vadimVladimir 0 Posted October 24, 2013 Author Share Posted October 24, 2013 Another noob question: Whats the min output voltage to pin be as "high" or "1" ? thanks again. Quote Link to post Share on other sites
roadrunner84 466 Posted October 25, 2013 Share Posted October 25, 2013 In the datasheets (not the user guid) is specified that when loading a pin with 6mA, then the voltage is at least Vcc - 0.3 (so for 3.6V Vcc the pin is at least 3.3V) Quote Link to post Share on other sites
vadimVladimir 0 Posted October 31, 2013 Author Share Posted October 31, 2013 Thanks again mr. Runner. But now i Quote Link to post Share on other sites
roadrunner84 466 Posted October 31, 2013 Share Posted October 31, 2013 No, your calculation does not account for the load of the MSP430. It is best to use two diodes in series, each casuing a drop of 0.7 V, 5V - 0.7V - 0.7V = 3.6V Alternatively use a resistor divider: connect a resistor (say 1k) between the PIC and the MSP430, then connect a 2k resistor from the MSP430 (same pin) to ground. (google voltage divider). vadimVladimir 1 Quote Link to post Share on other sites
vadimVladimir 0 Posted October 31, 2013 Author Share Posted October 31, 2013 No, your calculation does not account for the load of the MSP430. It is best to use two diodes in series, each casuing a drop of 0.7 V, 5V - 0.7V - 0.7V = 3.6V Alternatively use a resistor divider: connect a resistor (say 1k) between the PIC and the MSP430, then connect a 2k resistor from the MSP430 (same pin) to ground. (google voltage divider). Thanks Runner, what kind of diode ? 1n4148 ? 1n4007 ? I will google the tips. Have a nice day. Quote Link to post Share on other sites
roadrunner84 466 Posted October 31, 2013 Share Posted October 31, 2013 any, but the voltage divider solution is cheaper, and works very well if you know the load (only the msp430 pin) Quote Link to post Share on other sites
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