonH described, this can be done with a single op amp. This assumes that the source impedance is low enough. If not, a unity gain buffer can be added.<p>Look for an op amp that can work down to its negative supply rail. Examples of these would be the CA3130 or CA3160. These have a MOSFET output stage that can get very close to the supply rails for light loads. National Semiconductor also makes a few op amps of this type. Look for ones that feature "rail-to-rail" operation.<p>If close (a few tenths of a volt) is good enough, then you might be able to use the LM324 or something similar.

[ws6transam]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by jimandy:
What if you invert the 0-10v input, clip the resulting 10 - 0 v with a diode, then re-invert that? - This makes so much sense to me that I know there must be a hole in it wide enough for flaming arrows.<hr></blockquote><p>Thanks for the suggestion Jim (or is it Andy?) However, with the low output impedance of the first inverting amplifier, would there not be a significant current draw as the zener clamps the output to 5V while the op-amp tries to drive the output to 10 volts? Of course, a limiting resistor could be placed onto the output of the first stage, I suppose. <p>Thanks everyone for the ideas. I will certainly look at all of the options, and when I arrive at the solution, I'll be sure to pass it along!

[ws6transam]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by RonH:
Your function is Vout=Vin-5.
.....
(stuff edited here)
<hr></blockquote><p>Thanks Ron, I'm looking over this solution and doing the math in order to figure out the transfer function right now.<p>Let me see if I understand this: The original function of our circuit is indeed supposed to be Vout = Vin-5. <p>We have the input to the non-inverting op-amp set to Vin/2, via the voltage divider. We therefore have at the input of the amplifier the function Ein = Vin/2<p>Let's call the resistor from the 5V reference to the negative input Rin. Ein' which is the voltage developed across this resistor is equal to the input voltage of the amplifier minus the 5 volt reference voltage, which is Ein - 5, assuming an ideal op-amp. Therefore the current through this resistor is (Ein-5) / Rin, and we'll call it Iin.<p>Now, since the input current to the negative side of the ideal op-amp is zero, the current Iin must flow through the matched resistor to the output side of the op-amp. This matched resistor, which we'll call Rf is the same value as our Rin.<p>[edit: My math that was wrong.]<p>Here's the revised equations:<p>Ein' equals Ein - 5V<p>If equals Iin, which is the curent that flows through the second resistor into the amplifier output.<p>If = (Vout - Ein)/Rf.
Rf = Rin.
Substitited, If = (Vout - Ein)/Rin. <p>As defined,
In equals (Ein-5V)/Rin
If equals (Vout-Ein)/Rin
In equals If
Thus, (Ein-5V)/Rin = (Vout-Ein)/Rin
Multiply both sides of the equation by Rin
Ein - 5V = Vout - Ein
2Ein - 5V = Vout
Ein = Vin/2
Thus,
Vin - 5V = Vout
or, <p>Vout = Vin - 5V
which is our desired transfer function.<p>Thanks, Ron!<p>[ January 06, 2005: Message edited by: D Burk ]<p>[ January 06, 2005: Message edited by: D Burk ]</p>

[jimandy]

Oops . Thanks for filing in the missing word "zener" in my suggested solution and, yes, a limiting resistor would be necessary - just like in a zener voltage regulator circuit.
BTW, I suppose you could sub a pot in the first inverter opamp feedback path so you could twiddle with fine tuning.

[Ron H]

You can also do it by classic op amp gain equations:
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">code:</font><hr><pre>
attenuator non-inv. gain inv. gain
|| || ||
Vout=Vin*R2/(R1*R2)*(R3+R4)/R3 - Vref*R4/R3
</pre><hr></blockquote><p>Where R1 and R2 are the input voltage divider, R4 is the feedback resistor, and R3 is the resistor from Vref to the inverting input.<p>[ January 06, 2005: Message edited by: RonH ]<p>[ January 06, 2005: Message edited by: RonH ]</p>

[MrAl]

Hi again,<p>You dont need an op amp, just two resistors,
as long as the two grounds dont have to
be tied together.<p>Take care,
Al