On the BL2600, I have JP1/JP2 jumpered to use +K for the configurable I/O outputs and I have supplied 24V to the +K pin on J1 (also using +24V for board power if that matters).
I set DIO000-DIO007 to be outputs in the software and measured the voltage on them when in a high state. I expected to see +24V output but isntead I get 19.85V. I’ve tried other +K voltages and the output voltage on the DIO is always less than the +K voltage by a few volts.
I presume that must be due to some internal voltage drop.
Is it normal for the output voltage to be less than the +K input voltage?
The input circuit on the bl2600 consists of a 27k resistor between K+ and the input pin and a 100K resistor from the input pin to the 74hc541 input buffer. This creates a potential divider with K+ as the supply voltage, the pin of the 74hc541 as the “ground” and the input pin as the centre tap.
Under open circuit conditions, the pin of the 74hc541 is probably sitting near the VCC for the chip which is 3.3V so the voltage at the input pin of the BL2600 is approximately:
(K+ - 3.3) * (100/(100 + 27) + 3.3 = 19.6V if K+ = 24V
Regards,
Peter
There is another thing to remember about these I/O pins:
The voltage you see on the pin as a “high” state voltage is not really an output voltage, it is just a by product of the input resistance divider and cannot be used to supply any significant current.
These I/O are designed to work as sinking outputs which means that you would normally connect +K to one side of the load and connect the other side of the load to the I/O pin. In the “high” state the load is effectivly floating at +K with a small current possibly flowing due to the resistor divider (transistor is off). In the “low” state the transistor is switched on which effectivly connects the load to ground which resuilts in the full +K voltage across the load.
The “low” state is the active state with current flowing from +K (external to the BL2600) through the load and into the I/O pin, through the transistor to ground.
Regards,
Peter
Thanks Peter. I had planned on using these configurable I/O channels to drive some LEDs. (For example, Apem P/N Q8P1BXXG24E.) Perhaps I have it backwards then as I was planning on having the I/O pin provide voltage to one leg of the LED while the other leg of the LED was tied to gorund. So maybe I should reverse that and have one leg of the LED connected to +24V power and the other leg tied to the I/O pin so that when I set it to a “low” state it will go to ground and turn on the LED?
That should do the trick, and as the LED you are talking about is a 24V indicator you should not need to mess around with resistors etc.