0

I was taking measures from the outputs of an OpAmp when all the sudden measure 5 volts from all pins of the Arduino Mega for no reason. I disconnected the Arduino and loaded the Firmata and test. All the pins are reading high to 1023.. 5 volts are been output for no reason even when I have the Analog Ping to input.

Sketches code are loading and the lights look normal. I did noticed the processor chip getting hot and the voltage in the breadboard line to the Arduino Analog pins reading 7 volts.

Is this a sign of a serious damaged to my device?

Improve this question
8
  • 1
    Probably. It should not get hot. Was something connected to the pins that draws so much current that it could get hot? What are the power voltages for the OpAmp? Higher than 5v?
    – Jot
    Commented Jan 26, 2019 at 2:11
  • I doubt. A voltmeter the same output line from the OpAmp read 7 volts. But the reading returned to normal 35 ml after disconnecting the wiring to the arduino analog pin. Also, only 3 of the pins where connected to OpAmp device measuring input voltage, output and the ground. All the arduino pins ( except A0) are stuck in HIGH.
    – Jose Enrique Calderon
    Commented Jan 26, 2019 at 3:38
  • 1
    The voltage of the atmega2560 can be -0.5v to +5.5v when the atmega2560 is powered with 5.0v. A voltage of 7v might damage it. It is possible to blow only a part of the chip. When you are sure that al least one pin is blown, then you should get rid of it because it is possible that other parts inside the chip are damaged as well. I'm still not 100% sure that the chip is damaged, because when uploading a sketch, pin 0 and 1 must be working.
    – Jot
    Commented Jan 26, 2019 at 4:13
  • Can this chip me removed and replaced easily? Please point to any source to this repair surgery.
    – Jose Enrique Calderon
    Commented Jan 27, 2019 at 3:59
  • Ask someone with smd soldering skills, but it is a large chip which makes it very hard. What was the cause? How did the 7V get to the pins of the atmega2560? A 5v opamp with rail-to-rail output is no problem. When using higher voltages, a voltage divider or protection resistor should be used.
    – Jot
    Commented Jan 27, 2019 at 8:36

1 Answer 1

Reset to default
2

A high voltage at a pin can damage the microcontroller partially.

The ATtiny microcontrollers can sometimes survive a voltage of 7 V.
The ATmega microcontrollers are damaged by that voltage.
The ATmega2560 is older and even more critical with voltages, it can not work at 16 MHz below 4.5 V and gets easily damaged with voltages above 6 V.

When the microcontroller runs at 5.0 V, a input pin can have -0.5 V to +5.5 V.

When the microcontroller gets damaged, it can be damaged partially. It is impossible to determine which parts of the chip are still okay. Therefor a partially damaged microcontroller should not be used anymore.

Solution:

A simple solution is a protection resistor in the signal path to the analog input.

schematic

simulate this circuit – Schematic created using CircuitLab

According to application note AVR182, it is allowed to push 1 mA into a pin or pull 1 mA from a pin for the ATmega microcontrollers. That current flows via the internal ESD diodes to VCC or from GND.

With a protection resistor of 4k7 before an input, the microcontroller is protected up to: 1mA * 4k7 + 5.5V = 10.2 V.

The internal ESD diodes can have more than 1 mA, but the maximum is unspecified. It is also dangerous when for example the ATmega is put into sleep mode. Then the 1 mA can lift the VCC voltage and still damage the ATmega microcontroller.

When a voltage divider is used (two resistors, R1 and R2) and the internal reference is used, then the voltage divider can be designed to be protected against very high voltages. For example with a ATmega2560, internal reference of 2.56V, R1=47k, R2=10k. This can measure up to 14.6 V and be protected up to 78 V.

Improve this answer
2
  • My understanding is that the ATmega implements sleep mode by gating the clock, with Vcc still applied to the whole chip.
    – Edgar Bonet
    Commented Jan 27, 2019 at 20:54
  • 1
    @EdgarBonet yes, so a current pushed into a analog or digital pin can lift the vcc above 5v (if the voltage has nowhere else to go), and can damage the chip. Should I explain it better?
    – Jot
    Commented Jan 27, 2019 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.