For an op amp, there appears virtually equal potentials at the input inverting and non-inverting terminals; as result, the difference appears to be infinitely low and NOT ZERO.
The circuit from post#36 won't work with an LM358,
because the inputs are outside the common mode range.
Leo..
1 Like
I am supplying the output of a wheatstone bridge based pressure sensor to the amplifier.
Thanks.
Thanks. You have been a source of great help and encouragement. I have got the amplifier to work:
R2=121k
R1=1
when V1-V2 = 10.8mv, V0 = 1.29v (good)
but:
when V1-V2 =32mv2, V0 = 1.6 (wierd, i expected the output to be =0.0032*121= 3.87v)
An LM358 is not rail2rail, and has a limited output swing.
You might not get much more than 0 to 3.5volt out of it.
Try powering the opamp with 7-9volt (the sensor still powered with 5volt).
That way you also get more A/D values.
Add a 10k resistor between opamp output and Arduino pin, for pin protection.
Still think you should but that sensor+HX710 board.
Leo..
1 Like
If OP is not stuck using an LM358, in my opinion, an opamp like to the one @johnerrington has suggested would be the best solution. OP's pressure transducer appears to be a 40kpa FS with 5k bridge resistors. While the single opamp diff amp is not especially linear or low input impedance it will probably do in this application. It has been my experience that the gauge resistance matters with this kind of amplifier design and you have to account for it in the gain equation.
I suggest the above circuit will do the trick. Gain is approximately 67 with 5% resistor values and will allow OP to use the full range of the ADC along with rail to rail operation. If nulling the sensor zero is necessary that can be addressed. I'd probably use @johnerrington's MCP6021 as it has a handy ref pin that could be used to balance the transducer.
2 Likes
Remember that you can get this sensor with 24-bit A/D included for US $7.55 + shipping.
See post#22.
Leo..
1 Like
That's definitely the path I would take. The bare sensor just requires too much work to get acceptable results in my opinion. It seems that OP is bound and determined to use an opamp though.
Thanks for your helpful suggestions. I can't use the HX710 because it requires a minimum of two pins. I have only one pin to spare.
Thank you very much. I will try what you suggested.
I could but it won't be delivered to me on time (it takes at least 3 weeks for it to be delivered). Equally the challenge presents a learning opportunity. Thanks.
You're nit picking @GolamMostafa
An IDEAL op amp (ie one where you can do the mathematical analysis) is defined as having infinite loop gain.
Ideal op amps
An equivalent circuit of an operational amplifier that models some resistive non-ideal parameters.
An ideal op amp is usually considered to have the following characteristics:[4][5]
- Infinite open-loop gain G = v out / v in
- Infinite input impedance R in, and so zero input current
- Zero input offset voltage
- Infinite output voltage range
- Infinite bandwidth with zero phase shift and infinite slew rate
- Zero output impedance R out, and so infinite output current range
- Zero noise
- Infinite common-mode rejection ratio (CMRR)
- Infinite power supply rejection ratio.
These ideals can be summarized by the two golden rules:
- In a closed loop the output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
- The inputs draw no current.[6]: 177
what exactly is the difference between infinitely low and zero??
It is an academic matter on OA what I have learned in my 2nd year EEE Class some 46 years ago. The characteristics of OA are still the same as you have mentioned in post #52 with the following exceptions:
-
In a closed loop the output attempts to do whatever is necessary to make the voltage difference at the input terminals virtually equal. (There should be a difference to present the error voltage for the feedback to work?)
-
The inputs draw no noticeable current.
That is for a "real world" op amp; and can you analyse an op amp circuit for a non-ideal op amp?
Given the title of this thread I think dealing with a non-ideal op amp is beyond them - as you will see here
https://ee.eng.usm.my/eeacad/zul/EEE241/Non-ideal%20Operational%20Amplifier.pdf
If I am asked to find the value of Vo of the following circuit (Fig-1), I will assume the following parameter:
Current passing through feedback path is equal to the current passing through R1, and it is equal to ~42.5 mA. Am I correct?
Figure-1:
No. -2v/4700 = 0.425mA.
2 Likes
why 0.425 mA?
Should it not be minus?
Because volts divided by ohms is AMPS. Your number was 1000x too large. Ok yes strictly negative.
Who is supplying this current -- the virtual 5V source or the output stage of the OA?
The current flows through both. The OA output determines the current.