Adjusting Sensitivity of Photo Transistor?

[adamthole]

I have been using photo transistors lately as part of a laser beam detector device. The problem is that my transistor is not sensitive enough. The laser has to be directly into the transistor for it to turn on. In the datasheet, it says that I can use the base leg (that is currently unused) to adjust the sensitivity of it. I don't understand how I would do this. In the past I would always just connect a resistor to either P or N and the other end of the resistor to the base of the transistor to turn it on or off. I never did any half-way stuff.

Thanks!

 

[audioguru]

The base of a phototransistor is used to decrease its sensitivity.
A laser beam is very directional so of course must be shining directly on the phototransistor.

You could amplify the output of the phototransistor with an ordinary transistor. Then you might have sensitivity to temperature changes or ambient light.

 

[adamthole]

Actually I am already amplifying the output with a transistor. Perhaps 2? Or perhaps I just have an unsteady hand. I need to do a test with everything set up, not me trying to keep my hand steady.

And I forgot to post this earlier, but here is the datasheet of my photo transistor.

**broken link removed**

Thanks!

 

[audioguru]

Hi Adam,
Your phototransistor has a focussing lens in it. Therefore the laser must be directly in front of it for max sensitivity. If it is off to the side only 10 degrees, then the output is half. Maybe you could shine a laser beam on it with it directly in front of you, but far away, for target practice or something. :lol:

The phototransistor is most sensitive to infrared. At the red wavelength of a laser its output is almost only half.

 

[chemelec]

I would suggest you get a Better Photo-Transistor for the Visable Red Laser, Such as a 2N5777.

Gary

 

[adamthole]

Thanks for the help, I didn't really understand that stuff until now. I think my biggest problem is the angle part. I am too shaky to keep it in the right spot. After I get everything installed, it should be alright.

Thanks

 

[chemelec]

I think your Biggest Problem is the Wrong Photo Transistor.
The one your Using is best for UV Light.

Thats like using a Skate Board instead of a car to go get your weekly groceries.

 

[audioguru]

I think your Biggest Problem is the Wrong Photo Transistor.
The one your Using is best for UV Light.

Sorry Gary, no. This phototransistor responds best to light with a long wavelength, like infrared. Ultraviolet has a short wavelength. Visible light is in between. :lol:

 

[williB]

The base of a phototransistor is used to decrease its sensitivity.

That doesnt sound right to me..
because ,say that you set the bias current to some fixed level below the actual turn on point ..
then adding light will push it into the active region .
i dont see how you could possibly bias it to make it less sensetive.. :lol:

 

[chemelec]

audioguru: Sorry Gary, no. This phototransistor responds best to light with a long wavelength, like infrared. Ultraviolet has a short wavelength. Visible light is in between.

I Thought this guy was using a "Visible RED Laser Light"?

If So, it appears Off of the Sensitivity Region. According to that Graph that was also Posted by someone else????.

Correct me if I'm Wrong!

Gary

 

[williB]

Infrared peaks at 850nm which is where adamthole's detector is ,
adamthole did not say what wavelength he was using..

 

[audioguru]

Hi Willi,
A phototransistor has a "dark current" which is its leakage when dark. So it is already conducting and doesn't have a threshold. Adding some base current with a biasing resistor to the positive supply will only increase its idle current without increasing its sensitivity to light.
Using a resistor from its base to emitter will decrease its sensitivity because more light is required to create a voltage drop across the resistor, causing a threshold before the phototransistor begins to conduct.

 

[williB]

This is how it works ..
the collector current increases as the light intensity increases , as it should ..
and THE BASE CURRENT is proportional to the the light intensity

Hi Willi,
A phototransistor has a "dark current" which is its leakage when dark. So it is already conducting and doesn't have a threshold. Adding some base current with a biasing resistor to the positive supply will only increase its idle current without increasing its sensitivity to light.
Using a resistor from its base to emitter will decrease its sensitivity because more light is required to create a voltage drop across the resistor, causing a threshold before the phototransistor begins to conduct.

 

[adamthole]

I would suggest you get a Better Photo-Transistor for the Visable Red Laser, Such as a 2N5777.

Gary

Where would I get one of those, I checked my 3 main suppliers (Mouser, Digi-Key, and Jameco) and none of them had it, or atleast I didn't find it. When I did a search on google for it I just got all of those sites that list as many part numbers as possible, and are basically useless. Thanks for the info though. When I bought the phototransistor I am using from jameco I thought it was best for visable light. I'll have to memorize the color wavelengths.

Thanks!

 

[chiba]

You could use an LDR, they tend to be easier to use with a laser pointer.

 

[chemelec]

adamthole.

I thought that Digi-key had them, but they don't anymore.

I tried one of my suppliers, but there closed till monday.

Ether way, I will check with them next week because My stock on this part is also low and I need some more.

If I get some more of them, and you Email me about this, I could send you one for free. There not very expensive and its not worth the effort for you to send me the money.

Gary

 

[adamthole]

Thanks alot chemelec!

I made some adjustments to my circuit, and I got it to work alot better. The basic set up is that I have my photodiode's emitter connected to the base of a 2n3906 and the collector to ground. The emitter of the 2n3906 is connected to an HT12D chip. When the photodiode is active it will turn on 2n3906 and make that pin low. All other times that pin is held high via a resistor. In my first circuit I used a 1K resistor. In my second make of it, I changed the resistor to a 100K and it works much better. It still could be better, but it works much better than it did. I am looking forward to testing out the new photodiode.

Thanks everyone!

 

[ljcox]

Do you really need to use a laser?

It may be better to use a normal LED with a pair of lenses. One in front of the LED to focus the light into a beam and the other in front of the transistor to focus the light onto the base area.

Len

 

[audioguru]

Hi Adam,
Please sketch your circuit. It sounds like the phototransistor is an emitter-follower, that drives a PNP transistor that is also an emitter-follower. With a 1k emitter resistor, the phototransistor needed a lot of light to conduct.

Now with 100k as its emitter resistor, the phototransistor needs much less light to conduct.

Emitter-followers don't have any gain. Your circuit will operate with much higher sensitivity if you use the phototransistor as common-emitter which has a high gain.

With both transistors as emitter-followers, I don't see how "When the photodiode is active it will turn on 2n3906 and make that pin low." :?: :?:

 

[adamthole]

Here is my schematic from Eagle...

I did it that way because the phototransistor by itself couldn't pull the pin low. However it could turn on the 2n3906, which could then pull the pin low.