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Topic: UNO Rev. 3: What is the LM358 for ? (Read 3 times) previous topic - next topic

retrolefty


The schematics won't show the other pins for the LM358. What powers the IC? Since power source hasn't been selected yet, where does it get its power (5V, I guess) from?


Well it's called a auto-voltage selection circuit, but in reality it operates as a auto voltage cut-off circuit. If there is external power connected via the connector (or Vin pin) of at least the minimum value then that hardwires to the on-board voltage regulator which then generates +5vdc Vcc power which is hardwired to the board's Vcc bus. The op-amp comparator seeing this valid Vin voltage then switches it's output to shut-off the mosfet switch that in turns prevents the USB's +5vdc source from being routed to the board's Vcc buss. That make sense?

Lefty

CrossRoads

The LM358's power pins are shown to it's left with a power supply decoupling capacitor.
Designing & building electrical circuits for over 25 years. Check out the ATMega1284P based Bobuino and other '328P & '1284P creations & offerings at  www.crossroadsfencing.com/BobuinoRev17.
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Nantonos


If Vin/2 is greater than 3.3V, Vin is used for the power source.


So while DC in is supposed to be 7-12V it is actually 6.6-12V? (well, 6.61 or whtever)

AlxDroidDev

Thanks for the replies, my friends, and sorry for bothering you with even more questions!

I saw how the LM358 is being powered by a 5V source.

What I don't get is this: what selects what 5V source is to be used, since there are 3 possibilities: USB, Vin and both.

In my mind it is sounding like a chicken-and-egg paradox: the LM358 selects the power source, but in order to select a power source, it has to be powered by something that has not yet been selected.

I am here: http://arduino.cc/en/uploads/Main/Arduino_Uno_Rev3-schematic.pdf


Learn to live: Live to learn.
Showing off my work: http://arduino.cc/forum/index.php/topic,126197.0.html

retrolefty

#9
Oct 24, 2012, 04:22 pm Last Edit: Oct 24, 2012, 04:32 pm by retrolefty Reason: 1

Thanks for the replies, my friends, and sorry for bothering you with even more questions!

I saw how the LM358 is being powered by a 5V source.

The board's Vcc bus, the same bus that powers all the components requiring +5vdc on the board.

What I don't get is this: what selects what 5V source is to be used, since there are 3 possibilities: USB, Vin and both.

It's best to think of the two sources of +5vdc as either the USB's +5vdc or the +5vdc from the output of the on-board voltage regulator that is powered by Vin on it's input. There is really no 'switch' that select between two possible sources, but rather just a FET switch that turns off the USB's voltage if there happens to already be voltage on the board from the +5vdc and +3.3vdc as a result of their being a valid Vin voltage being applied. The purpose of the voltage selector circuit is to just prevent a situation where the on-board regulator is  +5vdc and USB +5vdc are being 'wired' together.

In my mind it is sounding like a chicken-and-egg paradox: the LM358 selects the power source, but in order to select a power source, it has to be powered by something that has not yet been selected.

There in nothing on the board that can prevent the output from the on-board +5vdc regulator from powering everything on the board that is using +5vdc. Think of it as the default power that can't be switched off as long as there is power being applied via the external power connector or the Vin pin. That is why if there is both external and USB power available at the same time, the external power is the default power being used and the USB power will be turned off via the FET 'switch'.

Again a more accurate and functional name for the circuit would be to call it the 'USB auto voltage cut-off circuit'.
Lefty

I am here: http://arduino.cc/en/uploads/Main/Arduino_Uno_Rev3-schematic.pdf




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