Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Instrumentation Op Amp Question

Status
Not open for further replies.
I have a chip that has an output of 10mV/degC, I'd really like to have 10mV/degF. I know the conversion is degF= 1.8 *degC + 32. I'm having some difficulty figuring how to implement the +32. With a two stage design, I can set the first IA to a gain of 1.8 and add the 32 in the second stage. My first thought is the 1.8*degC input is on in+ and place -320mV on in-. Is that right?

Can I do this in one stage? Apply the 10mV/degC signal to in+ and apply -320mV to in-? ?r do I need to compensate the 320mV by the gain and only apply 177.8 mV (320/1.8).
 

MikeMl

Well-Known Member
Most Helpful Member
Why not just buy an **broken link removed**?

Single op-amp stage with a gain of 0.8. That makes the gain from the non-inverting input to output 1+0.8 = 1.8. Assuming the Centigrade sensor (LM35?) puts out 0V at 0degC, then you would want the Fahrenheit sensor to output 0.32V. Since the gain from the resistor tied to the inverting input to output is -0.8, you would want to apply 0.32 / -(0.8) = -0.4V.
 
Why not just buy an **broken link removed**?

Single op-amp stage with a gain of 0.8. That makes the gain from the non-inverting input to output 1+0.8 = 1.8. Assuming the Centigrade sensor (LM35?) puts out 0V at 0degC, then you would want the Fahrenheit sensor to output 0.32V. Since the gain from the resistor tied to the inverting input to output is -0.8, you would want to apply 0.32 / -(0.8) = -0.4V.

Short answer. Lack of range. I need wider temperature range than the LM34 will provide. I'm not using an LM35, I have a K-type thermocouple and match linearizing amplifier (AD594) that provides the 10mV/DegC output.

I guess the basis of my question is, is the offset applied to the other input multiplied by the gain of the amp or does it pass straight through? Does the amp work to provide input times gain then add offset, or does it add the offset to the input and then apply gain to the total?

The second question I have is, in non-inverting mode (input to in+) the offset will be applied to in-. To increase the offset on the output, do I apply a positive or negative voltage to in-?
 

smanches

New Member
Where is the output of the op amp going? Can't you add the 32F in digital circuitry or code someplace? It's only needed for humans anyway, which assumes you are going to be displaying it. Can't you add it just before the display circuitry?
 

MikeMl

Well-Known Member
Most Helpful Member
...
I guess the basis of my question is, is the offset applied to the other input multiplied by the gain of the amp or does it pass straight through? Does the amp work to provide input times gain then add offset, or does it add the offset to the input and then apply gain to the total?

The second question I have is, in non-inverting mode (input to in+) the offset will be applied to in-. To increase the offset on the output, do I apply a positive or negative voltage to in-?

Read my answer again. I answered both questions.
 
Read my answer again. I answered both questions.

Mike,

The example you give doesn't seem to apply to an IA, so I'm having a hard time (due to my rustiness) understanding. There's no resistor from the inverting input to the output. The gain is set by applying a resistance between the two Rg pins.
 

crutschow

Well-Known Member
Most Helpful Member
Mike,

The example you give doesn't seem to apply to an IA, so I'm having a hard time (due to my rustiness) understanding. There's no resistor from the inverting input to the output. The gain is set by applying a resistance between the two Rg pins.
The confusion, I believe is in your use of the term "Instrumentation Op Amp". You apparently have an Differential Instrumentation Amp which is different from an op amp (although instrumentation amps can be made using several op amps). Mike's comments were referring to a standard op amp.

Instrumentation amps are usually used where you need a differential input with good common mode rejection and a fixed gain. The are not usually used for summing or level shifting signals (but they can be). Typically their gain is set with a single resistor between two pins which are separate from the input and output pins.
 
You are correct. I don't have a straight "op amp" I have an instrumentation amp, INA128 and 129 to be specific.

After much painful experimenting (because we aren't well setup for temporary prototyping of circuits) I have figured that the IA implements the equation 1.8*degC + (32/1.8). Thus the voltage required on the IN- pin is -0.177. -177mV*-1 (in-) *1.8 (Gain) =320mV, which at 10mV/degF is 32degF.
 
Status
Not open for further replies.

Latest threads

Top