Duemilanove power circuit - how does it work?
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Duemilanove power circuit - how does it work?

by blave on Tue Aug 25, 2009 9:27 pm

I understand everything on the Du. schematic except for the USB/power jack selection circuitry. I've looked here and at the arduino.cc site about this but haven't found anything. Just for my Edumacation, can someone explain how it works?

thanks,

Dave B.
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Re: Duemilanove power circuit - how does it work?

by Zener on Tue Aug 25, 2009 11:33 pm

That part of the schematic is hard to follow. You have to keep track of all the little flags such as VCC30, VIN, USBVCC and +5V

The way it works is:

The USBVCC supplies 5V. The FTDI chip makes 3.3V out of that (called VCC30 for some reason, I would think VCC33.....)

This VCC30 (3.3V) is put onto pin 2 of the LM358 (an op amp being used as a comparator there.) The VIN comes from external power to the jack, if any. It is divided in half by R10 and R11. So that means if it is above 6.6V at R10 then the output of the "comparator" (pin 1) will be high, and if below it will be low. The next stage of the op amp is set up as a buffer, or impedance buffer as I like to say sometimes. It just copies the voltage that goes into it. If the output of that is low then it turns on the P-channel fet T1 (usually transistors are named Q something, I guess not in Italy.)

So, if there is no external power to the jack (PWRIN) (no being less than about 7.1 V (6.6 + a diode drop)), then the fet is turned on and the USBVCC is passed along to the +5V line. If there is external power, the fet is turned off and IC4 takes over (voltage regulator).

Capisce?
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Re: Duemilanove power circuit - how does it work?

by floresta on Wed Aug 26, 2009 4:24 pm

Dave:

Here's a drawing that I made in order to figure it out myself. It agrees with Zener's description except he assumed a 0.5 volt diode drop and I used 0.7v.

Image

Zener:
... op amp is set up as a buffer, or impedance buffer as I like to say sometimes ...

It's also called a voltage follower.

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Re: Duemilanove power circuit - how does it work?

by adafruit on Wed Aug 26, 2009 6:57 pm

hey nice diagram!
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Re: Duemilanove power circuit - how does it work?

by floresta on Wed Aug 26, 2009 9:50 pm

Would you like to see the one I did for the Boarduino?

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Re: Duemilanove power circuit - how does it work?

by blave on Wed Aug 26, 2009 10:04 pm

Wow, I wish I'd posted my question here earlier. I spent quite a bit of time googling this, as I didn't want to bother anyone 8^) .

Thanks for the explanations, folks! As an EE I should've been able to figure this out, but I'm pretty rusty at circuit theory (which is why I've been messing around with Arduino/etc.!).

cheers,

Dave.
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Re: Duemilanove power circuit - how does it work?

by westfw on Thu Aug 27, 2009 3:06 am

I heard a rumor that the power circuit is based on the Maxim App Note:
http://www.maxim-ic.com/appnotes.cfm/an_pk/1136

So does the LM358 really accept high voltages on the inputs when powered from 5V, or are bad things likely to happen if the DC power source exceeds 11V or so (so that Vdc/2 > 5V) ?
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Re: Duemilanove power circuit - how does it work?

by gork on Thu Aug 27, 2009 9:12 am

What are the implications when the Arduino is powered via external 5V regulated supply and USB is also attached? In my application, I have 5V requirements beyond what the regulator on the Arduino can supply, but I still need USB attached. So far, there has not been an issue, so I'm wondering if this is expected behavior or I'm just lucky.

The current 5V circuit is supplied by a linear regulator with a banned heatsink dissipating about 6W, but I'm thinking of replacing it with a small switchmode power supply to reduce the heat output. Anything I should expect?
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Re: Duemilanove power circuit - how does it work?

by floresta on Thu Aug 27, 2009 9:55 am

westfw:

The datasheet shows a maximum supply voltage of +32 or +-16 and a maximum input voltage of -0.3 to +32. It makes no mention of any input voltage restrictions with respect to supply voltage. The equivalent circuit shows some sort of current source feeding each of the logic blocks so that is probably the reason that the input voltage can exceed the supply voltage without any smoke.

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Re: Duemilanove power circuit - how does it work?

by TomasReabe on Thu Sep 03, 2009 3:01 pm

Multiple questions: (sorry)
So how would the arduino respond if you used a ref. voltage (REF02 or something) for analog ref but the Vcc of the chip went low say 4.75V, 5% low. would the analog read and everything else still work just fine?

Wide input range DC supply:
from what I have found a switching voltage supply gives a very wide range of DC input, So if I want an arduino to work in my Car or Plane I need to work off of both 12v and 24V dc,(side note a 24Vdc lighter outlet in a plane can kill lots of your friends toys if you don't warn them.) so to find the best stable power supply to run an arduino off of either 12 or 24 dc what is the recomended method, I am also trying to hold a good ref voltage.

Thought 1: Use a switching power supply to get down to 9V dc, then use REF02 for Vref and a 7805 for VCC and acc. power.
Thought 2: I thought I saw a way to use an NPN transistor with REF02 to build a precision power supply.
Thought 3: Find a precision regulator with the V in in the range I need ( no luck so far)
Thought 4: Use a switching power supply to get down to 9V dc, then use REF02 for Vref and a mc33269d-5.0 for VCC and acc. power.
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Re: Duemilanove power circuit - how does it work?

by rikilshah on Wed Mar 27, 2013 12:44 am

@Zener Thanks dude, you made it easy to understand! :D
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Re: Duemilanove power circuit - how does it work?

by retrolefty on Tue Apr 02, 2013 12:53 pm

The circuit is a little more sutle (or complex if you like) then as you described:

The USBVCC supplies 5V. The FTDI chip makes 3.3V out of that (called VCC30 for some reason, I would think VCC33.....)


The problem with that analysis is that the FTDI cannot create 3.3vdc until the FTDI itself has 5.0vdc applied to it, and that doesn't come directly from the USBVCC but rather the boards Vcc 5vdc bus, so there is a chicken and egg problem with that statement.

Rather one has to look at the FET being used as the isolation device between the board's Vcc bus and the USBVCC source. The FET has a 'body diode' that will allow the USBVCC to power the board's Vcc bus (with a small diode voltage drop) so that the FTDI will get powered up and the opamp and it can then do the compare function to see if there is no external DC available it can then fully turn on the FET switch and thus 'shorting out' the internal fet diode and eliminate the voltage drop caused by the diode.

Now that begs the question is how complete is the USB power isolation when there is external DC power applied to the board and the FET is turned off to isolate the USBVCC from the board's Vcc bus, in that there is still the fet body diode to consider? If the USBVCC happens to be a little higher voltage then the on board 5 volt regulator wouldn't there be a possible current path to USBVCC even thought it's suppose to be isolated by the turned off FET?

The whole arduino auto-voltage switching design is in my opinion a somewhat flawed design and takes up more board space that could be used for more useful functional additions. A simple switch or jumper clip would be a better method and in fact is how the very first arduino boards handled USB/external DC power selection.

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by Michaeljohn on Fri May 03, 2013 3:13 am

boards Vcc 5vdc bus, so there is a chicken and egg problem with that statement.
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