Peak Forward DC Current of photodiode

[haku87]

Can anyone tell what it is meant by Peak Forward Current of photodiode..
I am using opb706A integrated IR sensor

 

[Styx]

it is just that the peak fwd current that that diode will withstand, there is usually a time associated with it.

All electronics are like fuses due to bond-wires and all that, pass enough current through a fuse and it... well fuses. Same is true for semiconductors

you can pass 1million amps through a small piece of wire for an extreamly short period of time with no damage its all about time


that diode would be able to take alot more for alot shorter period. Obviouly if it is hotter it can handle less (nearly all the time it is the temp that kills it be it too high temp near the silicon or the bond wires fuse)

 

[neon]

i am sorry to disagree Stix it is not about TIME it is about HEAT and power a simple diode can conduct 20 amperes if you can keep it cool. time is involved only because it cannot last forever. do you agree?

 

[Nigel Goodwin]

i am sorry to disagree Stix it is not about TIME it is about HEAT and power a simple diode can conduct 20 amperes if you can keep it cool. time is involved only because it cannot last forever. do you agree?

If we're getting pedantic?, it's not about heat, it's about POWER DISSIPATION - heat is a result, not a cause.

 

[Styx]

i am sorry to disagree Stix it is not about TIME it is about HEAT and power a simple diode can conduct 20 amperes if you can keep it cool. time is involved only because it cannot last forever. do you agree?

IF you want to be correctly padantic its all about energy :roll:

And I suggest you go and re-read my post FULLY

If I cool a piece of wire to -55C it will fuse after a period of time, if I had the wire at +125C it to would fuse after a period of time. The times will be different BUT it will still fuse after a period of time

 

[haku87]

Why?

1 us width
300pps
Current : 3A

That the time that come along.
Let say I only 'on' the IR transmitter for 100us with 10% duty.
Can I pass in 1A to it..

The output I put into a ADC.
It give me a reading of 19 - 120+ with 255 --> 5V
The signal is not big enough

I connect two sensor in series..
I used Ra (diode) = 18ohm, Rb(output 1) = 330k, Rc(output2) = 330K

I try using 33k for Rb and Rc but the maximum reading is just too low.. around 90 only

 

[Nigel Goodwin]

Re: Why?

1 us width
300pps
Current : 3A

That the time that come along.
Let say I only 'on' the IR transmitter for 100us with 10% duty.
Can I pass in 1A to it..

Yes, it's normal to pulse IR LED's at around 1A, but you need to ensure that the average current doesn't exceed the specs of the diode.

 

[haku87]

What do u mean by average current..
Any Formula?

 

[Nigel Goodwin]

What do u mean by average current..
Any Formula?

Average = current*mark/space

 

[haku87]

What is mark and space..
I turn on for 100us every 1ms
what u meant is average should not exceed Forward DC current.

How about the Load resistance. I know that rise time and fall time increase with load resistance. Is there a way to calculate how much load resistance is the best ?

Collector-Emitter Voltage 30V
Emitter Collector Voltage 5V
Collector DC Current

Dark Current 100nA, Vce = 5V, IF = 0;

 

[Styx]

you really want the RMS not the average current btw

 

[Nigel Goodwin]

What is mark and space..

Mark is ON, space is OFF.

I turn on for 100us every 1ms
what u meant is average should not exceed Forward DC current.

Styx is correct when he says you actually need the RMS value, and NOT the average (thank you Styx) - but as long as you're not pushing the limits the average is usually near enough. In your example above, with 1A pulses, the average is 100mA.

But you also need to consider the longer term average, IR is normally sent as bursts of 38KHz modulation - so if only 50% of the time consists of these bursts, the 100mA average now falls to 50mA.

 

[Roff]

Actually, duty cycle is mark/(mark+space).
And the rms value of a 10% duty cycle waveform is 3.16 times the average value.

 

[hyedenny]

Ron,

Dont you mean the average voltage or current of a sine wave is .318 times the peak-to-peak value??

I thought the RMS value is 1.11 times the average value, which would be "near enough" as Nigel pointed out.

Please explain if Im wrong.

 

[Roff]

Ron,

Dont you mean the average voltage or current of a sine wave is .318 times the peak-to-peak value??

I thought the RMS value is 1.11 times the average value, which would be "near enough" as Nigel pointed out.

Please explain if Im wrong.

Read the thread. It's about a 10% duty cycle current pulse waveform. I even mentioned 10% duty cycle in my post.

EDIT:
To actually calculate average power, you have to multiply rms current times rms voltage. For a nonlinear device like a diode, one or the other is not sufficient, as it is for a resistor, where you can calculate power if you know one or the other, and the resistance.

 

[hyedenny]

Read the thread. It's about a 10% duty cycle current pulse waveform. I even mentioned 10% duty cycle in my post.

Read my post. I even said "please! Where did you get the 3.16 number???


If youre unable to explain, could you please refer me to a text that would clarify the calculation? I realize that this whole thread is about a 10% duty cycle.

 

[Roff]

Read the thread. It's about a 10% duty cycle current pulse waveform. I even mentioned 10% duty cycle in my post.

Read my post. I even said "please! Where did you get the 3.16 number???


If youre unable to explain, could you please refer me to a text that would clarify the calculation? I realize that this whole thread is about a 10% duty cycle.

I was baffled as to why you thought I was talking about sine waves.

RMS means "the square root of the mean of the square". For example, let's say I have a 10Amp pulse with a 10% duty cycle (zero for 90% of the time).
The squared pulse is 100 (Amps squared) high. The mean (average) is 100 (Amps squared) times the duty cycle (10%), or 10 (Amps squared). The square root of 10 (Amps squared) is 3.16Amps, which is the RMS value.
I hate using so many parentheses, but I couldn't figure out a simpler way to write it. :cry:

 

[Styx]

:roll:

Known facts: 10% duty

MEAN => V*0.1
or I*0.1 if wanting amps


RMS == root of the mean of the square

RMS = SQRT( (V^2*t)/T)
Where T=Period and t = on-time => t/T == Duty

Abitary figure of 10 for V =>

MEAN = 1
RMS = 3.16

Therfore the ratio between RMS and MEAN for a square-wave of 10% duty is as Ron stated 3.16

 

[hyedenny]

OH, OK - thank you!
Styx, Did you mean that the MEAN=10 (not 1)??... and Ron, did you mean that the MEAN=10 (not the amps squared=10)?

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.

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.

 

[Roff]

<snip>Ron, did you mean that the MEAN=10 (not the amps squared=10)?<snip>.

Well, I said

The mean (average) is 100 (Amps squared) times the duty cycle (10%), or 10 (Amps squared).

The (Amps squared) is in there because that is the units of the mean, before you take the square root.

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.

This is all about RMS. The mean is just an intermediate result, after squaring and before square-rooting.
And where did the "50ma" come from?