# Phototransistor

Discussion in 'Sensors and Actuators' started by Rajinder, Jul 10, 2019.

1. ### Rajinder

419
7
Jan 30, 2016
Hi,
I am testing the circuit below.
My understanding is that light controls the current from the base to emitter, which controls the collector current.
I am powering from 3V.
I see a change of 0.2V (when it is dark)to around 2.8V at the output (light conditions).
I am trying to understand the theory to my actual results in terms of calculations. Can anyone assist?
Best regards
Rajinder

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2. ### Nanren888

195
36
Nov 8, 2015
Given the emitter resister and the voltages you quote, you can calculate the change in emitter current.
As an emitter follower it seems you have most of it worked out.
For an emitter follower, if you imagine that the current gain is high, most of the emitter current is from the collector.
In this case with not actual base connection, so no actual base current, ...
Not sure what you want that you can't derive immediately or google.

3. ### Bluejets

3,775
782
Oct 5, 2014
For a simple switch, the resistor is normally placed from positive rail to collector.

4. ### Rajinder

419
7
Jan 30, 2016
Thanks for the replies.
I just wanted to know how to calculate the gain ratio of such a circuit and relate this to the voltages i am seeing. Is there a simple calculation?
The phototransistor is
Kingbright KPA-3010P3C.

5. ### Audioguru

2,726
614
Sep 24, 2016
Since you do not feed base current into a photo-transistor then the "gain ratio" does not mean anything.
Instead, you look on its datasheet for its minimum brightness to output current ratio.

6. ### hevans1944Hop - AC8NS

4,378
2,046
Jun 21, 2012
Read the information provided in the datasheet, particularly the information shown in Figure 4 "Collector Current" which plots collector current as a function of illumination intensity.

I have no idea what you mean by "gain ratio" but this figure should help you calculate the output current as a function of input intensity. Does this help, or do you need something else?

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• ###### Kingbright Phototransistor Datasheet 2308777.pdf
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7. ### Rajinder

419
7
Jan 30, 2016
This is fine. Basically as the light intensity increases the collector current increases.
So i get around 2.8V when it is light and around 0.4v when it gets dark. So light mean more current multiplied by my 10K resistor. Hence more voltage.
Similarly less current multiplied by the 10K means less voltage. Is that correct?

I was getting confused that we had a hfe gain value for the transistor.
Thanks.

8. ### hevans1944Hop - AC8NS

4,378
2,046
Jun 21, 2012
This is plotted on non-linear scales in Figure 4 for both the illumination intensity and the resulting collector current as a function of that intensity. In your initial post you stated:
How do you expect to perform any calculations when you don't know the theory of how a photo-transistor controls collector current? Your "actual results" consist of two (and only two) measurements representing fully dark and fully illuminated. Where are the "measurements" you will need to "understand the theory to my actual results" when you only have two results?

The collector current varies continuously between some minimum value and some maximum value as the infrared illumination varies continuously from some minimum value to some maximum value. You can read the values between those two limits off the graph in Figure 4. Or set up a calibrated illumination source and measure them yourself... not an easy nor simple task, and it is fraught with systematic errors that you won't even recognize at your current level of knowledge.

Not even close to being correct. Resistors do not multiply current. And since there is ZERO base current, the transistor does not multiply current either. What the transistor DOES do is control the collector current as a function of the illumination intensity. The collector current causes a voltage drop across the resistors through which the collector current flows. Go read some good text books on solid state electronics theory to see how this might occur.

I bet you were, because nowhere in the datasheet is mentioned "a hfe gain value for the transistor". Why not? Because there is no base current ibe from which to form a ratio of ice to ibe to define hfe.

9. ### Rajinder

419
7
Jan 30, 2016
Thanks for your help. I know my questions seem silly sometimes. But i will do some solid reading. Thanks

10. ### hevans1944Hop - AC8NS

4,378
2,046
Jun 21, 2012
I think you will find solid state electronics theory both interesting and gratifying. Briefly, the photo-transistor effect is caused by the creation of electron/hole pairs by the absorption of photons in a semiconductor junction. Each photon can create on electron/hole pair, but not all absorbed photons do. Some are just absorbed and add a small amount of heat equal to their energy. The ratio of the number of electron/hole pairs created to the number of photons absorbed is the quantum efficiency of the device.

The electric field between the base and collector accelerates and adds energy to the electrons, thereby causing the collector current to increase with increasing light intensity. The holes diffuse through the thin layer of the junction and re-combine with electrons injected back into the emitter through the external emitter-collector circuit. Not all electron/hole pairs participate in the photo-transistor action. Some will re-combine inside the base-emitter junction, leading to reduced quantum efficiency,

Now, add some physics and some math to the above layman's description and you can begin making calculations and predictions of performance for photo-transistors you design and build. Or you can just consult manufacturer datasheets to find something that suits your purposes and objectives. Lots of variables involved here, some of which you can specify or select for and others that you just have to live with because it happens to the be state of the art at this time. But there are plenty of photo-transistors to choose from. May we ask why you chose this particular one? Or was it something you got cheap and just wanted to play with? I have found that is one of the best ways to learn about new things, especially new things I cannot afford. Keep on doing some solid reading. There are no silly questions, but the world abounds with silly answers. Keep on asking until understanding occurs, often as an "Ah, ha!" moment.

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11. ### Rajinder

419
7
Jan 30, 2016
Thanks for all your help. Could you recommend any books. I need something that has theory with some practical work on that theory. I think this would be the best solution for me.

12. ### hevans1944Hop - AC8NS

4,378
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Jun 21, 2012
All the texts that I studied from are fifty or more years old. You need to find something apropos to the 21st Century! The theory hasn't changed much, but the practical implementation has. The Internet is one of the best tools available to find what you need to study. One of the best sites on the Internet that I have found for exploring topics in physics is HyperPhysics located here.

When you visit this introductory page you will be offered an opportunity to purchase the latest version of the online tutorials and web links. Unless you lack broadband Internet access, i.e., have a slow connection to the Internet, I would NOT purchase either the DVD or the memory stick version. Instead, save the fifty bux toward the cost of a real text book that you can take notes in. There is nothing wrong with writing little reminders to yourself in the margins of the pages if that will help you remember and understand a concept. For that reason, I prefer books with large margins in which to write.

The HyperPhysics rabbit hole is very deep with thousands of side branches to explore, but just dive in and see where it takes you. Don't forget that your objective is to discover modern references to topics in physics. If you have a university library nearby that will allow you to check out books, that is a good way to "try before you buy" a text book. New text books are very expensive, but sometimes you can find bargains in used books. Beware of books that have been heavily highlighted. This is a sure indication that the owner was struggling to understand the topic being discussed, and probably failing to do so. Books without ANY margin notes are good too, because it indicates their owner really didn't use them much.

13. ### Nanren888

195
36
Nov 8, 2015
Actually a quick search of a lot of more recent electronics texts shows very few I looked at introduce phototransistors.