# driving an LED with a 7406 question

Discussion in 'Electronic Design' started by panfilero, Feb 13, 2007.

1. ### panfileroGuest

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

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 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?

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

2. ### GenomeGuest

Your 7406 has an open collector output so its output is a transistor... with
the emitter grounded, the base going into the guts of the IC and its
collector connected to the output pin.

It can only sink current, not source it.

The transistor has a maximum rating of 30V when off so you could, with an
appropriate limiting resistor connect your LED to 30V. You could connect a
number of LEDs in series along with an appropriate resistor and light them
all from one output.

The IOH of 0.25mA is probably the maximum specified 'leakage' current that
the transistor will sink when it is off with 30V applied to it. It's still
sinking it so it won't hurt your LED.

[I'm guessing, I haven't read the dirty sheet]

DNA

3. ### AnirbanGuest

It's not that (5V-3V)/R = 20mA.
You need to account for the 0.7V drop across the output transistor in
the 7406.
So it should be (5V-3V-0.7V)/R=20mA
Otherwise it is fine.

Aside, if the MCU can directly drive an LED, why use a 7406. Most MCUs
these days can. Tie the anode of the LED to Vcc=5V and the cathode to
the MCU pin. If needed, throw a 100ohm resistor in series. The cathode
should be tied to the MCU because most MCUs can sink more current than
they can source.

--- Anirban

4. ### panfileroGuest

7406 is high..... does this mean that IOH = 25 mA is coming out of the
7406 and headed towards the LED? and what is the VOH = 30 V about?
That seems really high? I thought I could treat the VOH like a
voltage drop inside the 7406, (like i did for the VOL = 0.7 V part)
but.... I guess I ignore the 30 V, cause there's no where for that to
be coming from.... so, why is the LED off now? I don't understand
how the LED is off, there's still that 5 Volts and that resistor....
aren't they pushing a current through the LED? does the lead from the
LED now connect to another reistance and voltage source equivalent
thing inside the 7406?

thanks
confused.

5. ### EeyoreGuest

What on earth are you using a 7406 for ?

Graham

6. ### EeyoreGuest

Yes, it's an inverting buffer.
No. In any case, the LED would be the wrong way round for that to work, if it
was the case, which it isn't.

The maximum voltage the open collector output can withstand.
http://en.wikipedia.org/wiki/Open_collector
It seems your confusion relates to not understanding what the data sheet's
telling you. And some problems understanding basic electricity and how circuits
work too !

Graham

That is the right approach, although Vol probably will depend on the
actual current.

One thing to note is that 20 mA will be very bright, brighter than what
is typically needed for an indicator. The best answer is probably to
try different values. You are in the right ball park.
Assuming the driver is powered by 5 V, the output will simply go to that
voltage. Since the other side of the LED is 5 V, there will be no
output current.
Not needed. With no significant potential difference the current will
be insignificant.

8. ### panfileroGuest

ok, thanks that does clear it up quite a bit... i was confused as how
we treat the VOL as a voltage drop but not the VOH.... what if I were
to remove the LED and put a transistor there in it's place with a load
in the emitter going to ground and the collector going to a 5 Volt
source.... like this:

o 5 Volts
|
|
|
/
\ R
/
\
| o 5 Volts
| |
| |
________ | |
| | | ----|
| | |\ | /
| | | \ | |/
| MCU |-----------| \---------|---------|
| | | / |\
| | | / \
| | |/ \/
| | |
------- 7406 |
|
-------
| |
| |
| |
| |
-------
|
|
---
\\\

If my VBE = 1 [V].... and I needed 750 mA to get my load to turn on,
and the load was small, like 4 Ohms.....could I do this? If I try to
figure it out like the LED one, then my 7406 has a voltage drop of 0.7
Volts. but that's in parallel now wtih the 1 V from the VBE drop and
the VLoad = (4)(.750) = 3 V drop from the load.... if my transistor
has a gain of hfe = 40, and the 7406 sinks IOL = 40 mA, then do I need
my base current to be..... IB = hfe*IC + 40mA ? that way the 40mA will
get sunk but the rest will excape in order to turn on my transistor
and turn on my load? I think this is correct, but then I don't know
what voltages to use in order to figure out what size resistor R I
would need to do this?

I don't know whether to use the VBE and VLoad.... or the VOL from the
7406?

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12. ### EeyoreGuest

There is nothing in the original post to indicate a reason for its use thank you
very much Mr Allegedly Polite !

Does anyone even still make a standard TTL 7406 ? It seems unlikely to me.

Graham

14. ### panfileroGuest

Who cares? It's what _he's_ using.

Thank you JF. I appreciates everyones concern over the components I'm
using, but I'm not looking for a more efficient design to this simple
problem, I was looking for some understanding to these kind of simple
circuits. This problem is just for helping myself understand things
not for efficiency or pratical use in any specific application.

Thank you

Well ? So ?

Graham

16. ### panfileroGuest

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

17. ### tsest(1)Guest

Yes

http://focus.ti.com/docs/prod/folders/print/sn7406.html

18. ### gGuest

The first thing I sometimes do, is take the LED and see how much
current
it needs to be bright. I'm not talking about lighting up the room
either.
I usually use between 5 an 10 ma. max for indicators. Back in the old
days
you needed the current for a little brightness.

greg

19. ### rickmanGuest

It can be hard enough to explain this sort of design in person, much
less by group postings, but I'll give it a try.

The 7406 or any other switching device has two sets of ratings; one
for input and one for output. The input specs tell you what is
required to drive the inputs to a valid state. The output specs tell
you what sort of load and voltage can be driven by the output.

The LED requires about 20 mA at about 2.2 volts to drive it. If you
use a 5 volt supply and use a 74LS06 as the control (I can't find a
data sheet on the 7406) you need to subtract the LED voltage and the
74LS06 output voltage ([email protected] mA = 0.4V) from the power supply to find
the voltage on the current limiting resistor. 5 - 2.2 -0.4 = 2.1V.
At 20 mA this will give you 2.4V / 20 mA = 120 ohms.

Since this is an open collector device, it will not source any current
to the LED, so you don't need to worry about that. The IOH of this
device is not given because it does not drive current when the output
is high. Regardless, IOH and IOL are maximum values or conditions for
measuring VOL and VOH. The outputs of digital devices are just
transistors connected to ground and power; they are not current
sources. They provide a voltage and the current drawn from the output

In the above example I only needed to know the current into the 74LS06
output because there is some resistance in the transistor connection
to ground so that the output voltage of the 74LS06 depends somewhat on
the current flowing. At 24 mA the output voltage is 0.4 V, at 48 mA
the output voltage is 0.5 V. You can see this is not a large
difference. Even a microAmps of current, the output voltage will
likely be around 0.2 volts. For most circuits you can estimate the
output low voltage to be about 0.4 volts for TTL type devices. CMOS
devices are much more like resistors on the output so the resulting
voltage varies directly with the current.

Does that help?

20. ### redbellyGuest

One approach is to try the larger resistor value (100 ohms) and
measure the current. If you're wrong, then you'll get too little
current. But it WON'T fry any of the components, and you can then get
a better idea what the correct resistor should be.

Sometimes a little experimenting helps more with learning electronics
than spending a lot of time getting the calculation correct.

Mark

p.s. to measure the current, I'd measure the resistance before putting
the resistor into the circuit, and then measure the voltage drop
across the resistor while running the LED. But that's just my
preference, I tend to avoid using ammeters directly.