P
panfilero
 Jan 1, 1970
 0
Hello, I have a question about driving an LED with a 7406. Basically
I have an MCU going into a 7406 inverter going into an LED going to a
resistor R and that going to a 5 V power supply. All in series, and
I'm trying to find the value for R that would make this thing work....
here's a sketch of what I'm talking about (this is my first time
skecthing on notepad)
o 5 Volts



/
\ R
/
\


__
\/ LED MV5353

 
  
  \ 
   \ 
 MCU  \
   /
   /
  /
 
 7406
This is how I began to tackle this problem: for MCU output = high I
looked on the spec sheet for the LED it said that the LED uses a
continuous forward current of 20 mA at a voltage drop of 3 Volts.
Then I went to the 7406 specs sheetand looked up IOL = 40 mA. So I
think to myself.... great, it can sink 40mA and I only need 20mA. so R
= (53)/20mA = 100 Ohms.... now I just have to check R with MCU out =
low....
I start to get stuck here, cause I begin to think.... should I include
VOL as a voltage drop when finding R? shold R = (53VOL)/20mA ? Cause
I'm thinking VOL is the voltage drop for the 7406.... but I look on
the spec sheet and there's 2 values for VOL = 0.7....
so R = (530.7)/20mA = 65 Ohms
so.... when MCU output = low.... I'm not sure what to do here.... I
look up VOH.... VOH = 30 Volts.... whoa! hmmm..... I look up IOH, IOH
= 0.25 mA....... and then i think, i've got 25mA going towards that
LED, but wait.... it's going the wrong way, the LED will stop it...
but I wonder how much current the LED can stop?
at this point I'm totally confused and am not sure if I'm looking at
things the right way anymore, the VOH,IOH,IOL,IOH stuff is confusing
to me..... can anyone tell me if I'm on the right track?
thank you
Joshua
so for MCU = low: I guess I need to find IOH? That's IOH = 0.25
mA.... I'm not sure what to do next, should I look up VOH
I have an MCU going into a 7406 inverter going into an LED going to a
resistor R and that going to a 5 V power supply. All in series, and
I'm trying to find the value for R that would make this thing work....
here's a sketch of what I'm talking about (this is my first time
skecthing on notepad)
o 5 Volts



/
\ R
/
\


__
\/ LED MV5353

 
  
  \ 
   \ 
 MCU  \
   /
   /
  /
 
 7406
This is how I began to tackle this problem: for MCU output = high I
looked on the spec sheet for the LED it said that the LED uses a
continuous forward current of 20 mA at a voltage drop of 3 Volts.
Then I went to the 7406 specs sheetand looked up IOL = 40 mA. So I
think to myself.... great, it can sink 40mA and I only need 20mA. so R
= (53)/20mA = 100 Ohms.... now I just have to check R with MCU out =
low....
I start to get stuck here, cause I begin to think.... should I include
VOL as a voltage drop when finding R? shold R = (53VOL)/20mA ? Cause
I'm thinking VOL is the voltage drop for the 7406.... but I look on
the spec sheet and there's 2 values for VOL = 0.7....
so R = (530.7)/20mA = 65 Ohms
so.... when MCU output = low.... I'm not sure what to do here.... I
look up VOH.... VOH = 30 Volts.... whoa! hmmm..... I look up IOH, IOH
= 0.25 mA....... and then i think, i've got 25mA going towards that
LED, but wait.... it's going the wrong way, the LED will stop it...
but I wonder how much current the LED can stop?
at this point I'm totally confused and am not sure if I'm looking at
things the right way anymore, the VOH,IOH,IOL,IOH stuff is confusing
to me..... can anyone tell me if I'm on the right track?
thank you
Joshua
so for MCU = low: I guess I need to find IOH? That's IOH = 0.25
mA.... I'm not sure what to do next, should I look up VOH