i have very limited knowledge on circuit and i need help to put this photodiode module to work. the circuit is on the PDF page 4 under S8745. from the diagram i understand that those enclosed within the dotted box with the window is inside the module, which i dont need to worry about. wat i dont understand is the pin 5, 10, 8 and both the capacitors are connected to ground. wat does that "ground" mean? since they all connected to ground can i connect them together?

the RL is an external voltage display. However how do i know wat Rf rating should i use?

your help is much appreciated.. thank you!

 

please anyone?

 

Yeah, all grounds are grounds. Notice the amplifier requires a dual power supply, meaning +15V, GND, and -15V.
Adding Rf lowers the gain of the amplifier, see graph on page 2.
What do you want to do with it?

 

Hey thanks kubeek.. i've looked over the internet wat does dual power supply means.. so from wat i understand is that if im using battery i would need 2 15V batteries to create the dual power supply, while the voltage between the 2 batt in series is the "ground"? or is there any other way to do it?

another thing is that if the op amp rating is 15V is 12V batt enough? cos thats wat i have atm.. XD

 

owh im using it to detect UV light from UV LED.. i guess i'll try with no external Rf first see whether the gain is enough..

 

You might be able to get by with a single supply if you are willing to add a few parts.
How will you be monitoring the output voltage?

 

owh im using it to detect UV light from UV LED

Detect as in measure the actual amount of UV hitting the sensor, or a simple on-off decision?

 

You might be able to get by with a single supply if you are willing to add a few parts.
How will you be monitoring the output voltage?

do u mind to instruct me how should i connect the circuit if thats so?

im monitoring the voltage using a digital voltage display.. the photodiode suppose to give diff voltage for diff light intensity detected..

 

Detect as in measure the actual amount of UV hitting the sensor, or a simple on-off decision?

im detecting the intensity of the light.. which means higher intensity would give higher voltage.. i've bought a voltage display which has a + and - input.. which i assume from the circuit the + would be the Rload (pin 4) and the - would be to the ground..

 

Im not so sure about that, the datasheet doesn´t state the minimum supply voltage, so if you use 30V supply and replace the 30k resistor with another 120k, then you basically get a +/-15V supply as in the datasheet. Also the output is inverting, so I think you will get 0 to -10V on the output for low to high light, instead 0 to +10V.

 

Im not so sure about that, the datasheet doesn´t state the minimum supply voltage, so if you use 30V supply and replace the 30k resistor with another 120k, then you basically get a +/-15V supply as in the datasheet. Also the output is inverting, so I think you will get 0 to -10V on the output for low to high light, instead 0 to +10V.

You are probably right about output polarity. I have deleted my post.

I think it would have worked with this unit, which has the diode reversed.

 

You are probably right about output polarity. I have deleted my post.

You could have left/altered the drawing, now I have to make it again
Use this to make a +/-15V supply from a 30V input. Note the output ground must NOT be connected to the input supply ground.

 

hey guys! really thanks a lot for the reply!

another stupid question.. i really only have very very basic knowledge about circuit.. so if i wanted to assemble a circuit as shown in the image u attached, wat kind of op amp should i use? and is there any 30V batt which is small? i tried google them but all look very big.. XD

 

For example TL071 will work.
Funny that you mention the battery thing now. If you can get some other photodiode that doesn´t have the opamp inside, you could make the same circuit as in the datasheet and run it for example from two 9V batteries.

 

ah ok.. i think i know wat i should do now.. thanks a lot for the help!

 

hey.. another question.. if i were to use the phodotiode without the op amp and assemble the circuit myself wat kind of op amp should i use? i would like to have voltage output..

once again thanks for your help!

 

Hey man i need your help again.. i assembled everything according to the circuit drawing but somehow i cant get the photodiode to work.. i even tried with another one ( i bought 2) fresh off the package but it doesn't work as well..

i've attached the drawing illustrating how i've connected my circuit on breadboard.. i used 3 12V battery as power source using the split supply circuit u've suggested.. i tested the output (before connecting to photodiode) with multimeter its about +16 and -16V, which is fine for the photodiode..

having no clue i tested the voltage between Vcc+, GND, Vcc- of the photodiode after i connected to the power and it appears that the Vcc+ - Vcc- is 31.6V, Vcc+ - GND is 1.5V, GND - Vcc- is 30V.. the voltage between the output and GND is forever 0.. which i cant explain why..

do you have any clue to get the photodiode to work? sorry im just an amateur trying to figure something out for my final year project..

 

Why is there an LED in series with the switch? It certainly can´t work like that.

 

Hi Kubeek.. thanks again for your reply

i got rid of the LED but it still doesn't work.. i tested the voltage between the output and the Ground its 16.5 and doesn't change no matter i shine light on it or not.. i've tested a couple of voltage point on the circuit as follow:

Vcc+ to Ground: 17V
Vcc- to Ground: -17V
Output to Ground: 16.5V
Vcc+ to Output: 1V
Vcc- to Output: -35V

Your advice would be most appreciated..

 

The ambient light might just be too much, have you tried adding the Rf resistor or covering the photodiode with black tape?