# LED Formard current

Discussion in 'Electronic Basics' started by GrahamIT, Aug 8, 2004.

1. ### GrahamITGuest

Think of an optocoupler, e.g. ISP847 from http://www.isocom.com the MAX
forward LED current as stated in the data sheet is 50mA, but all the current
transfer tests are done at 5mA except the Vce(sat) which is done at 20mA,
confusing or what!!!

The question is, which LED forward current and voltage would be the best
choice, so that the output photo transistor was switched hard on to full
saturation but the LED had as long a life as possible?

I've been under the impression that a transistors base current necessary to
only "just" saturate the transistor should be increased by a factor of 5 to
fully saturate. Therefore as 5mA through the LED would only just saturate
the phototransistor, an LED current of 25mA would fully saturate it, but
their Vce(sat) current was only 20mA

If you can make sence of the question, then you're a better man than me

2. ### Michael A. CovingtonGuest

LED current is not proportional to the base current of the phototransistor.
5 mA is probably plenty. If the Vce(sat) is specified at 20 mA LED current,
then use 20 mA.

3. ### John PopelishGuest

These two requirements need opposite extremes. You need to define how
much collector current the output must pass, and what voltage drop is
acceptable. Then using the data sheet, see if you can extrapolate
what LED current that will require and if it is low enough for
acceptable LED stability. This last part is hardest to guess from
most data sheets. But it is safe to assume that operating at the
absolute maximum LED current is not a good solution and that staying
at or below the current where most of the specifications are done will
give a reasonable life.
The terms 'just saturate' and 'fully saturate' are not very well
defined. What voltage drop can you tolerate?
The forward current transfer ration for the ISP847 is more than 50% at
5 mA LED current and 5 volts across the transistor (not saturated, at
all). This is 2.5 mA of collector current. The ISP847D guarantees
300% (15 mA collector current) under the same input and collector
voltage conditions.

At saturation (defined as no more than .2 volts collector to emitter,
and a 1 mA collector current, 20 mA LED current may be needed. This
represents a forward current transfer ratio of only 5%, a 10 times
reduction of the linear case. But it is one point only. If you could
tolerate .3 volts drop, a much smaller drive may do. You can get some
idea how the drive varies with saturation voltage from the graphical
data.

Based on what I can only assume is typical characteristics (since it
is so much better than the one data point guaranteed), the guaranteed
value is very conservative compared to the graph of saturation voltage
versus input current and for several collector currents.
You have to start with what you need. How much collector current must
the device pass, and what voltage drop is acceptable? Then you can
start the difficult process of extrapolating what LED current may be
needed based on the fragmentary info on the data sheet.

4. ### GarethGuest

Have a look at the datasheet, there is a graph with the title
"Collector-emitter Saturation Voltage vs. Forward Current". From this
graph you should be able to see what Vce(sat) will be for various LED
forward currents at 25 degrees C. Don't forget to leave some room for
variations in Vce(sat) with temperature.

--

5. ### GrahamITGuest

Sorry for my ignorance, when I said "just starting to saturate" I meant
"just starting to conduct", high voltage drop across transistor low
collector current etc.

What I actually need is for the output of the optocoupler to be TTL
compatible, because it is to feed other TTL logic circuits in a CNC
application I'm working on. This part is for the parallel port breakout
board so I don't fry my laptop if I get a short from the 36V 9A motor
supply.

The circuit so far is described as this.

Parallel port output > Tri-State buffer / line driver > Opto-Coupler > Drive
Logic with separate and independent Vcc and GND either side of the
Optocoupler. I could use another buffer on this side of the opto I suppose.

The tri-state buffer is needed for power up and down stability, so the CNC
machine can't move during a reboot, the software I'm using outputs a 12.5kHz
signal on a parallel port output pin when ready to send data to the CNC
machine and I'm feeding that into a 1mS timed charge pump to the Tri-state
enable lines, also because my laptop has 3.3V and 5V output voltages from
the parallel port, so I thought I'd standardise at 5V. so the breakout board
could be used on desktop PC's as well, without overdriving the opto LED's.

25 years ago I was an Army Radio Telecommunications Technician, it's
surprising how much you forget if you don't use it.

6. ### GarethGuest

There are optocouplers available with logic gate outputs which may be
easier to use, for example:

http://uk.farnell.com/jsp/endecaSearch/partDetail.jsp?SKU=143534&N=401
I have found that desktop PCs also have 3.3V parallel port outputs, well
mine does anyway I haven't carried out a detailed survey.

--

7. ### John PopelishGuest

Most TTL inputs see anything above about 2 volts as a logic high. See
the data sheet for a typical LSTTL gate:

So you can have a tiny bit of leakage and still have a logic high out
of the transistor. With a 4.7k pull up resistor (i mA to pull down to
..3 volts) you can have about a half mA passing through it and still
have greater than 2 volts out of the coupler.

Another good way to make a fast logic output that is leakage and noise
immune is to use two couplers, with their outputs stacked to form a
push pull totem pole. Drive their inputs alternately. Most TTL
inputs will treat an open circuit input (neither coupler on) as a
logic high. You can parallel the pull up coupler with a 10k resistor
for extra insurance (adds an extra half mA load to the pull down
coupler).

8. ### JamieGuest

Nah.
there is Max handling current and there is Max usage currents/
Normally the lowest!
basically its like this.
you really don't want to max out your LED because all your going to
do is shorten its life spanned, normally 50% of the max forward current
is a good rule of thumb to follow with LED's, then there's the chart
that some times is given in spec's to show you where at what point the
light output starts to drop off the linear scale which is also a good
point of
reference.
and as far as transistors go, yes it is a good idea to fully
saturate the bias plus alittle more to keep the passive resistance in
the transistor low so less heat build up will be generated. so if you use a
open collector scheme to drive the LED's you should either lower the
feed voltage through the common side of the LED's to prevent over
current or use a series resistor on each led ..
remember that you will lose approximately .6 volts via the transistor.
unless your planning on using Power Fets.
P.S.
there are some lower current fets that will generate enough
resistance when fully one to give you the drop you need for your leds
thus reducing the part count but remember the heating problem.

9. ### GrahamITGuest

Not a bad idea, but they're bloody expensive, I can get all 3 of my octal
chips for less than 4 of these and I'd need 8 of them for all my outputs.
The IBM PC parallel port spec. says TTL 5V or TTL 0V toggle on output ports,
I think the 3.3V comes from using TTL compatible CMOS devices that were not
around when the original interface spec. was made up.

I suppose I could use pullup resistors, but that always seems a bodge job to
me, think I'll stick to the tri-state buffer with schmitt trigger input for
safety > optocoupler > schmitt trigger buffer. That way is a bit belt and
braces but should work nicely, I'll let you know.