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schematic

simulate this circuit – Schematic created using CircuitLab

In the circuit the current i1 is flowing from external device. Now my doubt is what is the voltage developed at the analog input of the micro? I am asking this question since i don't know the resistance of the analog input. Can i simply neglect the resistance of the micro analog input, or i need to refer from the data sheet. Can i also neglect the 56K resistor in series? In that case is the voltage at the analog input is Is*4.7K? Please help

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The analog input's internal resistance can be safely considered so large as to be negligible. The actual leakage current into the input pin will probably be specified in the microcontroller's datasheet. Typical values will be in the \$\mu A\$ or \$nA\$ range and will change slightly when the internal circuitry samples the pin. For most cases, the voltage on either side of R3 will be about the same, so R3's presence can be neglected as well. Assuming all of the current from I1 goes through R1 to ground is valid for most cases.

If, however, R3 becomes sufficiently large, the internal resistance of the analog input can no longer be ignored. This is also true for sufficiently tiny values of I1. But those extremes are rare and can be mitigated with a well designed analog buffer (such as an op amp in voltage-following mode).

One other thing to note is that some microcontroller manufacturers have a limit on the series resistance into a analog input. For example, Microchip specifies a 10\$k\Omega\$ limit on the total resistance going into an analog input pin on their PIC microcontrollers. Your 56\$k\Omega\$ resistor above would cause the ADC function to fail on a PIC.

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  • \$\begingroup\$ Just curious to know how this 10kΩ or anyother series resistor will make the ADC circuit to fail. What is the internal component in ADC circuit which fails because of this series resistor? Also can i find this value in the datasheet? \$\endgroup\$
    – rajesh
    Jul 15, 2015 at 7:42
  • \$\begingroup\$ Yes, the datasheet is the place to find that information. As to why they chose 10kΩ, that's beyond the scope of a comment. But the short answer is because of timing requirements. The larger the resistor, the slower the ADC mechanism can sample the voltage. Make the resistor too large and the necessary timing requirements are violated. \$\endgroup\$
    – Dan Laks
    Jul 15, 2015 at 7:47

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