Maker Pro
Maker Pro

driving an LED with a 7406 question

J

John Fields

Jan 1, 1970
0
When the 7406 is off, can I treat it as a 0.7 Voltage load with 40 mA
running through it to ground?

---
Depends on what you mean by "off".

It might be simpler to look at the 7406 as a single transistor with
its base pulled up to Vcc and nothing connected to its collector.
(View in Courier)


Vcc
|
[10K]
| C
| <--O--+-----B
| E
O S1 |
| |
GND---+---------------+

Normally open switch S1 is also connected to the base and is used to
turn the transistor on and off. In the open position there will be
charge flowing from base to emitter and then to ground, and that
current will turn the transistor on, making the collector-to-emitter
resistance drop to a low value. When the switch is made, the base
current will be diverted to ground, through the switch, and the
transistor will turn off, causing the base-to-emitter resistance to
rise to a very high value. Notice that whether there is current
into the base or not, nothing comes _out_ of the collector.

Now let's change the circuit a little and see what happens.

30V
|
|R2
[750]
5V |
| +--->OUT
[10K] |
| C
| <--O--+-----B NPN
| E
O S1 |
| |
GND---+---------------+

Now, with S1 open the transistor will be on because charge will be
flowing through the base-emitter junction to ground. This will then
allow charge to flow from Vcc through R2 and through the
collector-to-emitter junction of the transistor to ground. If
there's enough current into the base to cause the transistor to be
saturated, then the maximum voltage dropped across the
collector-to-emitter junction will be 0.7 volts with 40 mA of
collector current (7406, remember?). So, since we have 0.7V dropped
across the transistor and we have a 30V collector supply the
collector current must be:


E 30V - 0.7V
I = --- = ------------ = 39mA
R 750R


Now let's close the switch and see what happens:


30V
|
|R2
[750]
5V |
| +--->OUT
[10K] |
| C
|<--O--+-----B NPN
| E
OS1 |
| |
GND----+--------------+

Since the base current is now diverted to ground the transistor will
be turned off and collector current will cease flowing since the
path to ground will no longer exist. When that happens, the voltage
on the collector will rise to the supply voltage, 30 volts, which is
what the 7406 must stand off when it's turned off.

So, to get back to your question, "When the 7406 is off, can I treat
it as a 0.7 Voltage load with 40 mA running through it to ground?",
the answer is "no".

When the 7406 is off you should treat it like a reverse biased diode
in series with your load, and when it's on you should treat it like
a forward biased diode in series with your load.


Now, to address your original post (I may be repeating someone else
since I haven't read through the whole thing, but anyway...


5V
|
[Rs]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /



Assuming that you want to drive a 20 mA LED with a Vf of 2.2V,

Looking at the 7406 data sheet shows that when it's sinking 16mA
worst case Vol will be 0.4V and when it's sinking 40mA worst case
Vol will be 0.7V. If we assume the change is linear, then we have a
24mA change in current causing a 300mV change in voltage, which is
12.5mV/mA.

So, for 20mA, worst case Vol would be 0.45V.

For Rs,

Vcc - (Vled + Vol) 5V - 2.65V
R = -------------------- = ------------- = 117.5 ohms
Iled 0.02A

The closest standard 5% value is 120 ohms, which would be fine, and
the power it would need to dissipate with 20mA through it would be:


P = IE = 0.02A * 2.35V = 0.047 watts.

So since there's a little less than 20 mA through it, a standard 1/4
watt resistor would be fine.

So, here's what you need:


5V
|
[120]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /

And the LED will light when the MCU I/O goes high or goes high-Z.
 
P

petrus bitbyter

Jan 1, 1970
0
panfilero said:
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
= (5-3)/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 = (5-3-VOL)/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 far you're on the right way. The value of VOL depends on the current. At
16mA the VOL is 0.4V as a maximum. At 40mA the VOL is 0.7V as a maximum. So
it's safe to use VOLmax=0.4V, so

R = (5-3-0.4)/20 = 80E

use 82E as the nearest standard value. In pratical situations 100E to 470E
is used depending on the amount of light required.
so R = (5-3-0.7)/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?

You're off track now. The specified VOH is the maximum voltage the output
can handle without being damaged. You're using only 5V.

The specified IOH is the maximum current that will leak into the SN7406
output when VOH=30V. As your VOH is only 5V the IOH will be neglectable.
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

When MCU low, the SN7406 output will be high. They are inverters remember?
When you look at the schematic on the datasheet, it means that the basis of
the outputtransistor will be low. So no current will flow from collector to
emitter except from the leakage mentioned earlier. This leakage will not be
enough the light the LED, so it's off.

When the MCU is high, the voltage on the basis of the SN7406 output
transistor will rise. The current that flows into the basis now will drive
the transistor into saturation. The current through collector and emitter
will be maximum, only limited by the load between collector and supply
voltage. According to the specifications the output voltage will sink to or
below 0.4V. No need to say the LED will be on.

petrus bitbyter
 
R

Rich Grise

Jan 1, 1970
0
.
So, here's what you need:


5V
|
[120]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /

And the LED will light when the MCU I/O goes high or goes high-Z.

Um, I think that if you expect the MCU I/O to go hi-Z, you'd want
a pullup so that the 7406 doesn't have a floating input.

Thanks,
Rich
 
J

John Fields

Jan 1, 1970
0
So, here's what you need:


5V
|
[120]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /

And the LED will light when the MCU I/O goes high or goes high-Z.

Um, I think that if you expect the MCU I/O to go hi-Z, you'd want
a pullup so that the 7406 doesn't have a floating input.

---
Um, It's TTL, not CMOS. The pullup isn't a BFD.

Here's a 7406, run it and watch what happens to the output when the
input floats:


Version 4
SHEET 1 880 680
WIRE -48 48 -464 48
WIRE 256 48 -48 48
WIRE 400 48 256 48
WIRE 512 48 400 48
WIRE -48 80 -48 48
WIRE 256 80 256 48
WIRE 400 112 400 48
WIRE 512 112 512 48
WIRE -48 192 -48 160
WIRE 160 208 128 208
WIRE 256 208 256 160
WIRE 256 208 224 208
WIRE 512 208 512 192
WIRE 592 208 512 208
WIRE 512 224 512 208
WIRE -128 256 -208 256
WIRE -96 256 -128 256
WIRE 64 256 0 256
WIRE 256 256 256 208
WIRE 400 272 400 192
WIRE 448 272 400 272
WIRE 400 304 400 272
WIRE 128 336 128 304
WIRE 256 352 256 336
WIRE 336 352 256 352
WIRE -128 384 -128 256
WIRE 256 384 256 352
WIRE -464 400 -464 48
WIRE 128 432 128 416
WIRE 192 432 128 432
WIRE 128 448 128 432
WIRE -464 560 -464 480
WIRE -128 560 -128 448
WIRE -128 560 -464 560
WIRE 128 560 128 528
WIRE 128 560 -128 560
WIRE 256 560 256 480
WIRE 256 560 128 560
WIRE 400 560 400 400
WIRE 400 560 256 560
WIRE 512 560 512 320
WIRE 512 560 400 560
WIRE -464 592 -464 560
FLAG -464 592 0
SYMBOL res -64 64 R0
SYMATTR InstName R1
SYMATTR Value 6k
SYMBOL npn 0 192 R90
WINDOW 0 58 34 Left 0
SYMATTR InstName Q1
SYMATTR Value 2N2222
SYMBOL diode -112 448 R180
WINDOW 0 51 30 Left 0
WINDOW 3 24 0 Left 0
SYMATTR InstName D1
SYMATTR Value 1N4148
SYMBOL npn 64 208 R0
SYMATTR InstName Q2
SYMATTR Value 2N2222
SYMBOL res 112 320 R0
WINDOW 0 -39 44 Left 0
WINDOW 3 -46 76 Left 0
SYMATTR InstName R2
SYMATTR Value 100
SYMBOL res 112 432 R0
WINDOW 0 -47 37 Left 0
WINDOW 3 -60 68 Left 0
SYMATTR InstName R3
SYMATTR Value 1000
SYMBOL npn 192 384 R0
SYMATTR InstName Q3
SYMATTR Value 2N2222
SYMBOL diode 224 192 R90
WINDOW 0 -36 36 VBottom 0
WINDOW 3 -32 36 VTop 0
SYMATTR InstName D2
SYMATTR Value 1N4148
SYMBOL res 240 64 R0
SYMATTR InstName R4
SYMATTR Value 1.4k
SYMBOL res 240 240 R0
SYMATTR InstName R5
SYMATTR Value 2k
SYMBOL npn 336 304 R0
SYMATTR InstName Q4
SYMATTR Value 2N2222
SYMBOL res 384 96 R0
SYMATTR InstName R6
SYMATTR Value 1.6k
SYMBOL npn 448 224 R0
SYMATTR InstName Q5
SYMATTR Value 2N2222
SYMBOL voltage -464 384 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL res 496 96 R0
SYMATTR InstName R7
SYMATTR Value 1,6k
TEXT -450 576 Left 0 !.tran .1 uic
TEXT -208 288 Left 0 ;IN
TEXT 552 232 Left 0 ;OUT
 
R

Rich Grise

Jan 1, 1970
0
So, here's what you need:


5V
|
[120]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /

And the LED will light when the MCU I/O goes high or goes high-Z.

Um, I think that if you expect the MCU I/O to go hi-Z, you'd want
a pullup so that the 7406 doesn't have a floating input.

Um, It's TTL, not CMOS. The pullup isn't a BFD.

Probably not, it's just that I'm chicken to let inputs float at all.
:)

Thanks!
Rich
 
R

rickman

Jan 1, 1970
0
---
Um, It's TTL, not CMOS. The pullup isn't a BFD.

Here's a 7406, run it and watch what happens to the output when the
input floats:

Version 4
SHEET 1 880 680
....snip...

Well, let's see... rude, obnoxious and ignorant. What a great
combination!!!

Dude, chill out a little bit. Simulations are not the real world and
unless you are dealing with no-neck, leather clad bikers with chains
and drawn knives, you are better off not trying your very best to PO
everyone you speak too. BTW, I know this from experience. :^)
 
J

John Fields

Jan 1, 1970
0
So, here's what you need:


5V
|
[120]
|
|A
7406 [LED]
| \ |
MCU I/O>---| >O---+
| /

And the LED will light when the MCU I/O goes high or goes high-Z.

Um, I think that if you expect the MCU I/O to go hi-Z, you'd want
a pullup so that the 7406 doesn't have a floating input.

Um, It's TTL, not CMOS. The pullup isn't a BFD.

Probably not, it's just that I'm chicken to let inputs float at all.
:)
 
J

John Fields

Jan 1, 1970
0
...snip...

Well, let's see... rude, obnoxious and ignorant. What a great
combination!!!

---
Ah, I see that another churl is itching to be dispatched.
---
unless you are dealing with no-neck, leather clad bikers with chains
and drawn knives, you are better off not trying your very best to PO
everyone you speak too. BTW, I know this from experience. :^)

---
**** you, asshole pussy.

Just because you said something stupid in real life/time and got
buttfucked for it in the environment you chose to embrace doesn't
mean that all of the rest of us are in the same boat.

Can you understand that?
 
R

Rich Grise

Jan 1, 1970
0
...snip...

Well, let's see... rude, obnoxious and ignorant.
What a great combination!!!

Huh? He was a little bit, um, abrupt, but I don't find his post "rude,
obnoxious and ignorant" at all. :)

Cheers!
Rich
 
C

CBFalconer

Jan 1, 1970
0
John said:
Ah, I see that another churl is itching to be dispatched.


**** you, asshole pussy.

Just because you said something stupid in real life/time and got
buttfucked for it in the environment you chose to embrace doesn't
mean that all of the rest of us are in the same boat.

Can you understand that?

PLONK.

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>

"A man who is right every time is not likely to do very much."
-- Francis Crick, co-discover of DNA
"There is nothing more amazing than stupidity in action."
-- Thomas Matthews
 
E

Eeyore

Jan 1, 1970
0
I'm sure it's working. It's just that he doesn't understand how it works.

Graham
 
E

Eeyore

Jan 1, 1970
0
You're rarely calm and collected.

You're usually profane.

Graham
 
R

rickman

Jan 1, 1970
0
---
Ah, I see that another churl is itching to be dispatched.
---


---
**** you, asshole pussy.

Just because you said something stupid in real life/time and got
buttfucked for it in the environment you chose to embrace doesn't
mean that all of the rest of us are in the same boat.

Can you understand that?

Boy, now I know why they came up with LOL! When I read this everyone
around me turned to see what I was laughing about!!!
 
J

John Fields

Jan 1, 1970
0
Boy, now I know why they came up with LOL! When I read this everyone
around me turned to see what I was laughing about!!!
 
Top