Is my design correct

I would like to take differential output from two photo darlingtons. I am biasing the photo darlingtons with 5 volt using LM7805. I am taking the output between the 10 k resistor and giving to opamp mcp602 with differential amplifier configuration without any gain. Is there any loading effect in my circuit? If yes does it require any buffer/voltage follower circuit in non inverting terminal? or how to improve my circuit? any other alternatives?

**broken link removed**

Re: is my design correct

Hi,

Your attachment link doesn't work for me.
Maybe take a screenshot and safe it on your PC as jpg or png.
Then use the "insert image" button to insert the picture.

Klaus

Re: is my design correct

KlausST said:

Hi,

Your attachment link doesn't work for me.
Maybe take a screenshot and safe it on your PC as jpg or png.
Then use the "insert image" button to insert the picture.

Klaus

Click to expand...

sorry for the trouble

Hi,

(The picture is a bit small, hard to recognize the part values)

without any gain

Click to expand...

This is not a technical term. Some may mis-interprete "no gain" with gain = 0.
But in your case "no gain" means gain = 1.

--> please in future use the term: "gain = 1"

****
Analyzing your circuit:

The first an mayb ebiggest problem is: You use an OPAMP supply of 9V, whil the OPAMP supply is spicified with 2.6V ... 6.0V.
--> You are beyond specification! This may kill the OPAMP.

Now look at the signal path of the left diode (non iverting path).
* diode anode at +5V
* from cathode there is a 10k to GND
* also from cathode there are 2x 1k (in series) to GND. (you may ignore the influence of the OPAMP, because the input current should be negligible)
--> the total diode load is: 10k in parallel to 2k = 1.6666k (this may be good or not, you need to know)
*****

Now let´s calculate the voltage at the noninverting OPAMP input (this determines the common mode voltage)
The voltage at the cathode (referenced to GND) is: diode_current x 1.666k
Because of the 1k/1k voltage divider --> the voltage at the OPAMP input if half of the cathode_voltage: IN+_voltage = diode_current / ( 2 x 1.6666k)

Now you see with no diode current there is 0V. This is critical because it is close to the OPAMP_supply_rail.
But according datasheet the OPAMP can handle common_mode_input_voltage down to -0.3V. So everything is OK so far.

Now assume the left diode carries some current.
--> the voltage at IN+ rises
--> the voltage at OPAMP_output rises.
Everything OK so far.

Now assume no diode carries current.
--> the OPAMP_output should be 0.000V
This is critical, because it is at the OPAMP supply rail.
According datasheet: "Linear Output Voltage Swing" is only down to 100mV, then it becomes unlinear and will saturate at about 20mV.
--> Trouble. You estimate 0mV, but there will be something between 20mV and 100mV

Now assume left diode carries no current, but right one carries some current. .
--> the OPAMP_output should be negative
This is critical, because it is below OPAMP supply rail.
According datasheet: "Linear Output Voltage Swing" is only down to 100mV, then it becomes unlinear and will saturate at about 20mV.
--> Trouble. You estimate negative voltage, but there will be something between around 20mV positive.

*****
Conclusion:
There are two major problems to solve now:
* Power supply for the OPAMP: Either use a suitable OPAMP or modify the supply voltage for the OPAMP.
* You need to define the expected output_voltage vs the "differential_input_current".

Next steps will follow...

Klaus

As explained by KlausST, the differential output can only work in one direction (Ileft >= Iright). But what's the intended function at all?

FvM said:

As explained by KlausST, the differential output can only work in one direction (Ileft >= Iright). But what's the intended function at all?

Click to expand...

KlausST given awesome explanation thank you. I am going to use left diode as reference diode another diode is test diode(coupling diode to fiber optic); reference diode output is always greater than test diode. so i am taking the voltage difference and measure the differential voltage v2-v1 from two diodes.

PFA, is this circuit is correct?

Hi,

You said: " v2-v1" .... as if your input is voltage:
* the photodiodes make current, not voltage
* unfortunately the voltage at diode2 (V2) is influenced by your Opamp circuit.
* the shown circuit makes: Vout = V1 + (V1 - V2), which is 2V1 - V2.

So indeed your post#6 doesn't give the answers to my questions.

Klaus

KlausST said:

Hi,

You said: " v2-v1" .... as if your input is voltage:
* the photodiodes make current, not voltage
* unfortunately the voltage at diode2 (V2) is influenced by your Opamp circuit.
* the shown circuit makes: Vout = V1 + (V1 - V2), which is 2V1 - V2.

So indeed your post#6 doesn't give the answers to my questions.

Klaus

Click to expand...

my expected output voltage is 0 to 4 volts. my reference PD output voltage will always be 3.8 to 4.1 volt, and test PD is swing between 0.9 to 3.1 volt. I am using photo darlington(PD) op560cView attachment op560a.pdf, LED is irl81aView attachment IRL 81A, Lead (Pb) Free Product - RoHS Compliant-335091.pdfView attachment ina118.pdf.

Can I use this circuit given in this website? light from my reference and test fiber optic sensor to get differenced output.

can i use the first configuration?

or can i use INA 118 with Rg=50kohm (gain = 1), by taking volltage from 10kohm resistor between PD.

or can you suggest any other simple circuit.