Diode vs LED

A

ADWSystems

Member
I am working on a project that requires an I/O module that is obsolete and no longer manufactured. I have the schematic for the input section of the module, the section I need.



The input is in the upper right, the output in the upper left.

I even have a parts list
IC5..IC12 is a 4N25 optocoupler
R11..R18 ar 6.8K
R19..R26 are 680
D1..D8 are 1N4148 diodes.

I thought it would be nice if there was an indicator on board to show when the input was active. Can I switch out the 1N4148 diode for a small LED?
 
Opto22 modules often have a LED connected in place of the diode.
Also you can pick these up on ebay cheap, it should not be hard to sub the SSR, there are many to pick from.
Max.
 
The forward voltage drop of an LED is three times that of a signal diode. If the output circuit has an extra 1.4 V of compliance, then it might work. The extra 1.4 V of diode voltage probably will decrease the current through that circuit loop, dimming the LED a bit. This can be compensated for by decreasing the 680 ohm resistor.

ak
 

I'm only following part of your logic. I don't follow the LED vF=3x signal diode. I have not seen LEDs with Vf that high in a long time.

For kicks, I just finished breadboarding the circuit. I inserted the LED between the pin 1 and diode. Measuring the voltage at pin 1, with the LED is 1.05,V, without the LED is 1.08V. With only a 30mV Vf on the LED, I should probably be worried about the Vf of the 1N4148 reintroducing too much voltage and blowing the opto. I also measure the output voltage (pin 5) and there was almost 300mV difference between with and without the LED.

Keeping in mind this is a very slow signal (<1Hz), are there any characteristics of the LED that are so much different than a signal diode that I should be aware of? Or would prevent the switch?
 
The minimum drop across an LED is for the red types which drop about 1.8V, which is about 1.8/.7 = 2.5 times the drop across a standard silicon junction diode. LEDs with other colors have higher forward drop.

Measuring the voltage from pin 1 to ground only tells you the voltage drop across the IC. It tells nothing about the diode drop.
You measure that across the diode anode to cathode.
The 4N25 input is sensitive to overcurrent so you can determine that by measuring the voltage across the 680Ω resistor and calculating the current.
 
ADWSystems said:
I have not seen LEDs with Vf that high in a long time.
Click to expand...

Garden-variety green LEDs clock in at 2.1V - 2.2V Vf. Reds are a bit lower, but rarely below 1.8V. Signal diodes like a 1N914 or 1N4148 measure between 0.6V and 0.7V depending on the test current.

ak
 
What is the drive voltage coming in from the right?
 
Not sure. The voltage rails are 0V and 12V. It appears the output is from an opamp in a comparator mode. So it could be 12V. Which I also did not pay attention to and tested with 5V.
 
The reverse voltage for a LED is usually about 5V. In any event, you should be able to compensate by changing the resistor, the opto (xfer ratio) and even the diode. In914's are a little on the high side for voltage drops. A Shotkey diode would be a lot less.

Alec asked a critical question. I'll ask another, what is the logic family used on the left and Vcc.
 
Vcc is 5V. High voltage input on the right is 10-11V.

The 1N4148 voltage drop is measured at 0.7V and the LED at 1.1V. The current changes from 15mA to 12mA when I include the LED. If I swap out the diode for the LED, I would just need to drop the 680 ohm resistor just a little to maintain the original 15mA.
 
You measured the current in. The minimum current gain of the OPTO is 20. Ic(Sat) = 50 mA. Anything bigger than 50 on th e output side means your good. All you need is 50/20 mA or a little over 2 MA. 12 is >>> 2.

The drops matter more when the voltage is lower. 1 drop, 1.2/12 * 100 is about 10% while 1.2/5 * 100 is 24%. Big difference between 12 and 5V.

yea, your right, R= (Vsupply-( sum of the max voltage drops)/(desired current). It doesn't appear significant. There is no HARM in upping it to the same as it was.

It may help you, if you needed it to work with a lower supply voltage but this system might actually be 12 to 13.8 V or thereabouts.
 
I was in a hurry and made a series of mistakes (Vf measure of the LED and power supply voltage selection to name two).

The power supply is 12VDC and ground. The op amp is an LM324 (no comment. isn't my board, isn't my design). The output reaches between 10 and 11V.

I agree with the calculation for R. I like the fact the LED will cut out a chunk of the power to be dissipated by the resistor. But is it correct the opto won't start to conduct until the input voltage is greater then LED Vf+Diode Vf+Opto Vf? Adding the LED or leaving the diode in will increase the turn on voltage by 0.7-1.1V. Right?
 
the OP said:
But is it correct the opto won't start to conduct until the input voltage is greater then LED Vf+Diode Vf+Opto Vf?
Click to expand...

Yea, but who cares? it's 10%. There is so much slop in all of those values.


the OP said:
Adding the LED or leaving the diode in will increase the turn on voltage by 0.7-1.1V. Right?
Click to expand...

LED w/o diode about +1.1-0.6 or about 0.5 V
LED with diode +1.1

My numbers might be off, but if it's TTL, OFF might be below 1.2 and ON above about 2.5 V (TTL logic levels) In saturation, the Opto Vce(SAT) matters. Let's say it's 0.6. 0.6 is less than 1.2 V, so it's ON.

The LED is already overdriven by like 600%. You need like ~2mA, and your feeding it 12 mA. I would probably drive it between 10 and 20 mA, hence 15 mA was chosen by the designer.

As was stated before, the color of the LED matters. Red has a lower drop.[/quote]

All those drops are a function of temperature.
 
The most you can hope for is saturating the optocoupler transistor. "Nice NORMAL" ratings for LEDS are about 10-20 mA with a not to exceed of 50 mA. Those would be the "rule of thumb" values not knowing anything. The diode could act as protection against reverse polarity, so that's probably the real issue as to why it's there.
 
Hi guys,

I am having problems with the 4N25 optocoupler. I want to test that pins 4 and 5 make a short circuit if I there is a current going through the emitting diode.

So I am connecting pin 1 to a 330 ohm resistor with the 5v of the Arduino and pin 2 to the ground. Now pins 4 and 5 should make a short circuit but they don't. and I really cannot figure where is the problem..

Regards,
Mohamed
 
The output of an opto coupler is not like relay contacts. You are unlikely to see any change on the output with a multimeter on the ohms range. Connect pin 4 to zero volts and pin 5 to +5 volts via a resistor between 10 K an 100K if you connect you meter set to the volts range between pins 4 and 5 you should see the voltage change when the current through the pins 1 and 2 is switched on and off.

Les.
 
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