Getting my LED's to light up from stepper interface board!!!

[jjellin]

Okay, please help me, I am a confused Mechanical Engineer!!! I have a stepper driver board, which has built in fault protection. When the board goes into fault protection, I want to light up an LED. Here's what I have to work with: The fault protection pin pulls low when fault is on, and floats when fault is off. According to the documentation it is an open drain n-channel transistor. From the manual for the INT-462 Interface Board by Intelligent Motion Systems:

"The INT-462 buffers the fault output signal through an open drain N-channel FET. The signal at the terminal strip is inverted, thus it is active when in a logic LOW state. "
Link:

http://www.imshome.com/Product Manual PDF/IB462H_12_07_2000.pdf

The tech support insisted that low means it outputs 5 volts. I reluctantly tried, as it was tried before, by the first engineer to call tech support, with no success, and then decided to pull out an EE book. Well I figured out to use a Pull up resistor, and now I have the LED always on until fault protection, then the LED goes off. I want to reverse that so the LED only comes on when fault protection is on. When floating, the fault protection pin outputs 100 mV and 25mA. My input to the Interface Board is 24V, and I used a 1k ohm pull up resistor to the fault protection Pin to get the led to light up (with a 470 ohm current limiting resistor on the LED). Below is a rough schematic of the current configuration that is always on until the fault pin goes low:

24V+ Pull Up
|
-
| | 1K Ohm
| |
-
|
|
Pin 12 ---------| <-- Roughly 12 Volts when not in fault
|
-
| | 470 Ohm & LED in Series
| |
-
|
GND



So how do I make the LED only come on when the Output line goes low. I remember there being some way to do it with a transitor, but I can't seem to figure it out.

Thanks for any help.

 

[pebe]

The tech support gave you wrong information. An active low means the transistor switches to ground when active. It will float (there will be no connection) in the inactive state. The voltage you are measuring in the floating state is due to current flowing through the 1K resistor from 24v.

You need to connect the LED in series with a resistor, between pin 12 and the Vdd or voltage supply to the chip. Note that may not be 24v! So choose a resistor value to suit.

 

[jjellin]

Are you saying connect the LED anode to Vdd (plus series resistor) and cathode to pin 12, cause I tried this and the LED was always on? Was I using the wrong Resistor Value? Is below what you are saying I should do? I'll try it again, but I'm sure the LED was always on.


24V+ (Vdd)
|
-
| | 1K Ohm
| |
-
|
|
-
| | LED
| |
-
|
Pin 12

My other diagram in the first post, with the pull up resistor does the inverse of what I want. LED on always (fault protection pin on float), and LED off when the fault protection pin goes LOW. I suppose I could use a relay to swap the on off state of the LED, or an Inverter, but I know there's got to be a cleaner easier way.
Thanks for help, I'll keep trying!!!

 

[hfavey]

Your chip is powered from 2 power supplies, one 5V for the logic and one 24V to power the motor.

In your first sketch, current will always flow from 24V to 5V supply through an internal ESD protection diode inside the chip. When the fault comes off, the transistor will activate, bypassing the LED and shutting this one off.

In your 2nd sketch, the LED is always on because of the same reason: you'll draw a certain amount of current from 24V to 5V through your 1k resistor, your LED and the internal ESD protection diode of the chip. When the fault comes off, then you draw current from 24V to GND through the resistor and the LED. Actually the LED should light brighter when in fault.

You need to use the same schematic as the 2nd one but connect to your 5V supply instead of 24V. Apparently (going quickly through the datasheet), 5V supply is available on pin 5 of the IB462H.

Check and let me know if it works better.
Regards,
Hugues

 

[jjellin]

I think I may be confusing you. The pin with +5V you mentioned above is on the IB-462H chip. I am interfacing to the INT-462 Interface Board shown on page 40 of the manual I linked to above. I found the five volts on the board and tried what was said above, but the LED never came on.

Here's what the tech is now telling me:
To connect an LED to the Fault output, simply connect one side to the Fault Pin, and the other side to Ground. When a Fault is detected, the unit will then sink up to 25mA through that Pin to Ground, effectively powering the LED.

Here's what I'm trying next:

24V+ Pull Up
|
-
| | 1K Ohm
| |
-
|
|
Pin 12 ---------|
|
-
| | 30 Ohm
| |
-
|
|
-
| | Relay (On - LED off)
| | Relay (Off - LED On)
-
|
GND

Here, when Pin 12 goes low due to Fault Protection, the relay should go off and the LED is on. When there is no fault protection, Pin 12 floats and the relay is on - LED off. So far the INT-462 Board has had no problem with my pull up technique, am I going to burn anything up doing this???

Thanks again guys

 

[jjellin]

Okay, that didn't work!!! The relay was always on. Why does the LED go On/Off, but the relay stays on. ... Just checked the voltage and its always at 24, in fault or not.

 

[hfavey]

I'm trying right now to understand the datasheet. There's this business with the supplies of the opto, but I don't think it's related.
Another thing, though: when you connect relays to electronic like here, you need to connect a diode to catch the flyback voltage of the coil when it re-opens, otherwise you take the risk to damage the output stage that drives the relay. I would rather use something like this:

VDD
----|
| |
- | | relay
diode ^ | |
| |
---|
|
| <- pin 12 is here, the rest below is in the board
/
transistor --|
GND

You don't need a pull-up resistor as the impedance of the relay will already be high enough to limit the current. I check again your datasheet and let you know if I have something new...

 

[pebe]

Quote:
'Here's what the tech is now telling me:
To connect an LED to the Fault output, simply connect one side to the Fault Pin, and the other side to Ground. When a Fault is detected, the unit will then sink up to 25mA through that Pin to Ground, effectively powering the LED.'

If the chip is wired in the usual way with ground=Vss, then what the tech is telling you is not correct. If the unit sinks a current of 25ma through that pin to ground, then the Fault output pin is obviously the drain of an N channel FET. When biassed into conduction, the FET will give a low resistance path to ground (or to wherever the source of the FET is connected). When biassed so it does not conduct, the drain will float. With the configuration the tech has given you, there is no source resistor to light the LED, so how can a low resistance across the LED make it conduct?

The only way the LED can light as the tech says, is if the ground is connected to Vdd.

 

[hfavey]

I agree with pebe last message. To me the correct connection would be as follow:

VDD (24V)
|
R = 2.2K (for 10mA current in the LED)
|
-
V LED, be careful of the orientation
|
|
pin 12 ---

 

[jjellin]

VDD (24V)
|
R = 2.2K (for 10mA current in the LED)
|
-
V LED, be careful of the orientation
|
|
pin 12 ---

I tried this with no luck. LED was always off. I haven't tried the relay as of yet. Maybe time to get the tech support out here and prove to me his way works!

 

[hfavey]

LED always off? Are you sure about the orientation of the LED? In case of doubt, just wire it the opposite way, you cannot damage anything. If you have a voltmeter handy, can you just measure the voltage on that pin with nothing connect to it for the two cases, no fault and fault. You then should have something vague (can be anything) when no fault and a strong 0V when in fault.

 

[jjellin]

I did test the pin and I get 0V in fault and like 80-120 mV when fault is off. I wired the LED with the anode (+, long lead) to the resistor, and the cathode (-, short lead) to pin 12. I'll try doing it the other way, but intuitively that seems backwards.

Thanks for the help guys!!!

 

[hfavey]

To make sure your wiring is correct, try the following:

-Wire your LED the same as our latest sch.
-short pin 12 to gnd externally with an extra wire (this is safe, no damage possible). The LED should turn on. If not, reverse the LED. Does it turn on?

The transistor in the board will do the exact same thing as what you do with the external wire (providing a path to gnd for the current through the LED). If it's working with the wire and not with the transistor, then maybe there's already a current limiting resistor on board?

 

[Thorn3]

Hello,
Example: In 555 timer applications, the output is always high, untill it times out, then it goes low. If I have an LED on the output, it will be on the whole time the timer is waiting to time out. If the LED is to light when the output goes low, use an N-channel transistor, which will conduct when the gate is low. So, what I did, was to connect the LED to +V, run it through a current limiting resistor, then through the transistor to GND. This way, each time the timer timed out, the low pulse would light the LED, then turn the LED off again when the timers output goes high again.
I would like to know what works for you though. This method has always worked in my applications.