led in circuit

[graphpix]

Hi, i have a lpt breakout opto board and you can see the output section of this circuit below. How can i place a led in this circuit to view signal output without reducing the output voltage 5V?

[gar]

071226-2001 EST USA

graphpix:

A Stanley EFR5366X (DigiKey 404-1090-ND) in series with your 4.7 K resistor will give lots of brightness. If you use a 470 ohm it will be much brighter.

Or a less bright LED in series with the opto input.

In each case the LED is on when current flows into the coupler.

(edit)
What is the load on the output? This will be a major factor on the output voltage. If it is a 2n7000 FET, then there is virtually no load. If it is a 100 ohm resistor to ground, then there won't be much output voltage. If it is 100 ohms to +5, then there is lots of output voltage, but the opto may not be able to pull it down. So it is true if the LED is in series with the 4.7 k resistor, then you might excessively reduce the output and caause a residual glow in the LED. You have to more completely define the circuit.
(end edit)

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[graphpix]

(edit)
What is the load on the output? This will be a major factor on the output voltage. If it is a 2n7000 FET, then there is virtually no load. If it is a 100 ohm resistor to ground, then there won't be much output voltage. If it is 100 ohms to +5, then there is lots of output voltage, but the opto may not be able to pull it down. So it is true if the LED is in series with the 4.7 k resistor, then you might excessively reduce the output and caause a residual glow in the LED. You have to more completely define the circuit.
(end edit)

output line directly connected to the simple l297-298 stepper driver board and l297's clock pin (18). driver board schematic below:

[gar]

071227-0816 EST USA

graphpix:

You need to get a datasheet for your for your IC and study the input characteristics of the inputs you expect to drive. Or experiment.

The Stanley LED I referenced has a drop of about 1.5 V with 1 M series resistance and 5 V source, or about 3.5 microamps current. A dim glow exists.

Assuming your input requires 100 microamps pullup, then a 4.7 K resistor has a drop of 0.47 V, and 470 ohms has 0.047 V. With a series LED you have to add the approximate 1.5 V.

If you use 470 ohms, and put the LED and a 27 K series resistor between the optic output and ground, then the approximate current thru the 470 ohm resistor is (5-1.5)/27,500. or 0.13 MA from the LED path. This is a drop of about 0.06 V from the LED circuit. To this you have to add the drop resulting from the pullup current to your I297. You can make various different asumptions and do your own calculations.

A separate question is whether a darlington opto isolator is fast enough for what you want to do?

.

[graphpix]

Thanks for detailed info gar;

"A separate question is whether a darlington opto isolator is fast enough for what you want to do?"

I don't know really if it is fast enough or not but i have lot's of this optocouplers and wanted to use them. I'm using 74ls04 for buffering lpt outputs and then connecting opto-couplers. Speed is not important for now, i'm trying to build a simple diy cnc machine, if it works i can spend some money for a professional parallel port opto-isolator board.

thanks for your help again.

[pminmo]

The problem you going to run into is the opto for the step and direction has to be fast unless you plan to write your own cnc software. You may also run into incorrect step problems because of the lack of a low inpedance drive to the l297.