5v pro mini pin voltage?

So, what are the operating voltages of the 5v arduino pro mini? I have tried looking up different things, and all the searches have been very unclear.

In case that was as clear as mud (as it probably was), my basic question is this:

Can I connect a 3.3v sd card module and a sim800l module to it safely, or will i need level converters?

Here's the link to my board: https://robu.in/product/atmega328p-pro-mini-5v-16mhz-compatible-with-arduino/

Become an expert in less than seven minutes:

Take your pick. I like using a voltage dividers and relying on the logic level thresholds.

Just don't put 5 volts out to an input on the 3.3 volt parts.

Someone can explain when you need to do more.

a7

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Yes, if you power the "5V Pro Mini" with 3.3V. The 16 MHz clock is out of specification, but it has always worked for me. I remove the voltage regulator and power the Pro Mini through the Vcc (or 5V) pin.

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So i just supply 3.3v to the 5v pin, and then I can treat it as a 3.3v device, correct?

@alto777 thanks for the links! I'll read through them, and as you put it: "become an expert in less than 7 minutes"

That is what I do. You will need a 3.3V USB-serial interface to program it.

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All right, thanks!!!

Hi

The voltages at the inputs/outputs of the microprocessor will depend on the voltage with which the microprocessor will be supplied.

The ATMEGA328PAU microprocessor can be supplied with a voltage between 1.8V minimum and 5.5V maximum.
Datasheet-ATMEGA328PAU.pdf

Therefore, we can change the LDO (U2 <=> MIC5205) that equips these cards so that the microprocessor will be powered with 3.3V or 5.2 Volts (or what you choose)

or we can deactivate the LDO (and the led) by removing the soldering of the SJ1 bridge to supply the microprocessor with an external power supply.

SJ1miniPro

I think as long as you remove the SD card while programming, you can use either a 3.3V or a 5V UART adapter so long as the adapter is powering the Pro Mini.

Also, it may depend on which regulator the Pro Mini clone has installed, but it may not be necessary to remove the regulator when powering at its ouput pin (the Vcc pin). With its power indicator LED and regulator removed, a Pro Mini can deep sleep at less than 1uA. If it's not much more than that with the regulator still there, you can just leave it be.

Edit: The Pro Mini clones I've bought don't have that regulator isolation jumper. Apparently the original Sparkfun Pro Mini had it, but I haven't seen it on clones.

Sleep current test:

#include <avr/sleep.h>
#include <avr/wdt.h>
int i;

void setup(){

  for (i = 0; i < 20; i++) {          // all pins to one rail or the other
    pinMode(i,OUTPUT);
    digitalWrite(i,LOW);
  }
  ADCSRA = 0;                         // disable ADC for power saving
  wdt_disable();                      // disable WDT for power saving
  set_sleep_mode (SLEEP_MODE_PWR_DOWN); // Deep sleep
  sleep_enable();
  sleep_bod_disable();                // disable brownout detector during sleep
  sleep_cpu();                        // now go to sleep

}

void loop(){
}

Sure, but all 3.3V modules must be disconnected during programming at 5V.

I forgot to mention that in the Arduino IDE you must select the 5V board for a 16 MHz Pro Mini. The 3.3V board assumes an 8 MHz clock.

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