opto Isolator

[labview1958]

I have designed a circuit that uses an opto isolator. Should the LED be in series or in parrallel with the resistor. The circuit is attached.<p>

[L. Daniel Rosa]

series

[rstofer]

In series as you have it drawn.<p>However, as it is drawn, the inputs are all active high. You need to be certain that the driving circuit can source the current and that the resistor is sized for the actual input voltage. In most cases where you pull up, the voltage will never get close to 5V.<p>It's a design issue but my preference would be to connect the resistor to +5 and the anode of the LED and pull the cathode to ground with the logic input. Most components can sink far more current than they can source.<p>Also, if the sending device is open collector, it will never be able to pull up the LEDs.

[labview1958]

I was refering to the LED and the resistor on the output side of the opto isolator. It looks to me the LED and the resistor are in parallel. should be in series?

[MrAl]

Hi there,<p>
The resistors on the *input* are correct.<p>The resistors on the *output* are NOT connected
correctly. They need to be in series with the
LED's there also.<p>Take care,
Al

[jwax]

Indeed- MrAl is correct.
Unless of course you want those output leds to live a very short, but active life!

[philba]

I swore I was gonna stay away from this one...<p>Even with the resistor and led in series, I don't know if this is going to do what you want. You are trying to do two things - drive LEDs and some signals with one circuit. That usually needs some care to get both conditions correct. With the output pullup and led in series you will have, effectively, a very strong pull up. 15V and a 2Vf LED at 10mA will require a 1300 ohm "pullup/dropping" resistor. That will take a pretty strong signal from the opto to over come the pullup effect and pull the signal to zero. Note also, you will need to use 1/4 watt resistors if it does work. I would not have an LED If of more than 10mA - at 20mA you will require a 1/2 watt resistor (620 ohms) and may not be "pull-downable". <p>Also, with the very strong pullups, I wonder how quickly the output transistors will switch. it should have good noise immunity, though.<p>Unfortunately, I don't have a spice model of an opto to experiment with... <p>On the point rick brought up - you probably should sink rather than source on the input side. That's always a good idea with the PC Parallel port. Also because your circuit is inverting the signal. I suspect that isn't what you want.<p>Personally, I'd drive LEDs with transistors or maybe a darlington array in the ULN2xxx family. This does make me the question why 15V in the first place.<p>[ September 03, 2005: Message edited by: philba ]</p>

[philba]

one other thing related to the 15V question. It looks like you are trying to drive a stepper controller with that circuit. Most stepper controller's logic circuits are 5V - I'd be very careful and read the controller specs very closely. Pay attention to max voltage on the step and direction inputs.<p>[ September 03, 2005: Message edited by: philba ]</p>

[Gorgon]

Hi Labview1958,
If you would list the component types and values when you ask for an opinion, it would be much easier to give you advice.<p>As Philba said, if you are driving 5v logic inputs you will have problems. The solution is a bit depending on what you are driving. If the input do have active pullup resistors you could just connect 4 diodes (1n4148 types or better low forward voltage types) with the cathode connected to the optocoupler output and the anode to the logic inputs. This will isolate the logic input from the 15V. You may have problems with the voltage input level, due to the optocoupler Vce in addition to the Vforward for the diode. You need to check the specs of the inpust you want to control. <p>Philba, I finally got my pixma4000 printer. Looks great so far.<p>TOK

[labview1958]

On the input of the opto the resistors are 100 ohms. On the output side the resistors are 4.7 kilohms. The input side of the opto is connected to the NI-6025E DAQ card. I cannot find any symbol for the input side of the opto in EAGLE. Thus I am using +V. On the output side the opto is connected to a 4-phase unipolar stepper motor drive board RS 217-3611. Again I cannot find appropriate symbol, thus I am using -V. I am going to change the opto output resistor from parallel to series.

[philba]

Of course you won't find boards or assembly symbols in eagle (or any other package). I use pinheaders or other connectors to indicate connections off the board.<p>I found that board at RS components but it doesn't say what the maximums are. It does, however, say that the controll inputs are "cmos and ttl" compatible. I read that to mean 5V (or maybe 6V max). Your circuit may not blow the stepper controller but I wouldn't risk it at all.<p>You really should rethink your design. Keep the control logic side at 5V. Here's a snippet from an isolator board I am working on.
<p>[ September 03, 2005: Message edited by: philba ]</p>

[labview1958]

What is Vccio and Gndio? Actually my Ni-6025E Daq card signals are already digital +5V 7 0V. I should eliminate Vccio but retain Gndio. Thus Vccio should be changed to Vgndio. Is my reasoning correct?

[philba]

well the schematic tells you what Vccio and Gndio are. <p>No, you misunderstand how this works. There are two ways to light up an LED. One way is sourcing - the cathode is tied to gnd and you apply a positive voltage to the anode. The other way is sinking - the anode is tied to +V and you apply zero volts to the cathode. If your DAQ card can supply enough current then it won't matter whether you source or sink but there is another issue.<p>The problem, as I said before, is that the way your opto is working it forms an inverter. you are sourcing from the PC Port - that means a logic 1 from the PC is high but the transistor side of the opto will be 0 (i.e. low). If your software can handle that then there is no problem . <p>My board assumes a Parallel PC Port and thus operates in sink mode. The way the opto works, a 1 (high) on the port will cause the LED to not light up (anode and cathode are at the same voltage level) and the transistor doesn't conduct. Thus, the output remains pulled high. If a 0 (low) is output, the LED lights up and causes the transistor to conduct, thus pulling the output voltage low.

[labview1958]

Here is my revised circuit;<p><p>Can it still be improved?

[philba]

don't know - its completely unreadable.