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Author Topic: FDN340P Drawn incorrectly on schematic?  (Read 2271 times)
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Hi there,

I was just looking at the Leonardo schematic: http://arduino.cc/en/uploads/Main/arduino-leonardo-schematic_3b.pdf

Is is just me, or is the FDN340P at the bottom of the schematic drawn the wrong way around? Should the Source of this P-Channel MOSFET not be connected to the USB's 5V? In the schematic they have connected the Drain to 5V from the USB instead.

It might just be me, but would love clarification on this!

Thanks


* FDN340P.JPG (14.63 KB, 359x185 - viewed 70 times.)
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How does it compare to the Duemilanove, Uno, Mega, and other boards with auto power switching?
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That was one of the first things I checked. All of the boards you mention have it drawn the same way - which I why I thought it couldn't be an error. However that goes against the operation of a P-channel MOSFET as a switch - which is how it is being used here.

Argh!!! Someone please clarify!

Thanks
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However that goes against the operation of a P-channel MOSFET as a switch

Why? If gate is low, then switch is ON.  So 5Volts will be coming from VUSB.

If a DC jack is plugged in (and VIN will now be some value) and if VIN is more than 3.3V, then mosfet switches off (disconnects from VUSB). 5Volts will now be coming regulator IC1.
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http://www.fairchildsemi.com/ds/FD/FDN340P.pdf

Here's the best way to view it - with a MOSFET, when the Gate is active, current can flow from D to S, or S to D, and when it's not the two sides are isolated.
So with the comparator driving the P-channel, when the Gate is High (Vin/2 >3.3V) the device is off, the two sides are isolated and 5V comes from the 5V regulator.
When the Gate is Low (Vin/2 < 3.3V) the device is On and the USB is allowed to flow thru as the 5V source.
As a switch that's acceptable operation. As a linear amplifier it probably wouldn't be the preferred mode of operation.
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Thanks for the replies.

I understand and agree with the description of the operation you have both given, however the dispute I have is with specific terminals of the P MOSFET and how they are connected. When using a P MOSFET as a switch, the load (or circuitry to be switched) should be connected to the Drain of the MOSFET which has not been done in this schematic (they have connected the load to the Source).

The MOSFET switches using the Vgs (Voltage between the Gate and Source). I.e in this case the voltage between the output of the comparator and what should be the voltage of the USB 5V (however this is NOT what is drawn).

Does that make sense?

Thanks
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The MOSFET switches using the Vgs (Voltage between the Gate and Source). I.e in this case the voltage between the output of the comparator and what should be the voltage of the USB 5V (however this is NOT what is drawn).

What I understand you're saying above is you think T1 selects whether the voltage comes from VUSB or GATE?
Like an SPDT switch?
No.

The mosfet is just being used as an on/off switch, i.e. whether to disconnect or connect VUSB, and allow it to flow or not to U3.

The output of the comparator (gate) just "flicks" this switch on or off. 
(when comparator is low, powered from VUSB. When comparator is high, disconnected from VUSB.
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Someone once posted on this forum in detail explanation about the specific usage of this FET switch in the arduino auto-voltage switching application and it involved the body diode of all things? So while it might be an unconventional set up it appears to perform the function that is required.

Lefty
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There is a big discussion of it in the old forum, I didn't attempt to find it tho.
Operation is as vasquo says - USB not connected to 5V when comparator output is high, and USB becomes the source for USB when comparator output is low (generally, powered from PC and nothing connected to barrel jack).
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Hi,

No I do not think it acts like a SPDT switch! Like I said, I am happy with the 'descriptive' purpose and operation of the MOSFET - i.e it is switching USB 5V either on or off. What I am not happy with is the way the terminals of the MOSFTET are drawn in the schematic.

Please see my attached image.

Circuit A is what has been drawn on the schematic.

I believe that what should have been drawn is Circuit B.

Any comments?

Thanks


* MOSFTET.jpg (68.75 KB, 800x185 - viewed 100 times.)
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Hi,

No I do not think it acts like a SPDT switch! Like I said, I am happy with the 'descriptive' purpose and operation of the MOSFET - i.e it is switching USB 5V either on or off. What I am not happy with is the way the terminals of the MOSFTET are drawn in the schematic.

Please see my attached image.

Circuit A is what has been drawn on the schematic.

I believe that what should have been drawn is Circuit B.

Any comments?

Thanks

Yes, don't guess or postulate. Get the datasheet for the specific FET and see it's terminal pin out and then look on your board to see if it's wired as drawn or not. The fact that it works as presently installed on arduino boards is not in question, just how the FET might be being used. Mosfets are able to conduct current in either direction to and from drain and source, that is also known.

Lefty
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Ahhhh i see what you mean. Why didn't you say so in the 1st post. Your beef is with the drawn diode! smiley

Looks like the internal diode for the mosfet symbol is drawn the other way around.

Per datasheet, figure B mosfet symbol is correct.
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Cool, thank you vasquo! Good to know I'm not going insane!

yes my beef is the way the diode is drawn and hence ALSO the drain and source terminals.

In their schematic they have drawn the circuit symbol for a P MOSFET such that the Drain terminal of the mosftet is connected to the USB 5V when in fact they should have drawn it such that the source terminal is connected to USB 5V. Basically the MOSFET symbol is drawn back to front in their diagram.

I thought it was just a simple error in the schematic drawing, however I was doubtful given this error has been present since the Duemilanove schematics back in 2009.... I can't believe this hasn't been picked up on to date?
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Ahhhh i see what you mean. Why didn't you say so in the 1st post. Your beef is with the drawn diode! smiley

No, diode is built into fet and is drawn correctly in both examples and just like the datasheet shows cathode of diode connected to source terminal. His issue is how the fet is wired into the larger circuit, which side should be drain and source terminals.

Looks like the internal diode for the mosfet symbol is drawn the other way around.

Per datasheet, figure B mosfet symbol is correct.
« Last Edit: February 12, 2013, 04:52:08 pm by retrolefty » Logged

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The drain/source is drawn correctly and works correctly.
It's just the drawn diode that is wrong in the schematic.... see datasheet for correct diode orientation.


FYI, in Eagle, you can draw the schematic any which way (heck, you can just show the mosfet as a box with 3 legs and a smiley face) and as long as you assign and label the pins, and connect them to the correct device pins, it will work.  
« Last Edit: February 12, 2013, 04:53:58 pm by vasquo » Logged

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