Hello all,
I'm using a strain gauge with a 5V power supply across the wheatstone bridge. at max load, I should be getting 2.5mV. Therefore, I decided to build an amplifier to range from 05V, with 5V being 2.5 mV. I have an LM 324 which I'm using as the op amp and I believe that those are great for DC and low frequency applications.
Attached is an image of the current circuit I have bread boarded. First, I figured I would need a 2000 V/V gain with my amplifier. I also want to mock up the 2.5 mV output of my strain gauge, so I used a voltage divider on the 12V to get me 2.5mV.
Math:
Gain: 2000=(1+R3/R4); R3=1999*R4.
Voltage divider: 2.5mV=(R2/(R2+R1)*12V; R1=4799*R2.
With these, I found resistor values which are close to what I want.
R1 = 1.05M
R2 = 220
R3 = 200k
R4 = 100
When I power on the circuit, my output is 10.5V. I checked the negative input of the op amp and I found that it is 5 mV while the positive terminal is 2.5 mV. I thought op amps want to have the negative and positive terminals at equal voltages, so why is the negative terminal twice as high as the positive?
I went back and read the AAC's write up of divided feedback (http://www.allaboutcircuits.com/vol_3/chpt_8/5.html) and that basically reaffirmed what I thought about the positive and negative terminals wanting be equal voltage.
What am I do wrong? Is my basic understanding of how the op amp works incorrect?
Matt
EDIT:
I forgot to add the true gain and the voltage from the divider
Voltage divider = 2.514 mV
Gain = 2001 V/V
I'm using a strain gauge with a 5V power supply across the wheatstone bridge. at max load, I should be getting 2.5mV. Therefore, I decided to build an amplifier to range from 05V, with 5V being 2.5 mV. I have an LM 324 which I'm using as the op amp and I believe that those are great for DC and low frequency applications.
Attached is an image of the current circuit I have bread boarded. First, I figured I would need a 2000 V/V gain with my amplifier. I also want to mock up the 2.5 mV output of my strain gauge, so I used a voltage divider on the 12V to get me 2.5mV.
Math:
Gain: 2000=(1+R3/R4); R3=1999*R4.
Voltage divider: 2.5mV=(R2/(R2+R1)*12V; R1=4799*R2.
With these, I found resistor values which are close to what I want.
R1 = 1.05M
R2 = 220
R3 = 200k
R4 = 100
When I power on the circuit, my output is 10.5V. I checked the negative input of the op amp and I found that it is 5 mV while the positive terminal is 2.5 mV. I thought op amps want to have the negative and positive terminals at equal voltages, so why is the negative terminal twice as high as the positive?
I went back and read the AAC's write up of divided feedback (http://www.allaboutcircuits.com/vol_3/chpt_8/5.html) and that basically reaffirmed what I thought about the positive and negative terminals wanting be equal voltage.
What am I do wrong? Is my basic understanding of how the op amp works incorrect?
Matt
EDIT:
I forgot to add the true gain and the voltage from the divider
Voltage divider = 2.514 mV
Gain = 2001 V/V
Attachments

12.2 KB Views: 59
Last edited: