# Differential amplifier

Discussion in 'General Electronics Discussion' started by johnchew, Oct 23, 2015.

1. ### johnchew

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Oct 23, 2015
I am currently doing a mini project for one of the module in school, where i encountered problem with the accuracy of the differential amplifier output. The signal of interest is from a amplified and filtered IR sensor, the sensor outputs a nominal voltage of about 3V and varies in the mV range when object moves toward or away from it. Therefore I have tried to used a differential amplifier to remove the dc offset, however, by feeding the sensing signal into the positive input and 3Vdc into the negative input, I still got a output voltage of about 1V with a unity gain configuration (just to test out the circuit).

The differential amplifier that i used is connected as the following. And the output voltage should be govern by
Vo = R3/R1 (V1-V2) but that is not what I'm getting. I'm hopping to get a varying range of few voltmeter that I could use to control the duty cycle of a buzzer.

also, Is there a way to control the frequency of my buzzer using the output voltage that I will be getting?

Thank you.

2. ### Ratch

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Mar 10, 2013
Doesn't R2 and R4 have anything to say about what the voltage will be? I think you should reconsider what the voltage formula is.

Ratch

3. ### johnchew

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Oct 23, 2015
I understand that R2 and R4 works as a voltage divider. Currently I'm connecting 4 similar resistor as my R1, R2, R3, R4. By calculation, it shld give me Vo = - V1+V2 or am I missing something over here?

5,164
1,087
Dec 18, 2013
What are the resistor values you are using?

5,164
1,087
Dec 18, 2013
Are you using a split supply and what opamp are you using?

6. ### johnchew

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Oct 23, 2015
I'm using 5.6k resistors also prior to the resistor I already have my signals pass through a voltage follower thus it shld also remove any impedance matching issues

5,164
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Dec 18, 2013

8. ### johnchew

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Oct 23, 2015
I'm using LM385 cos that's what I been given. And for this project I can only use component given by my school

#### Attached Files:

• ###### Form of Component List.pdf
File size:
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120

5,164
1,087
Dec 18, 2013
You mean LM358?

10. ### Ratch

1,094
335
Mar 10, 2013
I agree that if all the resistors are equal, then Vo = - V1+V2.

Ratch

11. ### johnchew

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Oct 23, 2015
Yup it is lm358

12. ### BobK

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Jan 5, 2010
You have not told us what wrong output you are getting with what inputs. And you have not answered the question about whether you are using a split supply. How do you expect us to help?

Bob

13. ### johnchew

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Oct 23, 2015
Sorry I'm using a single 5v supply to power my op amp

5,164
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Dec 18, 2013

15. ### johnchew

7
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Oct 23, 2015
Sorry all, here are the details regarding the circuit inputs/outputs.

-The op-amps I am using are LM358. Currently I am only using a single supply. +9V to pin8 and Ground to pin4 as I do not need negative outputs.
-The differential amp I have built with the LM358 is the same as the picture I have posted in the first post.
-I have use the following configuration for the differential amp, R1=R2=R3=R4, all resistor are 5.6k ohms.
-V1 is the output from a voltage buffer (follower) with a voltage of 3.200V and V1 is fixed.
-V2 is the output from another voltage buffer with a voltage range of 3.2xxV to 5.5xxV.
-V2 is an amplified signal from an IR detector.
-For testing, I would stand infront of the IR detector and move either closer or further away from the IR detector. V2 changes accordingly.
-The signal V2 is non-linear with distance. Meaning to say, within 15cm, V2 changes from 3.5xxV to 5.5xxV. However beyond 50cm, only the last two digits, 3.2xxV changes.
-As I am doing a distance measurement device for the blind, I am interested in the distance of 0.5m to 2m away from the IR detector itself.
-The readings for V2 I get from the IR detector is not always exactly the same, but it is quite similar during each test.
-I would stand at 2m infront of the IR detector, and V2 would be around 3.211V, as I move closer to the IR, say 1.5m, V2 would be 3.230V, then I move ever closer to the IR, say 1m, V2 would be 3.270V.
-And finally within the 15cm range, V2 increases exponential to 5.5xxV as my hand moves closer and closer to cover the IR.
-Hence, I am interested in only the last two digit of V2.
-I set up a differential amp to get V2-V1, in which V1 is 3.200V.
-Example, at 1.5m, V2-V1, i would get 3.230-3.200=0.030V.
-I would then amplify this 0.030V by 100times to get a useful output to drive my buzzer.
-However the differential amplifier is giving me a output of, V2-V1, 3.230-3.200 = 1.xxxV

5,164
1,087
Dec 18, 2013
Ok so I expect the opamp you are using cant resolve the low voltage you need. You may need to look at a rail to rail opamp. The LM358 is quite old now.

davenn likes this.
17. ### BobK

7,682
1,688
Jan 5, 2010
The LM358 cannot reach 0V when driven by a single supply. It has a miniumum output of something like 40mV. Since the voltages you are trying to get are in that range it is not suitable.

However, you still should not have gotten a voltage of over 1V for the case you mention. Did you actually measure both inputs and the outputs with stable voltages on them? This seems wrong to me.

Also, I question how accurate this IR is for distance sensing in the range you specify. It's output voltage would have to be accurate to about 0.1% in order to get any resolution between 3.21 and 3.27.

Actually, if all you are trying to do is discriminate a single distance, a comparator is a better circuit to use.

Bob

18. ### Laplace

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185
Apr 4, 2010
In case the school did not provide a comparator in the bag of components to use, the LM358 makes a fair comparator in a circuit such as this. You might consider generating the reference voltage directly from the sensor itself by using a very long time constant RC filter, if sensor drift is a problem.