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
How does it compare to the Duemilanove, Uno, Mega, and other boards with auto power switching?
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
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.
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.
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
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.
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
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).
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
Lydecker:
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
Ahhhh i see what you mean. Why didn't you say so in the 1st post. Your beef is with the drawn diode! 
Looks like the internal diode for the mosfet symbol is drawn the other way around.
Per datasheet, figure B mosfet symbol is correct.
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?
vasquo:
Ahhhh i see what you mean. Why didn't you say so in the 1st post. Your beef is with the drawn diode!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.
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.
Lydecker:
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?
That's because it's not in error, just your failure to see how the device operates in the circuit being used. Have you compared the physical mounting of the fet in your board Vs the datasheet terminals?
Lefty
I also struggled with this and thought D and S should have been swapped in the schematic (and on the board). But there is a good, simple explanation here of why it is correct as drawn:
http://www.engineeredentropy.com/2013/01/arduino-power-supply-selector/#more-214
In summary, the MOSFET is actually being used as a diode, not so much as a switch. But when the gate is low the channel conducts and the diode forward drop is essentially eliminated.
If you swapped D and S and put it in a more conventional PMOS switch arrangement then the body diode will allow flow from the Arduino Vreg back to USBVCC when the gate is high and the channel is blocked.
Jim
JimG:
I also struggled with this and thought D and S should have been swapped in the schematic (and on the board). But there is a good, simple explanation here of why it is correct as drawn:
http://www.engineeredentropy.com/2013/01/arduino-power-supply-selector/#more-214In summary, the MOSFET is actually being used as a diode, not so much as a switch. But when the gate is low the channel conducts and the diode forward drop is essentially eliminated.
If you swapped D and S and put it in a more conventional PMOS switch arrangement then the body diode will allow flow from the Arduino Vreg back to USBVCC when the gate is high and the channel is blocked.
Jim
Hi Jim,
So when there is no external supply, the comparator output goes low, the MOSFET conducts, the drain-source channel shorts out the body diode, and USBVCC flows through to Vreg ("+5V" on the schematic).
Question: The MOSFET conducts only if Vg is sufficiently lower than Vs/Vreg. But isn't Vs/Vreg a floating node in this case? So how would bringing the gate low activate the MOSFET if Vreg is not at a well-defined voltage?
Eric
Summary
The circuit works correctly with the PMOS Field Effect Transistor source and drain connected as shown in the schematic. The diodes in the PMOS FDN340P unit are the issue of concern. Do not worry, the diodes will not cause trouble.
Details
In the schematic, the source is on the right with the p type source shorted to the n-type substrate inside the unit. A protective diode is also in the unit. The other diode is not shown in the transistor symbol, but it exists where the p-type drain is metallurgically connected to the n-type substrate during silicon wafer manufacturing.
If the drain p-type material goes above the n-type substrate by 0.7 volts, the hidden diode will turn on with the forward bias.
The drawn diode in the unit duplicates the hidden diode from drain to substrate. This highlights for the engineer that : since the substrate is shorted to the source, there is a parasitic diode wired from drain to substrate that is shorted to the source terminal. The engineer wil recognize that applying a large voltage from drain to source will forward bias the parasitic diode. It is a bad idea to forward bias the substrate! That can provide a latch-up current problem if the geometries on the silicon are not robust.
The drawn diode symbol in the unit may be intended to prevent inductive spikes from causing damage to the MOSFET. It also indicates the polarity of the hidden diode from drain to substrate. The drawn diode may be a Schottky diode which turns on at 0.45 volts to prevent the pn junction diode from reaching 0.70 volts.
The datasheet Figure 6 implies that a Schottky diode is present, since 1mA flows at 0.45 volts at 25 degrees c. Intelligent Power and Sensing Technologies | onsemi
Conclusion
The schematic has it wired right. The diode seems to be a Schottky diode to prevent the pn diode from being forward biased, preventing latch up.
Reversal
If the source and drain get reversed in this circuit, a problem can occur when the +5 volt node will supply the USB with current if the USB cable touches a ground or unpowered computer.
I'm really sorry to post to a really old topic by Lydecker is right. The mosfet is the wrong way around, not just on the schematic, but the board too, the FDN340p is literally doing nothing on the board.
When Gate is off current flows, and when gate is on current flows.
Here is an image of the circuit on a breadboard. Arduino's circuit, and the proper way.
I used an FDN340p, not a different mosfet, I just I soldered header pins onto its pins.
Edit: Just read page 2, so if Jim is still around, just wondering why they would bother with the comparator going to the gate, if they were using it as a diode? Serious question. Only thing I can think of is that there is a tiny bit more resistance when its pulled high, but it still doesn't seem like enough to warrant the comparator.
