3
\$\begingroup\$

I'm trying to understand the following section of the Arduino Uno's schematic:

Comparator section of Arduino schematic

(Here's a link to Arduino Uno R3 complete schematic)

The Arduino's site recommends that Vin should be between 7 and 12V. However, I was reading the datasheet for the comparator (LM358) and it seems that the voltage range is -0.2 to 5.7V, where 5.7V is absolute maximum for the input pin.

As a result, wouldn't 12V damage the chip since 12V would input 6V into the comparator?

I tried asking this question in another forum. However, no one really knew the answer why, and tech support at SparkFun was also stumped.

Thanks for the help in advance!

\$\endgroup\$
  • \$\begingroup\$ The comparator is on board? \$\endgroup\$ – Adam Nov 4 '14 at 23:21
  • \$\begingroup\$ Yup, according the the Arduino Uno's schematic : arduino.cc/en/uploads/Main/Arduino_Uno_Rev3-schematic.pdf. It is on board to turn off the P-mosfet (from my understanding), incase you plugged in an external supply while having USB plugged in. \$\endgroup\$ – yangXdynasty Nov 4 '14 at 23:24
5
\$\begingroup\$

The very same schematic was used for the original Uno32 from chipKIT. As it happened, someone managed to blow it up by applying 15V to the barrel jack. That was perfectly within the specs of the regulator, but the comparator ejected all the Magic Smoke™ and ceased to work.

After some diagnosis of their board I did the same calculations as you and came to the same conclusion. 15V, while within the limits of the regulator, overwhelmed the input of the op-amp.

I subsequently recommended a pair of clamping diodes on that input (now in Rev C of the Uno32), and since then there have been no reports of dead boards from too much input voltage.

enter image description here

(Note: I don't know if it's just a labelling problem in the PDF version of the schematic, but the VIN on the upper diode should actually be Vcc so it's clamped to the same supply voltage as the comparator.)

\$\endgroup\$
  • \$\begingroup\$ Thanks! I have a quick question. I'm not really knowledgable in clamping circuits, so I just want to clarify if lets say the VCMP = 6V would it be clamped as a 5.7V input to the comparator? \$\endgroup\$ – yangXdynasty Nov 5 '14 at 0:17
  • \$\begingroup\$ That is correct - or lower if you use shottky diodes. Check the data sheet for the comparator to see what's a safe voltage to go over Vcc by. \$\endgroup\$ – Majenko Nov 5 '14 at 0:18
  • \$\begingroup\$ Thanks again for the help! 5.7V is the absolute max, so I guess I should use shottky diodes to provide a bit of margin for error. \$\endgroup\$ – yangXdynasty Nov 5 '14 at 0:31
2
\$\begingroup\$

If you take a look at the datasheet, you can see Vin is after a diode voltage drop D1.

The M7 has a maximum forward voltage drop of 1.1V.

Being conservative and saying it has a 0.7V drop.

(12V - 0.7V) /2 = 5.65V. Which is inside the maximum voltage for the comparator. enter image description here

\$\endgroup\$
  • \$\begingroup\$ I thought that X1 was the power jack or some power from an adaptor, which got me confused, because I was more concerned for the Vin of the header pin. \$\endgroup\$ – yangXdynasty Nov 4 '14 at 23:33
  • 1
    \$\begingroup\$ You are correct, I don't think applying a 12V voltage to the Vin pin header will keep within spec of the comparator - take a look at Majenko's answer \$\endgroup\$ – tgun926 Nov 4 '14 at 23:46
0
\$\begingroup\$

Observe the LMV358 supply voltage is 5V on pins 8 and 4 Observe the LMV358 supply voltage is 5V on pins 8 and 4

Since the supply voltage of the LMV358 is 5V, Vout(pin1) is within 0-5V based on the value on CMP signal (pin3). There is a voltage divider of Vin/2 that is compared by 3.3V. So Vin can be higher than 7V up to 10V; However a 12V (Vin) will work fine since CMP=6V thus it will saturate the op Amp to 5V.

Extra info: If CMP is zero (Vin=0V), the FET T1 is ON so the USB bus will power the Arduino board.Otherwise, T1 is OFF and the diode on T1 will never conduct since USB_VCC is close to 5V reference.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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