# Strange op-amp configuration

1. Jul 19, 2011

### Jesus P

Im doing my thesis its about measure haze at ambient, so I found very little information about DIY schematics to work with photodiodes, at last my tutor found something, but when I was doing the circuit I found that this doesn't worked! well I made some mods and for a mistake I found a configuration that worked well!! BUT!!! which configuration of op-amp is this??? heve you seen something like this documented??? I didn't but it works very well with any diode except for LEDs, if any one find something interesting in this circuit please let me know. Thanks

2. Jul 20, 2011

### skeptic2

Can you tell us a little more about how the circuit works? When you shine light on the photodiode, what happens to the opamp output? By the way, the symbol you used for the photo diode indicates it is an LED. What is the part number of the photodiode? What is the ground connected to?

3. Jul 20, 2011

### Staff: Mentor

Your configuration is close to the traditional photodiode opamp current to voltage converter circuit. I used Google Images and those search terms to get these diagrams:

It looks like your "photodiode" is backwards, BTW.

4. Jul 20, 2011

### Jesus P

Hi, no it is no a LED, I just used that symbol to figure out the diagram, it is a photodiode from inside an old mouse used to capt the movement of the ball, it is infrared, The circuit is totally strange because... , yes, it seems to be a traditional current to voltage but it is not, the output is proportional to the light that reaches the photodiode and I get the voltage measurement from pin 6 and pin 2 (see the diagram) and the ground goes to the pin 3 in the op-amp

5. Jul 20, 2011

### waht

This circuit is a standard current to voltage converter. The photodiode produces a small amount of current when is exposed to light, and the current is converted to a voltage that can be easily measured or further processed.

6. Jul 20, 2011

### Jesus P

Well, maybe I have to say that this configuration is actually running well, and I cant do anithing with that what you are showing to me... thats my problem,, Every standar circuit that I do fails, so I did find this configuration by mistake

7. Jul 21, 2011

### Staff: Mentor

Please turn the photodiode around the correct way, and re-post the circuit diagram. This is a very standard circuit, and should work well.

BTW, there is an advantage to having a large reverse bias voltage across the photodiode. Can you research that assertion some, and tell us what that advantage might be? Your circuit does not provide that reverse bias voltage...

8. Jul 21, 2011

### Jesus P

is that what Im traying to say, this schematic is working but I havet documented it jet, Im just starting the measurements, but the schematic is exactly as I have in my protoboar working, the output at sunlight is at about 4v and in dark is 0.250v

9. Jul 21, 2011

### Staff: Mentor

Only if you've turned the photodiode around. The photocurrent flows from cathode to anode, so you cannot get a positive output voltage from the photocurrent with the photodiode oriented as shown in your schematic.

10. Jul 21, 2011

### uart

It's just a standard current to voltage converter with the diode working in photovoltaic mode. It's basically a short circuit photovoltaic cell. The diode orientation is correct for this mode of operation.

The output voltage is negative but the OP is apparently unaware of the convention of measuring voltage relative to the zero volt. He is reporting that he's measuring the voltage between pins 2 and 6 and not specifying which of those he's taking as the reference for polarity, which is adding to the confusion. OP please measure your voltage output at pin 6 relative to ground (zero volts). You'll find that your output voltage is in fact negative when the diode is illuminated.

11. Jul 21, 2011

### Jesus P

Im measuring between pin 2(-) and 6(+) but when I measure ouput pint relative to ground I have no output

12. Jul 21, 2011

### waht

Verify that you have +12V and -12V on the power supply, or some other voltage in that range. Double check all wiring. Try different values of R1 from 50K to 1M. Also, try reversing the photodiode polarity and see what you get. Measure all voltages with respect to ground.

13. Jul 21, 2011

### Jesus P

done. done. done. one week at that. Trying everything

14. Jul 21, 2011

### Staff: Mentor

Is the opamp socketed in your circuit? If so, try a 2nd opamp to see if maybe the first one is bad.

15. Jul 21, 2011

### Jesus P

done. :-)

16. Jul 22, 2011

### uart

Then what is the voltage between pin 2 and ground. This one genuinely should be zero.

By "ground" I meant the zero volt connected to pin 3 btw.

17. Jul 24, 2011

### Jesus P

Ive already done that, and nothing, iv tried several 741 cofigurations and nothing.. please if any one can try this http://laser.physics.sunysb.edu/~tanya/report2/" [Broken] or any configuration of current to voltaje converter to work with a LED letme know which one did u used. Im talkin do it in a protoboard not in paper, physical. and please do my configuration too an telme if I am wrong but this one works.

Last edited by a moderator: May 5, 2017
18. Jul 24, 2011

### waht

You mentioned that the photodiode is a pull from a mouse. I'm not sure what kind of diode they use there. Maybe if you try a different diode you might get a different result. Forrest Mims uses a regular LED in the circuit.

19. Jul 24, 2011

### uart

Hi Jesus P. So far you've told us the following :

$V_{6,2}$ = signal (0.25V dark, 4V illuminated)

$V_{6,0}$ = no signal (0V)

$V_{2,0}$ = no signal (0V)

Yet basic physics tells us that : $V_{6,2} = V_{6,0} - V_{2,0}$.

So basic physics tells you that this doesn't add up. Does this mean that you've discovered an incredible circuit that defies the laws of physics? Sadly no, this type of thing is nearly always just a side effect of basic "noobness" and measurement error. In particular when you probe pin2 you are dealing with a high impedance point that normally shouldn't have anything connected except the basic circuit elements required (D1 and R1 in this case). You have to be very careful connecting anything to pin2 and not induce some anomalous behavior in your circuit. You may also have an open circuit somewhere that your measurement circuit is indirectly closing in some way.

20. Jul 25, 2011

### skeptic2

Uart, back to my question of post #2, what is ground connected to? He doesn't show it as being connected to zero volts. As far as his schematic is concerned, ground is just a node, nothing more. In fact, his answer from post #4 seems to indicate it is not connected to zero volts.