Hi, What is the maximum voltage a level than can be given as an input on a pin on nrf51-dk? I've measured VDD at 2.91V. Is it from VDD-0.3 to VDD? If yes, is there a way to give 5v as input to a pin? Thanks in advance for the answer :)
Hi, What is the maximum voltage a level than can be given as an input on a pin on nrf51-dk? I've measured VDD at 2.91V. Is it from VDD-0.3 to VDD? If yes, is there a way to give 5v as input to a pin? Thanks in advance for the answer :)
Hi
Have a look at the Hardware reference files for the nRF51 DK. This is the power supply layout:
As you can see, the 5 volts at the USB port is actually the only input voltage with a 3.3V voltage regulator (U3). The battery connector and header P21 is not regulated at all. All three voltage inputs are protected by a Low Forward Voltage Drop diode SD103ATW. The voltage drops across these diodes are ~0.3V. This is why you only see ~3.0V on VDD_nRF when you apply 3.3V on one of the inputs.
So to answer your question; you can only supply 5 volts to the USB port. The maximum supply voltage of the nRF51 is 3.6V (read the Product Specification). Hence, you can not attach a power supply with voltages greater than 3.9V on the unregulated input ports. This assumes that you have not bypassed the protection diodes so your VDD_nRF voltage will be 3.9V-0.3V = 3.6V.
Thanks for the answer. So, I should be alright if I give a signal with voltage level 3.3V to the digital pin of nrf. My other question was, is there any way in which I can give signal of voltage level 5V to the digital pin of nrf without damaging the board and without using step down circuit? Are the digital pins tolerant to VDD to VDD-0.3 only or is there any other range?
There is a protection diode between the GPIO input and Vdd on the nRF. The forward voltage drop over this diode is ~0.3V and hence, it will start leading current at Vdd + ~0.3V. It is possible to put a resistor in series with your input signal and the GPIO pin to limit the current that will flow through the diode when you connect a 5V signal.