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Fr Robert Ballecer, SJ I watched show 349, MAD SCIENTIST BUBBLER, today I noted your use of t...

G+_Jim Hofmann

By G+_Jim Hofmann,
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G+_Jim Hofmann Newbie

G+_Jim Hofmann

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Fr. Robert Ballecer, SJ I watched show 349, MAD SCIENTIST BUBBLER, today. I noted your use of two diodes for power isolation for both the positive and ground leads. I've always held that ground is ground. By putting a diode in the ground lead raised the Ardunio ground by +0.6 volts above the UBec ground. It probably doesn't make a difference and is just overkill in this case but I've always had issues with grounds with different potentials.

 

Is there another reason for using an additional diode in the Ardunio ground lead?

 

Thanks, Jim

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G+_Telford Dorr Newbie

G+_Telford Dorr

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X2 on that. Except under special circumstances, ground should always be ground, as all digital / analog signals are referenced to it. In this case, the signals to the LED ring are offset 0.6 volts by the diode in the ground lead, which can lead to signal integrity issues. The diode in the positive supply lead to the Nano provides adequate power isolation.

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G+_Telford Dorr Newbie

G+_Telford Dorr

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That's true. It's considered good practice to have 'analog ground' connect to digital ground at only one point. This avoids ground loops (small voltage drops caused by other signal current flowing to ground via multiple paths), and the resulting noise thus introduced. This is especially important if you have any circuitry (op-amps, filters, attenuators etc.) between your analog signal source and the A/D converter. Now for serious analog accuracy, you use differential techniques: two A/D inputs - one measuring the signal, and one measuring the 'ground' at the signal source. The true signal reading is the difference between the two, as the common mode voltage tends to cancel out.

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G+_Kevin Marr Newbie

G+_Kevin Marr

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Technically, there should be a diode in the usb power path to prevent the UBEC from driving current into the usb port, but I'm guessing the usb port has enough over voltage protection.

With the UBEC and usb connected, there are two supplies trying to drive the nano and they should both be diode or'd into the nano.

 

From wikipedia Diode-or_circuit Diode-OR.png:

 

15048%20-%20Diode-OR.png>

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G+_Telford Dorr Newbie

G+_Telford Dorr

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According to my Nano schematic, they have a diode (D1, a MBR0520 schottky diode) on the board to keep any voltage from either Vin or +5v out of VUSB. So the Arduino designers agree with you. I guess great minds think alike... ;-)

 

That said, there is an issue with the Vin pin and the UA78M05 regulator chip: you do not want a load of any kind (like an unplugged wall wart) connected to Vin while power is supplied to the board from either +5v or VUSB, as this will cause the UA78M05 to be reverse biased (Vout > Vin), and that's 'no bueno'...

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G+_Telford Dorr Newbie

G+_Telford Dorr

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You're correct - the LEDs would be more than the regulator could take - BUT (according to the datasheet), a (real) UA78M05 is supposed to be overload protected. So, while it may get absurdly hot, you shouldn't be able to cook it. Now the datasheet also says that there's supposed to be a 330nf capacitor from the regulator Vin to ground for "stability", and the Nano board doesn't have one. Since you've apparently experienced regulator failures, I wonder if the lack of a five cent capacitor leads to regulator failure? Curious...

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