Need help with a spec in a datasheet.
James Newman
Posts: 133
http://www.vishay.com/docs/83608/83608.pdf
This is a datasheet for an optocoupler I have alot of. I was about to use some with the prop, and was happy to read that the input is 'IC' compatible... however it states later in electrical characteristics "Forward voltage IF = 10 mA VF 1.2 1.5 V" 1.2 being typical and 1.5 being max. Is that the maximum voltage I can apply from base to emitter, or is that the maximum voltage required to fully turn it on? I would expect it to be listed under 'Absolute Maximum Ratings' instead if that was the former.
Someone with more datasheet expertise should enlighten me please.
This is a datasheet for an optocoupler I have alot of. I was about to use some with the prop, and was happy to read that the input is 'IC' compatible... however it states later in electrical characteristics "Forward voltage IF = 10 mA VF 1.2 1.5 V" 1.2 being typical and 1.5 being max. Is that the maximum voltage I can apply from base to emitter, or is that the maximum voltage required to fully turn it on? I would expect it to be listed under 'Absolute Maximum Ratings' instead if that was the former.
Someone with more datasheet expertise should enlighten me please.
Comments
Most LEDS i have worked with, don't state maximum voltage, just current.
More voltage means you need a higher current limiting resistor.
Any PROPER datasheet engineer would know that LOGIC level = 0 to 5v. (if you look at history)
(in GENERAL, yes yes, cmos, 3.3v logic, etc.. if any engineer thinks that LOGIC voltage is something other than the standard 5v, then the proper thing to do is specify what they mean by LOGIC LEVELS.) If they don't specifically state what they mean by LOGIC LEVEL, then it SHOULD mean 5v.
Post Edited (Clock Loop) : 9/1/2009 2:04:47 AM GMT
·····(3.3V - VF) / IF = (3.3 - 1.2) / 0.010 = 210
So a 220R series resistor will give you the current drive you need from the Prop.
-Phil
Uhh, this looks like your saying that the LED's will conduct even if you apply 10ma at 1v. Which it won't.
1.5v(1.2v) is how much the voltage will DROP when 10ma is applied.
If I fed 5v @ 10ma into the LED, if you read the voltage on one of the legs (depending on where you put the resistor) you will read 5v - 1.2v = 3.8v after drop Or the higher value, 5v - 1.5v = 3.5v after drop
The LED will not conduct with less than 1.2v applied. The VF is how much is required to bias the LED, and is also how much the voltage will drop when forward biased.
I think I got all that right. Im sure I mangled something in there. And oh, lets not argue voltage versus current, I hate that game. Its all relative.
Post Edited (Clock Loop) : 9/1/2009 2:15:03 AM GMT
Think of the input of an optoisolator as a garden-variety LED.· You can think of it that way because it is a garden-variety LED.· Phil's post tells you how to drive a garden-variety LED from a Prop output.
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· -- Carl, nn5i@arrl.net
Well with my "mangled" glasses, I see it differently. SO, fer all yew people with mangled glasses out there,
I offer another way to interpret the data.
The data sheet characteristics tell you more about the behaviour of the device - so here the VF characteristic tells you that with a current of 10 mA flowing the LED, (or opto-isolator in your case) will have a volts drop of typically 1.2 V.
Perhaps 10 mA is a good value to use to turn on your opto – if so then using a 3.3V source, (i.e. a Prop port), you only need to connect a series resistor - the value of which was calculated for you by Phil.
This is your fundamental mistake. The led will drop roughly its VF whether it's drawing 1ma or blowing up at 1A. You can't "apply 10ma at 1v". You apply a voltage and the LED draws a current. You can limit the current by using a (calculated value) series resistor. A led will conduct quite happily at 1V. It may not draw a lot of current or emit a lot of light, but it certainly conducts.
Ohms law is incredibly simple, but it's known as a *law* for a reason. A little like gravity or thermodynamics, it's just not negotiable.
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lt's not particularly silly, is it?
You can even drive them directly from 120vac, though in that case using a capacitor rather than a resistor to limit the current results in less heat.· This is quite typical of what you'll find in cheap plug-in devices like those air fresheners.
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we can think of the diode as having an infinite resistance at voltages below vf and then above vf diodes have whatever resistance is necessary to maintain vf at whatever current is applied to them. then they will get damaged if the current applied to them is too high, so you have to limit the current in the whole system by adding a resistor. to do this, as is shown in phil's equation you subtract vf from the voltage that you are supplying to the system, which is 3.3 volts, then that gives you the voltage that the resistor sees 2.1 volts. since the current going through the resistor and the led will be the same you just apply ohm's law v=ir to find r where v=2.1 and i =.01 .
bradc,
diodes are non ohmic devices. on page 3 of the data-sheet there is a graph of voltage versus current, if it followed v=ir it would be linear, but instead it is a flat line at voltages below vf, and a curve near it. a diode that followed ohms law would be a resistor.
Thanks again [noparse]:)[/noparse]
Now I just need to get my head around caps better too.... mainly in determining values... but I think I'm getting there.
I never suggested a diode was an ohmic device. I specifically referred to a calculated series resistor to limit the current in the device. I'm sorry about the reference to ohms law, I thought it was quite obvious I was using it to calculate the series resistor.
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lt's not particularly silly, is it?
(is the glass half empty or half full)
Truer words were never spoken. I like to phrase it thusly : "Know what you don't know".
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lt's not particularly silly, is it?