should I isolate the 2 power supplies completely by using a lets say 6 volt rectified source for the signaling side and the existing 12 volt source for the motor side.

A separate power supply is good.

You can also try to get the +5V from PC power supply and then there will be no need for a separate power supply. You can even remove the 7805 in this case.

 

Good point
Thanks

 

By the way, if I decide to get the 5 volts from the PC, should I bypass the diode and feed it straight to Vin in your illustration and short Vin to Vout or should the diode still be in use as it is; ie pin2 of J3.

 

By the way, if I decide to get the 5 volts from the PC, should I bypass the diode and feed it straight to Vin in your illustration and short Vin to Vout or should the diode still be in use as it is; ie pin2 of J3.

Remove diode, remove 7805 or 78L05, then short Vin to Vout on PCB.

 

Hi
I am back. Can you tell me what is the use of this diode in this circuit in the first place anyway. Is it for protection, if yes how and what does it protect?

Thanks
Bluebirdiran

 

Hi
I am back. Can you tell me what is the use of this diode in this circuit in the first place anyway. Is it for protection, if yes how and what does it protect?

Thanks
Bluebirdiran

The diode is for the protection of incorrectly connected external DC supply from damaging the rest of the circuit.

 

This came to me as PM.

Thanks a lot for all your help. I have been fiddling around with this circuit and have realized that when it is connected to the PC and the signals are coming from the parallel port, the state of the voltage at the output does change as it should.

Good. That's what the circuit should be doing as intended.

But when I disconnect it from the PC and try to apply a similar dc voltage of between 3.5 to 5 volts at the inputs, the "on" state at the output never changes. I cannot get it to go to low. What is the difference between the signal that the PC sends and the signal that I apply manually everything else staying the same.

The output of the circuit would acts more like a switch to 0V.

To test a switch, you will need to connect a load in series with a power source to the output. Then you can apply 3.5V to 5V at the input to control the ON/OFF of the load.

 

Most Motor driver boards such as the one for step motor controllers will take a 0 to 5 volt ttl input, hardly will draw any current from the LPT port, so why the ULN2803's?? A much more simple buffer could be made for interfacing to the LPT port.... it seems though that if the LPT port already has a couple of blown output pins that the controller IC will have sustained damaged rendering the rest of it possibly unstable..... Have you tried a different PC/LPT port?



My .02

 

.... it seems though that if the LPT port already has a couple of blown output pins that the controller IC will have sustained damaged rendering the rest of it possibly unstable.....

Agree.

But the OP's problem is the buffer works correctly when connects to the LPT port but didn't when it is disconnected from the PC.

 

There was a load in the shape of a LED in series with a 560Ohm resistor. When signals came from the PC the lead went on and off alternatively but when I applied a 3.5 to 5 volt to the buffer the LED stayed on all the time at it was to begin with.
Pls. explain if you have the time.
Thanks
Bluebird

 

when I applied a 3.5 to 5 volt to the buffer the LED stayed on all the time at it was to begin with.

If you mentioned this LED earlier, the comment to your problem will differ.

If you apply 3.5 to 5V to the input, of course the LED will turn ON because this is how one turns ON the LED via the buffer.

If you apply 0V to the input, e.g. by shorting the input to 0V, the LED will turn OFF.

 

You are so right, I have just done exactly that and the LED did turn off.

Thanks a lot once again.

Bluebird

 

I am sorry to sound like such a dumb, but why is the input pins at 5 volts while nothing is connected to them. To me it would be more logical if the pins were low and when 5v was applied to them they would go high and make the output high too. What is the necessity for this upside down approach? I know there is a sound logic behind this but what it is, I don't know and would very much like to know.
Thanks
Bluebird

 

I am sorry to sound like such a dumb, but why is the input pins at 5 volts while nothing is connected to them.

I don't know why you have found the inputs at 5V when nothing is connected to them as the datasheet indicates there are two resistors (7.2KΩ+3KΩ) on the input to 0V which SHOULD pull the input LOW.

 

I have just mearured both the input and the output when there is nothing connected to the input. The input shows 2.1v and the output shows 4.3 with the LED on. When I touch the input pin to the ground the LED at the output goes out.

 

Mr. Chung, I owe you an appology. Not only have I confused myself, I am probably confusing you too I suppose. The situation that I described above was for when the pull up resistor array was still in the circiut. After taking it out, everything went to normal. ie when the input was driven high the output went high and vise versa.
Sorry again

 

No problem.

I'm glad you've finally nailed it. Keep up the CNC work.

 

I have just done my first piece of real flame cutting on 10mm steel plate. It feels wonderful, I must say. A special thanks you.
I will contact you again when I am in trouble. I hope you will find the time to help again.
By the way, I am going to buy a solid state relay sensitive enough to be activated by the parallel port directly. If you are familiar with such things can you give me the commercial part number.It should be driven from the LPT directly and the out put should be able to conduct 220v ac at less than .5A, such as a electric valve. East Asian products would be more affordable here.

Thanks a lot
Bluebirdiran