How to combine a voltage offset with an amplification ?

Hello,

I need to amplify the ouput of a Hall effect sensor but in a non-symetrical fashion. The sensor is in a situation where I want to amplify its maximum - which currently reaches something like 3.75 V - to 5 V. However I do not need to keep the sensor's minimum (0 V) as the smallest sensed value in my circuit won't go below, say, 1.5 V. So what I need is a circuit to translate the 1.5 V - 3.75 V range into 0-5 V. Therefore I need to introduce some offset and I would like to how I do it and if I need to do it before or after amplification.

Thank you in advance.

This is exactly the situation where the opamp inverting or differential amplifier circuits are used. The differential (4 resistor) circuit
is the most adaptable / flexible.

I don't get it. A differential amplifier senses the difference between two voltage sources. What would be the two sources here ?

No, I mean the opamp differential amplifier circuit - Operational amplifier applications - Wikipedia

[ note that the resistor they show (Rg) going to ground can go to any reference voltage you like
and generate an offset voltage at the output ]

I understand the differential op-amp allows one to introduce an offset but how do I connect this circuit to a single sensor ?

amundsen:
Hello,

I need to amplify the ouput of a Hall effect sensor but in a non-symetrical fashion. The sensor is in a situation where I want to amplify its maximum - which currently reaches something like 3.75 V - to 5 V. However I do not need to keep the sensor's minimum (0 V) as the smallest sensed value in my circuit won't go below, say, 1.5 V. So what I need is a circuit to translate the 1.5 V - 3.75 V range into 0-5 V. Therefore I need to introduce some offset and I would like to how I do it and if I need to do it before or after amplification.

Thank you in advance.

Many op-amp circuits can best be described by the math function you are having them simulate. In your case you need to have an offset input voltage of -1.5vdc and the amp gain set to 2.222, this will result in a input voltage range of 0-2.25 and a output voltage range of 0-5 vdc.

There are lots of op-amp tutorials around that you can see the method of how to set the gain of a stage and how to add offsets.

Lefty

Thank you everyone. I have found another discussion about the same subject here :

amundsen:
Thank you everyone. I have found another discussion about the same subject here :

operational amplifier - Analog voltage level conversion (level shift) - Electrical Engineering Stack Exchange

That ckt looks good, and the website looks useful. Now that you've found it, notice the
ckt is virtually identical to the one Mark indicated,

One thing you want to pay attention to, is that there is a gain term in the offset pathway,
as well as in the signal pathway. This shows up as the -(Rf/R1)*V1 term on Mark's page.
IOW, if you set the amp up with some gain > 1, you'll need to knock down the offset
voltage to compensate [ie, lower the R2 value from your page], else your opAmp output
voltage will saturate. You'll see.

I simulated Olin's circuit using an LT1677 rail-to-rail op-amp and it linearly maps 1.5V - 3.5V input to a full 0 - 5V output range. Anything less than 1.5V input is 0V output and anything greater than 3.5V input is 5V output.

If you look closely, you'll see that it isn't quite like the circuit in the wiki.

EDIT: After tinkering with Olin's circuit, I think I have it figured out. At the beginning, the output is low because the inverting (-) input is at 2.5V reduced by the voltage divider made by the 150K and 100K resistors. This places the inverting input bias at 1.5V because 1V is dropped by the 100k resistor and the remaining 1.5V is dropped by the 150k resistor. Until the non-inverting (+) input is raised above 1.5V, nothing basically happens. Once it is above 1.5V, the output begins to rise at a rate that is 150% faster than the input; this is due to the 2.5X multiplication effected by the 150k and 100k resistors. The input voltage ramp crosses the output voltage ramp right at 2.5V and proceeds all the way to 5V as the input reaches 3.5V. Pretty slick, but that's the kind guy Olin is. I've been witnessing his work for more than a decade on the PICLIST. He's not exactly a Dale Carnegie personality, but he really knows what he is doing.

EDIT2: The "formula" for the operation seems to be Ov = (Iv * 2.5) - 3.75, where Iv is the input voltage and Ov is the output voltage.

Thanks afremont for simulating Olin's circuit and finding the formula!