Hello, I am new to the electronics and having a very weak base. I want to build a IR indicator by using the IR LED and the photodiode. I had tried to connect the my circuit according to the figure below. However, I get no reading change for the V(out) when I tried to block the photodiode from IR LED by using black color cover.
The IR LED is function well.

Btw, do anybody still have datasheet for the 5mm IR LED and 3mm photodiode? I tried google it but get nothing same with my components.

I am not sure about the operating current and voltage for the photodiode.
Is that R2 in the correct value?
Is that longer leg for the photodiode indicating anode for it?
If I put the LED to the V(out), is that the LED will light up when I block each other? (LED parallel with the photodiode)

Thank for answers.

 

I just picked the first google result of this search as an example:
"3mm photodiode datasheet"
Reverse light current is 3μA so the voltage across the resistor would be 0.000003 A * 3000 Ω = 0.009 V
To get a bigger voltage you need a larger resistance.
It makes more sense to measure the voltage across the resistor as it has the largest percentage change and you can use a low range on the voltmeter. You can swap their positions if you need to amplify the signal with an op amp.

 

If I want to light up the LED at the photodiode side, do I need to amplify its voltage and current?

If the distance between IR LED and photodiode is about 1 cm, is that alright? I don't know about the minimum and maximum distance of them. I afraid the photodiode will affected by the heat as well.

Thank in advance.

 

Have you tried flipping the photodiode around?

 

Yup, I tried =(
sadly, no difference... Will photodiode burned easily?

 

If you have a digital multimeter with a diode check position you can check to see if your photodiode is OK,as it will just look like a diode.
The DMM may be sensitive enough to show the difference between an illuminated & non-illuminated photodiode in this position.

 

If you can get your hands on a phototransistor instead of a photodiode it makes things easier. The currents are in the milliamp range instead of microamps so the rest of the circuit becomes simpler.
The photodiode should be fine, it should only be damaged by too much voltage.

 

Alright, I will try to get a phototransistor soon, the shops are closing today as it's national day in our country.

Anyway, thank for helping.


Added information...

Is this a good condition photodiode?

 

Looks to me it's good.

Allen

 

Well, I still fail to figured out where is the problem lied =(

V(supply) measured = 4.95V.
The voltage across the photodiode is about 4.75-4.80V either blocked or unblocked from IR LED. =(

 

Reverse light current is 3μA so the voltage across the resistor would be 0.000003 A * 3000 Ω = 0.009 V
To get a bigger voltage you need a larger resistance.
It makes more sense to measure the voltage across the resistor as it has the largest percentage change and you can use a low range on the voltmeter. You can swap their positions if you need to amplify the signal with an op amp.

Did you increase the 3KΩ resistor to 10KΩ as suggested by markd77?
After changing, try to measure the voltage again and see if the difference between On/Off conditions are greater.

If you can get your hands on a phototransistor instead of a photodiode it makes things easier. The currents are in the milliamp range instead of microamps so the rest of the circuit becomes simpler.
The photodiode should be fine, it should only be damaged by too much voltage.

If the above still doesnt work, try to look for a photo-transistor. It would be far more sensitive than a photo-diode.

Allen

 

Yes, I changed the R2 to 10KΩ and 1MΩ.
V(out) for 10KΩ = 4.75-4.80V
V(out) for 1MΩ = 2.31-2.37V

Does the photodiode sensitive to the visible light or it only can detect IR?

 

Why don't you understand that the photo-diode has a tiny output signal that must be amplified?
It leaks some current when dark so when amplified it will have an output signal that might be too high when dark in your reverse-biased circuit.

A photo-diode can also be used without a bias voltage as a photo-cell that produces a tiny voltage when lighted. Then it has no leakage so has no output when amplified in the dark.

 

You can get slotted Opto-coupler like the ones in the attached to replace your IR led and opto-diode pair. These components can be found in old printers, DVD drives and floppy disk drives.

I think you'll have better chances with them doing your experiments.

They are also available here

http://my.element14.com/jsp/search/browse.jsp?N=2105+203598+110002105+110202249+110124082+110198143&Ntk=gensearch&Ntt=opto&Ntx=mode+matchallpartial&No=0&getResults=true&appliedparametrics=true&locale=en_MY&divisionLocale=en_MY&catalogId=&skipManufacturer

Allen

 

Sorry for my ignorance on the tiny signal of photodiode.

I just want to make a simple detector that can detect the obstacle so that it will give the respond that I assigned. I had make some researches on how to make a simple switch by using the light. I tried the LDR but afraid it will affected by the operation of time (eg. day and night). So I planned to move to the IR.

I had studied from the youtude and found that the circuit is simple and easy to understand. So I didn't ever think about op-amp and plan to try it.

IR LED SWITCH (YOUTUBE)

It shows that when no obstacle in between, the V(out) should be 0V (which I unable to obtain); when there is an obstacle, the V(out) should be 5V.

*ps:
I obtained the 3mm IR pairs from the electronics shop but it just stated IR TX-RX in my receipt so I am not sure whether their wavelength are in same or not. After fail to obtain the same result as the video, I move to 5mm IR LED. I want to purchase 5mm TX as well, but they don't have.
I had tried to request for the datasheet but the shop owner don't have either.


Added action taken: Trying to disassembly the old floppy disk drive.

 

The little kid in the video must be using a photo-transistor, not a photo-diode.
A photo-diode does not conduct enough current to turn off the LED when it is lighted.

An IR photo-diode or photo-transistor has a black coating that blocks some but not all visible light.

An IR or visible LED can have a narrow focussed beam which makes a cheap weak one appear bright or it can be a modern strong one with a wide-angle very bright beam.

 

you can get photo-transistor at ESCOL, Johore.

http://www.escol.com.my/

 

It's my mistake to use the photodiode, thank for pointing out =)
I will try to obtain the phototransistor to replace it.

Reply to Allen,
What is this actually?

Btw, can the IR receiver in the VCD player be used in my case? I got the remote as well so that can get it IR LED.

Allen, thank for your informations, but I am in Kedah, I will try to find the phototransistor from the nearest electronics shop first as the shipping fee quite high if I only purchase 1 item.

 

The thing in the red rectangle is the Read/Write head of the floppy disk drive.

The red arrow is pointing to the opto-coupler.

The 3 pin component on the other pictures may be a photo transistor or optical receiver.

Allen

 

Yes, I changed the R2 to 10KΩ and 1MΩ.
V(out) for 10KΩ = 4.75-4.80V
V(out) for 1MΩ = 2.31-2.37V

Does the photodiode sensitive to the visible light or it only can detect IR?

It should only detect IR, but there is plenty of IR in sunlight and some other light sources so you should try to shield the sensor from everything apart from the IR LED as much as possible.
Another source of slotted IR pairs is mice. The old ones with the ball usually have two on the rollers that touch the ball and one on the scrollwheel. New mice usually just have one for the scrollwheel.
The gap is pretty narrow and each one has two sensors, very close to each other.