0
\$\begingroup\$

I have an XBee Series 2 and I was trying to detect an input from a signal which was 12v, so I used a voltage divider with R1=220k and R2=24k, I checked the output voltage with multipmeter and it was 1.2v, which is what XBee S2 input pin is meant to get, however the XBee was not able to detect that input, could you please tell me why was it? could it be because the output current (Io) from the voltage divider was so low that it couldn't detect it? if so what register values should I use.

Also my second question is what if I want to power XBee from the output of that voltage divider, what register values should I used if I am to power it from the same voltage divider?

Basically I want to use XBee (S2) to get power from a Texecom Premier 24 alarm system, and also want it to read the status of the output pins (digi output) and send signals to a remote server wirelessly, but since XBee data input pins cannot read 12v signal which is why I had to use a voltage divider but still it was not been able to detect the signal, which is why I'm thinking may be the register values I used were not the correct ones for this setup.

I'll really appreciate if someone could help me with this, as I seem to be getting no where on my own because of having very limited knowledge of electronics

Thanks

\$\endgroup\$
  • \$\begingroup\$ How do you come up with 1.2V for input to the XBee? I don't know what model exactly you're using but I looked at the XBee 865/868 User Guide (ftp1.digi.com/support/documentation/90002126_H.pdf) and I see that supply voltage (Vdd) is 2.7 - 3.6V and GPIO high threshold voltage is min. 0.7 x Vdd. Assuming you're using 3.3V for Vdd, the minimum high input would be 2.3V. I'd suggest setting the VD to produce 3.0V, say 30K and 10K ohm. If you can point me at the user guide for the specific module you're using, I can check the values there. \$\endgroup\$ – DoxyLover Nov 3 '14 at 5:34
  • \$\begingroup\$ Hi - I'm using XBee S2, the one shown here: sparkfun.com/products/10414 I have a question sorry for being so ignorant, what is VDD and is it the same as VCC? if so why in electronics they have different abbreviations? \$\endgroup\$ – Imran Imtiaz Nov 4 '14 at 19:56
  • \$\begingroup\$ Vdd == Vcc. Vcc is a hold over from bipolar-junction transistors (BJT) where "C" stands for collector. Vdd refers to FET drain ("D"). Similarly, Vee == Vss (generally) == ground. "E" for BJT emitter and "S" for FET source. \$\endgroup\$ – DoxyLover Nov 4 '14 at 20:14
0
\$\begingroup\$

Question 1: The 1.2V is the maximal value for inputs that must be sampled (ADC). If you have configured the GPIO correctly for digital input, the input value must be bigger than 0.8*Vcc: enter image description here

Your 1.2V input is lower than 0.8*VCC, if you use any VCC voltage between 3.0V-3.4V. You must change the voltage divider configuration to make it work properly. Example: The resistor divider to 3.0V output -> R1/R2=3, try to use not too high resistors values. (R2 between input and ground, R1 between 12V and input)

Question 2: "my second question is what if I want to power XBee from the output of that voltage divider.." if the output of your voltage divider is 1.2V you cannot power the XBEE, as the datasheet shows, the device requires voltage supply between 3.0V - 3.4V. Take into account that the maximal recommended input voltage for the pins is 3.3V ( ..."Regardless of which XBee you chose, you must not exceed 3.3V on any pin or you will damage the radio module."..). Therefore be sure that your 12Voltage are regulated to prevent damages.

I should say, I don’t understand the reason why do you want to power the XBEE from the voltage divider, the voltage divider is to adjust the signal to the digital input of the XBEE, if you use this to power the XBEE, when the signal is low it means the XBEE is turned off.

\$\endgroup\$
  • \$\begingroup\$ Hi, thanks this is really helpful. Do I have to worry about the source impedance (also what is it? I read it somewhere but don't understand), the problem is, there could be so many combinations of resistor values which would give 3.2V when used in a voltage divider, like I can use resistors in mega ohms or just ohms, so how do I decide that what resistor values would give the optimised results? I really appreciate your answers, thanks again for that. \$\endgroup\$ – Imran Imtiaz Nov 4 '14 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.