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I am having this Microcontroller

In this Microcontroller, I am using the ADC peripheral and normal GPIO port.

My questions:

  1. On Page 58, table 42, There is a parameter called as Supply Current per ADC which is mentioned as 1mA. What does it mean? It also mentions that it depends on the conversion rate. Does it mean, if I connect a voltage (within the mentioned limits of the pin) with a series resistor to that ADC pin, that pin will consume 1mA? So, if I connect two ADC inputs to the ADC peripheral, it will consume 2mA?

  2. Suppose, I configure a port as Input port. How much maximum current will the Microcontroller pin draw? Injection current will come only when I impress a voltage above that the supply voltage of the GPIO. And Leakage current, will come into picture only when the device is operating in low power state or shutdown. So, If I donot use the microcontroller in the above two conditions, (applying voltage to the pin within limits and in normal working mode), how much will the microcontroller pin take for a normal GPI input?

  3. This Microcontroll has two supplies. ADC Reference supply high and ADC Reference supply low? Usually, I have seen only ADC reference supply as a single reference supply pin. But here, there is reference high and reference low? Any idea on why is this implemented and any use case of it?

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1) It means the current consumption of the ADC circuit within the IC, not from the input pin. Each ADC that you enable consumes 1 mA from the IC power supply.

2) That is the leakage current. 0.5 µA max., in or out. But if you use the ADC to sample a voltage, it gulps current in to charge its sample-hold capacitor.

3) It is implemented to enable the user to select both Vref- and Vref+ for the ADC. Users might want full-scale readings between 0.5 V and 2.8 V instead of between 0 V and 3.3 V.

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  • \$\begingroup\$ Thank you for the answer. From you answer, 1. ADC current consumption from the Power supply rail. But When you say "Each ADC that you enable", do you mean, each input channel or the entire ADC peripheral? And 2. I thought the leakage current occurs only during shutdown or low power conditions. Could you just provide a simple numerical example for the input current if its a GPIO and if it was a ADC pin? How to calculate the input current for the ADC pin when you say,"It gulps the current to charge the capacitor". Please help. Thank you \$\endgroup\$
    – user220456
    Apr 28, 2020 at 9:48
  • \$\begingroup\$ No, I mean each ADC peripheral when I say each ADC, it does not mean each channel. No, leakage current happens always, it is leakage. No, I can't provide other numerical examples, the datasheet says the IO leakage. I have no idea what you are sampling and what output impedance it has, you have not provided that info. I am not familiar with your MCU either, but surely the manual provides ADC specifications to get some kind of number. \$\endgroup\$
    – Justme
    Apr 28, 2020 at 10:08
  • \$\begingroup\$ Ok. Thank you. I have a voltage of 4V. This is connected to the ADC pin via a 11.5k resistor. So, now, as this is connected to the ADC pin, the maximum current through the resistor initially would be 4V/11.5kohm, right? I am calculating like this because there is an internal capacitor at the ADC pin like you mentioned. So, initially, it will act as a short. So, the max current would be 4V/11.5kohm. Correct? \$\endgroup\$
    – user220456
    Apr 28, 2020 at 10:29
  • \$\begingroup\$ Yes, that would be the worst case. Except that internal mux and switch resistances are already in the kilo-ohm range so it will be less. But why do you even need to calculate this current? Why do you even have a resistor there? Besides the resistance is over two times the recommended maximum source impedance. Look at the ADC electrical specs. \$\endgroup\$
    – Justme
    Apr 28, 2020 at 10:44
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    \$\begingroup\$ Ok, so it has near zero output impedance. It will work with series resistances between 0 and 5k, so if you want to have some protection you can use a 4k7 resistor to limit current in case something goes horribly wrong and the sampled voltage exceeds MCU supply voltage. \$\endgroup\$
    – Justme
    Apr 28, 2020 at 11:51

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