Is this op-amp configuration the solution for measuring haze at ambient?

[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 anyone find something interesting in this circuit please let me know. Thanks

 

[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?

 

[berkeman]

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 anyone find something interesting in this circuit please let me know. Thanks

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:

http://www.google.com/search?tbm=is...converter+photodiode&gbv=2&aq=f&aqi=&aql=&oq=

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

 

[Jesus P]

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?

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

 

[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.

 

[Jesus P]

Well, maybe I have to say that this configuration is actually running well, and I can't do anithing with that what you are showing to me... that's my problem,, Every standar circuit that I do fails, so I did find this configuration by mistake

 

[berkeman]

Well, maybe I have to say that this configuration is actually running well, and I can't do anithing with that what you are showing to me... that's my problem,, Every standar circuit that I do fails, so I did find this configuration by mistake

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...

 

[Jesus P]

is that what I am traying to say, this schematic is working but I havet documented it jet, I am 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

 

[berkeman]

is that what I am traying to say, this schematic is working but I havet documented it jet, I am 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

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.

 

[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.

 

[Jesus P]

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.

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

 

[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.

 

[Jesus P]

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.

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

 

[berkeman]

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.

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

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

 

[Jesus P]

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

done. :-)

 

[uart]

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

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.

 

[Jesus P]

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

Ive already done that, and nothing, iv tried several 741 cofigurations and nothing.. please if anyone can try this http://laser.physics.sunysb.edu/~tanya/report2/" or any configuration of current to voltaje converter to work with a LED letme know which one did u used. I am 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.

 

[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.

 

[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.

 

[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.

 

[uart]

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.

Good point. Normally you'd assume that it was connected to the center tap of the +/- 12V supply, but given the anomalous behavior it's probably best not to assume anything. :)

 

[Jesus P]

ahh, now a got your cuestion, yes it is conected to 0v, its a dual supply, and yes, I now that I haven't discovered nothing new, but... what is wrong with a theoretical current to voltage circuit that I can't get it to work properly, I know electronics or almost enough to know when a 741 pin is conected and where it goes. I know that basic physics (IN PAPER) says that

V6,2 = signal (0.25V dark, 4V illuminated)

V6,0 = no signal (0V)

V2,0 = no signal (0V)

Yet basic physics tells us that : V6,2=V6,0−V2,0.

I started with the construcction of eve current to voltage converters configuration known by internet and books, and I can't manage to make it work. And my cueston is Why a very very ... very simple circuit dosent works? I have 8 diffrent 741 all of them works (ive tested them as followers and normal amplifiers) But every time I try to make it work as a LED sensor amplifier I don't get... nothnig 0v 0v 0v and no reaction, I have ans Oscilloscope and a Fluke (very expensive) tester and I know to use them :-) I know that what UART says "is an abnormal and very unstable circuit that which I made" but is the only one that I can make to work.

Please forgive me for my bad English and really thanks for the help! really! but can anyone manage to get a simple LED sensor amplifier to work?

I found a site where the author says that she made it but I constructed 4 diferent of them and again NOTHING... it is very easy.

and here is the site

http://laser.physics.sunysb.edu/~tanya/report2/"

Can anyone get it to work (not in paper I mean real world) and tell me if this configuration is not a fake??. Thanks again to every one. best regards :-)

 

[Jesus P]

I think I found something! the problem with normal current to voltage configuration is not due theory or something like that (of course not"!) but, it is related to the op-amp, it gets saturated too fast! even when I put it in a follower configuration, it gets saturation point and dosent follow the signal further, only for a few volts, I NEED TO FORGET TO USE 741 for serious measures! :-(

Iv tested that behavior with and oscope a function generator and a fluke multmetter