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neon said: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?
neon said: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?
haku87 said: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..
haku87 said:What do u mean by average current..
Any Formula?
haku87 said:What is mark and space..
I turn on for 100us every 1ms
what u meant is average should not exceed Forward DC current.
Read the thread. It's about a 10% duty cycle current pulse waveform. I even mentioned 10% duty cycle in my post.hyedenny said: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.
Ron H said:Read the thread. It's about a 10% duty cycle current pulse waveform. I even mentioned 10% duty cycle in my post.
I was baffled as to why you thought I was talking about sine waves.hyedenny said:Ron H said: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.
Well, I saidhyedenny said:<snip>Ron, did you mean that the MEAN=10 (not the amps squared=10)?<snip>.
The (Amps squared) is in there because that is the units of the mean, before you take the square root.The mean (average) is 100 (Amps squared) times the duty cycle (10%), or 10 (Amps squared).
This is all about RMS. The mean is just an intermediate result, after squaring and before square-rooting.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.