How to remove unwanted voltage?

[sid2286]

Hello,

I have a input from -200mV to 1.5V, I want to use the input only from 100mV to 1.5V.
I don't want the input below 100mV, how do I remove the unwanted input?

completely clueless how the circuit would be?
Thanks,
Sid

 

[Harald Kapp]

That depends on what out put the circuit has to deliver for voltages <100mV. There are several transfer function I can imagine:

1. Vout=0 for Vin < +100mV, Vout=Vin for Vin >= 100mV
2. Vout=100mV for Vin < +100mV, Vout=Vin for Vin >= 100mV
3. Vout= 0.8235*Vin+0,2647V Which will scale Vin such that for Vin=200mV -> Vout=100mV and Vin=1.5V->Vout=1.5V

Which transfer function do you require?

 

[KrisBlueNZ]

You need to explain your requirements a lot better than that.

What does this input come from?
What are you using it for?
Can you explain the project in general?
How much current will be flowing in the circuit?
What power supply rail(s) do you have available?
Can you provide links to the existing components and products in your system?

 

[sid2286]

well, the circuit is as follows:
I have a sensor output given to the differential amplifier from which the another opamps amplifies it.
the output from the amplifier opamp is -200mV to 1.5mV.

Now I want to read only from 100mV onwards to 1.5V,
though I set 100mV using the amplifying stage, but due to the drift in sensor (due to temp etc) it keeps falling below 100mV.

what I want is transfer function no. 2
Vout=100mV for Vin < +100mV, Vout=Vin for Vin >= 100mV.

please suggest,
Sid

 

[Harald Kapp]

You can use an opamp as "ideal diode" plus two additional reference voltages:

Notes:

• V3 can be derived from V1 by a simple resistive divider 1/3.
• Vout wil be negative. Add an inverting opamp for your desired positive output.

Harald

 

[sid2286]

Can you please explain how it works.
I am no expert, but would like to understand a bit.

without Diode it looks like a subtractor? what does diode do here?
also V1 and V2? is that the Vin you are considering ?

 

[KrisBlueNZ]

There is not much detail in your description.

If the voltage from the differential amplifier is drifting below +100 mV because of drift in the sensor due to temperature, clamping the voltage to a minimum of +100 mV will not correct for this drift. Clamping the voltage will cause part of the sensor's detection range to be ignored. If that's what you want, then fine. If not, you will need to explain your project fully, including a description of the sensor and how and why it is affected by temperature and any other factors that cause drift, so we can try to come up with a solution that will do what you need.

 

[Harald Kapp]

The precision rectifier is described here.

I combined two effects:
Offsetting the threshold for the diode circuit by adding -100mV to the + input of the opamp plus adding -300mV to the input voltage via R1 and R3.

You can replace V2,R1 plus V1,R3 by a voltage source V=(V1+V1)/2 and a resistance R=R1||R3=5K. The gain of the ideal diode wil be g=-R2/R=-2. I'll add an image for reference.

Assume Vin=V2=-200mV. Then (V1+V2)/2 = -250mV. This s negative compared to V3, therefore the output of the opamp will be positive. D1 will be off and D2 will close the feedback path around the opamp such that at the inputs of the opamp V(-)-V(+)=0V. As long as there is no load at the output (!) R2 will show the -100mV that are present at the - input of the opamp at the output.
Note that if you add a load resistor that is noteably smaller than e.g. 100kOhm, you will see a drop in output voltage. This is why Kris asked for more details about your requirements. The inverter you'll have to add for a positive output voltage needs to have a very high impedance. It is best to insert a noninverting buffer at the output of the ideal diode circuit to minimize the load.

Assume Vin=V2=+100mV. Then (V1+V2)/2 = --100mV. This is the same voltage as V3. The output of the opamp will be 0V (ideally, in practice it will be either >0V or <0V, depending on tolerances). D1 will still be off and the above explanation for the output voltage applies.

Assume Vin=V2 >100mV, e.g. Vin=1.5V -> (Vin+V2)/2=0.6V. This is positive with respect to V3 and the output of the opamp will go positive. The feedback loop is closed via R2 such that the output voltage is Vout= [(Vin+V2)/2-V3]*2-2*V3 = [0.6V-(-0.1V)]*2-(-0,1V)=0.7V*2+0,1V= 1.5V

By mixing the two effects it is not very easy to understand the circuit - put possible.