11-16-2009 05:15 PM
I am using 7833R FPGA with SCB-68 box. After creating a simple logic to read DI status, we found that the application shows DI status "ON" when the DI channel is kept open. It shows OFF as soon as we short the channel. Why it is so? In general DI should show ON when it is short and OFF when open.
Thanks and Regards,
Manas
11-17-2009 09:34 AM
The digital inputs on the 7833R do not include any pull up or pull down resistor and the behavior you are seeing is correct (as expected by design).
When you short the input it will pull the input channel to 0V which is defined as low on the DI (LVTTL logic). When you leave the input open the state of the line is not defined and the DI may read high or low. Commonly the DI will float high in this case and the DI will ready high/On in software. With the digital inputs on the 7833R you should always apply a known voltage (either high or low) to guarantee that you are reading an expected state. In your example you should connect a pull-up resistor (~10kOhm) to 3.3 or 5V. When left open your input will be defined high. When you short the input to ground it will be a defined low. If you want the reverse logic (active low) simply apply the Boolean inverse function in your code right after reading the input state.