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Simple Diode Experiment

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shay87

New Member
For my school task I must design an experiment based on Electronics and more specifically diodes and resistors, my teacher has kindly given us suitable equipment breadboards, photodiodes , voltmeters, LED's, Resistors and various power suppliers.

That is my design, I'm going to vary the voltage (with + and - ) inputs and the LED with a IRLED, In my discussion I am going to talk about Threshold Voltage and how it varied in both forward and reverse bias. All I need to do now is formulate an aim and a method, I have come up with a general aim, but I want to see what you guys think would be suitable for the experiment and I need a hand writing the method, also I need to make up a graph somehow?
 

schmitt trigger

Well-Known Member
You could plot a graph of led current vs emiter current...in case you are wondering, in optoisolators this is called CTR. (Current transfer ratio)
 

Torben

Well-Known Member
Hi, and welcome to the forum!

First off, either the schematic is not complete or else the it's doomed, I'm afraid. The IR LED needs its cathode connected to something--in this case, ground (0V) from the 12V power supply.

Second, 10Ω is way too small for the current-limiting resistor on the IR LED. That will give 1.2A current to the poor little LED and it will become an SED (smoke emitting diode). A couple of milliseconds later it will permanently become a DED (dark emitting diode). The resistor will also let out its magic smoke unless it's rated at something like 15-20 Watts. I would suggest starting out with a 1K resistor (to be quite conservative) or an 820Ω.

What is the purpose of the receiver side? As posted it doesn't appear to do anything--will you just use a multimeter to determine whether it's receiving anything from the transmitter?

Sorry that these are both negative points--I'm not trying to be harsh or anything. Just trying to help you get something going which will work.


Regards,

Torben
 

MikeMl

Well-Known Member
Most Helpful Member
And with a 10Ω current limiting resistor in series with the LED you are sure to blow it up if you turn the power supply adjustment knob above 3V!

The current limiting resistor must be sized so as to limit the current through the LED to its maximum allowed forward current when the supply is 12V.

R=E/I=(12-Vf)/Imax, where Vf is the forward voltage across the LED when the maximum allowed current Imax is flowing through it.

Vf for an IR LED that I am familiar with is 2.4V, while its Imax is 50mA.

so, R = (12-2.4)/0.05 = 192Ω, so round it up to 200Ω

Look up the Data Sheet for the specific LED you are using and recalculate the resistance.
 
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shay87

New Member
Thanks for the help guys, as you can see electronics aint my strongest field.
:eek:
What is the purpose of the receiver side? As posted it doesn't appear to do anything--will you just use a multimeter to determine whether it's receiving anything from the transmitter?

Sorry that these are both negative points--I'm not trying to be harsh or anything. Just trying to help you get something going which will work.


Regards,

Torben
I was thinking about doing something like that. Do you have any other ideas? i just need something easy to assemble, I want to keep the LED side of the set up though
 

Torben

Well-Known Member
Thanks for the help guys, as you can see electronics aint my strongest field.
:eek:

I was thinking about doing something like that. Do you have any other ideas? i just need something easy to assemble, I want to keep the LED side of the set up though
First I'll address the LED side of it. Don't keep what you have. (Well, the parts are there but the connections are incomplete and the resistor is far too low in value.) Read what MikeMl and I wrote. Do you have the specs of the IR LED you're using? Because as drawn, the circuit will do nothing since the LED has no current return path. And if you add a current return path to the power supply's 0V line as you should, and the power supply really is 12V, your LED will explode. You need to connect the LED's cathode and you really need to raise the value of the current limiting resistor from 10Ω to something much higher.

For the receiver side, I would add another transistor and use that to light a visible LED when the phototransistor detects something from the IR LED.

You will also need to be careful to shield the phototransistor from ambient light--the light from the windows/overhead lights/whatever will interfere and make it trigger even when it's not getting anything from the IR LED.

How much time do you have for this project?


Regards,

Torben
 

Mikebits

Well-Known Member
Ag taught me a lesson on this one. Many IR led's are rated around 1 amp of current. They are not like a standard Led.
 

shay87

New Member
First I'll address the LED side of it. Don't keep what you have. (Well, the parts are there but the connections are incomplete and the resistor is far too low in value.) Read what MikeMl and I wrote. Do you have the specs of the IR LED you're using? Because as drawn, the circuit will do nothing since the LED has no current return path. And if you add a current return path to the power supply's 0V line as you should, and the power supply really is 12V, your LED will explode. You need to connect the LED's cathode and you really need to raise the value of the current limiting resistor from 10Ω to something much higher.

For the receiver side, I would add another transistor and use that to light a visible LED when the phototransistor detects something from the IR LED.

You will also need to be careful to shield the phototransistor from ambient light--the light from the windows/overhead lights/whatever will interfere and make it trigger even when it's not getting anything from the IR LED.

How much time do you have for this project?


Regards,

Torben
I have until next monday, umm thats cool with the extra LED but how could I use that to formulate a graph and how should I set up the circuit to include the extra diode
 

Mikebits

Well-Known Member
Instead of varying the voltage to the IR LED, perhaps you might consider switching the LED on and off, sort of like sending a smoke signal. This data format would be easier to work with. You could then detect IR Led on and off much more easily. Something like a __--_---____---
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Ag taught me a lesson on this one. Many IR led's are rated around 1 amp of current. They are not like a standard Led.
That's rather misleading - they are rated at 1A+ current in VERY small pulses - essentially no different to a normal LED, if you pulse them at 1A for 1mS, then OFF for 99mS, the average current is only 10mA. This is how multiplexing works - it's the average current that matters.
 

Mikebits

Well-Known Member
No, many IR leds are rated continuous of around 350 ma and surge of plus 1 A. And I would think in most apps, the led would be pulsed rather than full on.
 

audioguru

Well-Known Member
Most Helpful Member
Show us the datasheet of an IR LED rated at 350mA continuously.
Maybe 50mA but certainly not 350mA.
Its forward voltage is about 1.3V and if its current is 350mA then it would dissipate 0.46W which will melt it.
 

Mikebits

Well-Known Member
Show us the datasheet of an IR LED rated at 350mA continuously.
Maybe 50mA but certainly not 350mA.
Its forward voltage is about 1.3V and if its current is 350mA then it would dissipate 0.46W which will melt it.
Well actually I thought you were the one that told me that in a previous thread, so apparently I misunderstood you. :eek:
 

unclejed613

Well-Known Member
Most Helpful Member
you could use a variable "constant" current source instead of the resistor. i'll post the schematic for you later...... so you will be able to directly control the LED current linearly and predictably, and it will make it easier to plot the graph...
 
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