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mcp3204/mcp3208 question — Parallax Forums

mcp3204/mcp3208 question

archer1424archer1424 Posts: 7
edited 2013-05-15 17:33 in General Discussion
I am using a propeller mc to read an analog voltage from a pressure transducer. I have an mcp3204 powered and referenced from a regulated 5V source. However, my analog signal will vary from about 1V - 8V. The data sheet says the range of the inputs is 0 to vdd (5v), but I will be exceeding this with my sensor voltage.

question:
Will the mcp3204 chip fry if my input is larger than my ref voltage?

if so- What can I do to limit the voltage? I'm actually only concerned with voltages between 1V and 5V anyways. Could I use a zener diode (5v breakdown) to shunt the signal if it passes 5 V?

if not- then what?



Thanks
Dan

Comments

  • SRLMSRLM Posts: 5,045
    edited 2013-05-07 14:16
    You don't want to go past the absolute maximum ratings, unless you know exactly what you are doing.

    In this case, you can use a pair of resistors to make a voltage divider. I would use two 10Kohm resistors for your divider (so that you have some overhead on the maximum signal).
  • Bobb FwedBobb Fwed Posts: 1,119
    edited 2013-05-07 14:17
    Anything above 0.6V above Vcc (5V) will just free flow through an internal diode. The datasheet should have a rating for the amount of power that can safely go through these diodes. The most obvious solution is to put a voltage divider so the 1V-8V is translated to a max of 5V.
    As you said, you can put a resistor and a zener, that would work. Though near the breakdown of the zener, the results may be a little skewed (though that may not matter to you).
  • Mark_TMark_T Posts: 1,981
    edited 2013-05-07 17:33
    Once you use a resistor-divider you have to check the output impedance of the pressure transducer since the divider will now be loading it,
    hopefully your transducer has a low output impedance (if not you have to factor its output impedance into your divider calculations).

    The divider also forms a simple protection circuit by limiting current even if the input is too high for the divider ratio... The protection diode
    in the ADC will have a maximum current it can tolerate and with a series resistor that means over-voltage tolerance until its high enough
    to overload the diode. With no divider there is no resistor to limit current into the diode.
  • archer1424archer1424 Posts: 7
    edited 2013-05-07 18:45
    Mark_T wrote: »
    Once you use a resistor-divider you have to check the output impedance of the pressure transducer since the divider will now be loading it,
    hopefully your transducer has a low output impedance (if not you have to factor its output impedance into your divider calculations).

    The divider also forms a simple protection circuit by limiting current even if the input is too high for the divider ratio... The protection diode
    in the ADC will have a maximum current it can tolerate and with a series resistor that means over-voltage tolerance until its high enough
    to overload the diode. With no divider there is no resistor to limit current into the diode.


    Do you mean it would be better if my output had a 'higher' impedance instead of 'low impedance? If not, why? Can the impedance be checked with a resistance measurement from a DMM?

    by "factoring in", do you mean that the output impedance would effectively be a resistor in parallel with R2 of a voltage divider? If that's the case I would just get the equivalent resistance and then calculate the voltage from the divider like normal?


    Thanks for the help and education

    -Dan
  • kwinnkwinn Posts: 8,697
    edited 2013-05-07 23:09
    archer1424 wrote: »
    Do you mean it would be better if my output had a 'higher' impedance instead of 'low impedance? If not, why? Can the impedance be checked with a resistance measurement from a DMM?

    No and no. A lower output impedance is better. That means it can provide enough current to drive the ADC input to the correct voltage. And no, you cannot measure the output impedance with a DMM. You could calculate it by measuring the output voltages using 2 different resistors as loads.
    by "factoring in", do you mean that the output impedance would effectively be a resistor in parallel with R2 of a voltage divider? If that's the case I would just get the equivalent resistance and then calculate the voltage from the divider like normal?

    Thanks for the help and education

    -Dan

    No. The output impedance of the pressure transducer would be in series with the resistor divider. Taking the output impedance of the Transducer (R1) and the input impedance (R4) of the MCP into consideration means the voltage divider (assuming R2/R3 are 10K) would be (R1 + 10K) / ((R4 x 10K) / (R4 + 10K))
  • zapmasterzapmaster Posts: 54
    edited 2013-05-15 17:33
    I have made a first out system for a ski lift. i had to look at 24vdc i used a voltage devider and a zeaner 5 volt diode to prevent damage during the inductive kick. working for the past 3 years
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