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Peak Forward DC Current of photodiode

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hyedenny said:
One more question: Isnt the RMS current through the diode more significant in this case? It would seem that using the average vs RMS isnt at all "near enough."

I think this is getting VERY confusing?, and certainly I'm getting confused :?

Giving it some serious thought (after having had a kebab for tea!), I don't see the relevance of RMS in this case?.

The diode isn't specified for RMS maximum current, so as I see it there's no reason to concern yourself with it?. 1A of current for 10% of the time works out as an average of 100mA, or for 50% of the time works out as 500mA and so on.
 
Nigel Goodwin said:
hyedenny said:
One more question: Isnt the RMS current through the diode more significant in this case? It would seem that using the average vs RMS isnt at all "near enough."

I think this is getting VERY confusing?, and certainly I'm getting confused :?

Giving it some serious thought (after having had a kebab for tea!), I don't see the relevance of RMS in this case?.

The diode isn't specified for RMS maximum current, so as I see it there's no reason to concern yourself with it?. 1A of current for 10% of the time works out as an average of 100mA, or for 50% of the time works out as 500mA and so on.
But it is specified for maximum power: 75mw-1.25mw/°C, but that isn't as easy to predict, because of the nonlinear nature of the beast.
I have to confess I started this quagmire just to clarify the fact that there is a significant difference between average and RMS for a pulsed waveform.
 
Ron H said:
I have to confess I started this quagmire just to clarify the fact that there is a significant difference between average and RMS for a pulsed waveform.

Sorry, I'm not a priest, I don't take confessions :lol:

To try and make it easier to understand, how about this:

You have a 1KW electric fire, and turn it ON for an hour, energy consumed 1KWH - heat produced accordingly. You then turn the same fire ON for half an hour, and leave it OFF for the second half hour, energy consumed is 0.5KWH - heat produced is half of that previously produced.

The only difference between that and the LED is the frequency of switching.
 
Ok good, so Im NOT crazy..... yet.

Kebabs sound good!!! ....but with TEA????!!!! Hmmmm - food for thought. I guess thats another forum though.
 
Nigel Goodwin said:
Ron H said:
I have to confess I started this quagmire just to clarify the fact that there is a significant difference between average and RMS for a pulsed waveform.

Sorry, I'm not a priest, I don't take confessions :lol:

To try and make it easier to understand, how about this:

You have a 1KW electric fire, and turn it ON for an hour, energy consumed 1KWH - heat produced accordingly. You then turn the same fire ON for half an hour, and leave it OFF for the second half hour, energy consumed is 0.5KWH - heat produced is half of that previously produced.

The only difference between that and the LED is the frequency of switching.
Just keep in mind that we are talking about power here, and not current, which was the original focus.
 
hyedenny said:
OH, OK - thank you!
Styx, Did you mean that the MEAN=10 (not 1)??...
IF I ment MEAN=10 I would of said MEAN =10
There might be two MEAN in this case the first being the "mean" of the pulse-height which if you go read my post I took an arbitarty level of 10. With a PWM (which is effectivly averages ie MEAN-level) 10*0.1 = 1

hyedenny said:
So its: sqrrt[((10^2)(10))/100]=3.16

I thought RMS was normally assumed to refer to sinusoidal conditions (Horowitz & Hill, pg 16; Electronic Circuits Tooley, pg 68 ), thus my reference to sine waves and the .318 (1/pi) conversion factor. I knew that this wasnt a case of sine waves though.
RMS can be applied to ANY waveform its just in engineering the figure of 0.707 (or 1/SQRT(2) ) is drilled into us as the RMS of a SIN wave since a sine wave is a regular waveform. It is also a prime example by the point you get to starting to know abt RMS you would have worked with Sinwaves alot and would of mastered Trig. SINwaves are a GREAT waveform because the MEAN =0 (the squarewave in this example doesnt).

Now since the question usually goes "an electric heater is connected to the mains that is sinusoidal, how much power is transmitted"
Since up until that point we had always been working on MEAN from the knowledge at that point it looks like an average amps of ZERO and an average volts of ZERO is no power??
THEN RMS is explained and we go AHHHHHHH

RMS gets applied to anything.
Take a Quazi squarewave EASY (120deg on 60deg off +VE and -VE, important waveorm in brushless DC)
RMS = SQRT(2/3)*V

hyedenny said:
One more question: Isnt the RMS current through the diode more significant in this case? It would seem that using the average vs RMS isnt at all "near enough." The 50mA average would become 158mA RMS.
Im going to have to do some reading today! Thanks again.
Well this is where me and othere here will have a difference is opinion to me it is ALWAYS RMS when it comes to power and since power is involved here then it is RMS.

Since the original question was abt peak-pulse of current the datasheet will say how wide the pulse can be and the period of repitition.
IF you work out the RMS of that pulse you can then have a higher PEAK current but for less time

ALWAYS RMS
 
It kinda confusing.. Let say if I connect the diode in series with a resistor and a driver(LN2003). with 5V applied to it.. the input of the driver is a clock signal.

What is the resistor value to use? I mean how to calculate out the value.
I try using 36 ohm. But the output of the ADC range from 19 to 120+ only. The maximum should be 255. Therefore, I think that the current is no enough. My lecturer told me that 50mA is already very big to drive the diode which he take from datasheet but I think otherwise.

I saw the peak forward current and realised that the maximum should be around 3 since I not turning it on all the time. Only 10% duty cycle.

Could anyone show me how to calculate the resistor value of the diode current limiting resistor

And also the value of the load resistor?

do u really need the rms of the clock signal. I am using a rectangular wave. ('1' for 100 us and '0' for 900 us)
 
Post the circuit that you're talking about, I'm rather confused by your description?.

The series resistor is simply calculated from ohms law, you know the current you want, and you need to calculate the voltage dropped across the resistor - basically the supply less the forward voltage drop across the LED.

I see no need to worry about RMS, I don't see as it has any bearing at all here - see my electric fire example for the reasoning!.
 
here is the diagram?
The problem lie in I dun know wat current should I allow it to flow in the diode. I afraid it will burn the diode.
 

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haku87 said:
here is the diagram?
The problem lie in I dun know wat current should I allow it to flow in the diode. I afraid it will burn the diode.

What are you trying to do anyway?, the diagram gives the impression of a very short distance between the LED and photo-transistor.

Assuming you're looking for remote control type distances?, feel free to 'steal' the circuit off my tutorials.
 
Line tracking..

Distance from sensor and the ground is around 0.5 - 1 cm only.. SO dun need IR modulation.

Trying to experiment how signal change when going over black to white line.

the diode and transistor are integrated and are very close to each other
 
haku87 said:
Line tracking..

Distance from sensor and the ground is around 0.5 - 1 cm only.. SO dun need IR modulation.

Trying to experiment how signal change when going over black to white line.

the diode and transistor are integrated and are very close to each other
I'm not a robotics expert, or an IR expert, but (hyperbole follows) you could probably follow the white line on the highway in a helicopter if you pulse the diode with 3 amps!
 
As Ron H says, you only need VERY low power to do that, 1A pulses give 10m-20m remote control ranges, you only need to feed it with a few mA. There are loads of line follower and micro-mouse robot designs on the net, just check what they use!.

You can also buy reflective IR sensors (LED and photo-transistor in one package), try checking the datasheets for those!.
 
Styx,
uh..... OK.

RonH,
Thank you for clearing that up - and for the use of coherent and comprehensible english, grammar, syntax, spelling, and math.
 
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