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I've just started working through the Freenove ultimate starter kit for the Pi Pico, and I'm having trouble understanding why it has so many resistors in this circuit.

I'm following the tutorial document at https://github.com/Freenove/Freenove_Ultimate_Starter_Kit_for_Raspberry_Pi_Pico/blob/master/Python/Python_Tutorial.pdf and have made it to project 2.1. The relevant circuit diagram is attached. Circuit diagram

I understand it uses a 220ohm resistor to prevent the LED burning out.

It also uses 2 10k resistors connected to the button - these are the ones that I don't understand. I'm trying to clarify a number of things:

  • Am I correct in thinking that 2 10k resistors in series are the equivalent of a single 20k resistor?
  • Is this done so that we're not passing too much current to the GP13 pin?

I'm having trouble finding an answer as to what the acceptable inputs for the GPIO pins are - the closest I've found is What are the min/max voltage/current values the gpio pins can handle? which recommends no more than 0.5mA. But that's an old post, and not specifically about the Pico.

  • Assuming the GPIO pin could take the full 3.3V @ whatever Amps, would I need either of these resistors?
  • Could I remove safely remove just R2?

This is my first introduction to electronics, so I'm very much in the camp of not even knowing what topics I need to understand. Any advice would be greatly appreciated.

AngelsJinx
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1 Answers1

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R2 is a pullup resistor, so the top of S1 is pulled up to 3.3V when open, pulled down to 0V when S1 is closed. R3 limits the maximum current that can be fed into GP13; this is to guard against the pin being set as an output, and the button being pressed, in which case the output would be short-circuited to ground. The 10K value limits the current to well under 1mA, so the output won't be damaged.

When set as an input, GP13 consumes negligible current, so there won't be any voltage drop across R3, and GP13 will receive the full switch voltage.

You are along the right lines in saying that R1 prevents LED1 being burned out, but the important question is: why is it necessary? The answer is that an LED is a constant-voltage device (around 1.8V for a red LED), and it doesn't work with voltages higher or lower than that - higher and too much current will flow, lower and it won't light up. So the series resistor allows the LED to work at a voltage it is comfortable with, and the current (= brightness) is set by the resistor value using ohm's law, I = V / R = (3.3 - 1.8) / 220 = 6.8 mA, which will be a reasonable brightness without overloading the processor output pin.

As others have suggested, this isn't necessarily the best example of the circuit designer's art...

jayben
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