Hello Gents,

I really need your help on this. I have an accelerometer whose output varies from .41V to 4.59V and I need to use an amplifier (opamp) to achieve 0-2.5V, but I am not getting favorable results. I will be using op184 and also have a 2.5V stable voltage reference in case i need it. Can someone please please help? I have been using multisim and have used inverting and noninverting amplifiers with various gains but i just cannot get results

 

Use a resistor voltage divider to multiply by .598
That gets you in a range of .2452 volts to 2.7452 volts
Then use a DC offset of .2452 volts
The result is 0 to 2.500015 volts

The numbers come out so perfect that this looks like a homework problem.

 

Can you post a link to the datasheet of the accelerometer? We need to know its output resistance.

 

ADXL278. Im using the 55g one.

 

Here is one way.

 

I would look hard at just using a 2:1 divider. Here's why:

Your end points become .205 to 2.295, or 2.04 in a 2.5 V span, or 81% providing useful data.

If your signal drops below .2V or above 2.3V you have an indication of problems.

Finally, it is simpler, cheaper, and more stable then an op amp circuit.

 

why not use a comparator along with a rail to rail op-amp.
So that whenever vin < 2.5 V,Vout = 0;otherwise Vout =2.5V.
He never mentioned about the kind of output,only that it should be 0-2.5

 

He never mentioned about the kind of output

Actually he did.

 

why not use a comparator along with a rail to rail op-amp.
So that whenever vin < 2.5 V,Vout = 0;otherwise Vout =2.5V.
He never mentioned about the kind of output,only that it should be 0-2.5

I don't think he is looking for greater-than/less-than. I think he wants to send the op amp output to an ADC.

 

Ron,

Overall, you have hit the issue on the nail. This is exactly what I needed for an ADC input. Maybe u can help me understand the logic of it all as well. I really want to know how uou derived this since I also have an LM35d question where i want to measure -40 to 175c and change to 0-2.5v out.

 

Ron,

Overall, you have hit the issue on the nail. This is exactly what I needed for an ADC input. Maybe u can help me understand the logic of it all as well. I really want to know how uou derived this since I also have an LM35d question where i want to measure -40 to 175c and change to 0-2.5v out.

Your desired output is related to the input as follows:
\(Vout=Vin(\frac{2.5}{4.59-0.41})-0.24522\)

My op amp circuit output as a function of its inputs is as follows:
\(Vout=V1(\frac{R3+R4}{R3})(\frac{R2}{R1+R2})-V2(\frac{R4}{R3})\)

Let V1=Vin, and V2=2.5

Now,
\(Vout=Vin(\frac{R3+R4}{R3})(\frac{R2}{R1+R2})-2.5(\frac{R4}{R3})\)

We can see by inspection that
\(2.5(\frac{R4}{R3})=0.24522\)

Therefore,
\(R4=.098086(R3)\)


Substituting,
\(Vout=Vin(1.09896)(\frac{R2}{R1+R2})-0.24522\)

\(Vin(1.09896)(\frac{R2}{R1+R2})=Vin(\frac{2.5}{4.59-0.41})\)

\(\frac{R2}{R1+R2}=0.54423\)

\(R2=1.1941(R1)\)

The last step is to use an online resistor ratio calculator such as this one to choose the best values of R1 and R2, and R3 and R4.