Body diode in MOSFET

Circuit

Senior Member
I am proposing to switch 12V AC at around 500mA using this a PICAXE in this configuration;
AC switching.png
This involves using the body diode of the MOSFET to conduct and I find it difficult to get full specs on the body diode from many MOSFET datasheets. Clearly, the maximum current capability of the body diode is much less than that of the junction and therefore it would seem sensible to add an external Schottky diode across the MOSFET. My concern is that this may not actually help if the primary current flow is through the MOSFET body diode which may still overload. Can any of the experts advise regarding this configuration?
 

Buzby

Senior Member
I'm not a MOSFET expert, but I always thought the body diode was not a 'real' diode, but an artifact of the construction process.

I don't think it's ever supposed to be used as a real diode, but as a thing to be aware of. ( Like the capacitance inherent in a diode, not usually a problem, but needed to be considered at high switching speeds. )
 

rq3

Senior Member
I am proposing to switch 12V AC at around 500mA using this a PICAXE in this configuration;
View attachment 22075
This involves using the body diode of the MOSFET to conduct and I find it difficult to get full specs on the body diode from many MOSFET datasheets. Clearly, the maximum current capability of the body diode is much less than that of the junction and therefore it would seem sensible to add an external Schottky diode across the MOSFET. My concern is that this may not actually help if the primary current flow is through the MOSFET body diode which may still overload. Can any of the experts advise regarding this configuration?
The body diode of (most) mosfets will have as much current capability as the active mosfet region itself. This is because it is not an intentionally fabricated diode, but because it is simply the intrinsic drain-source substrate acting as a diode when biased the "wrong" way.

It is often a useful and neglected "feature" of mosfets, especially power mosfets, as it can serve as an inductive load free-wheel diode, as a reverse current blocking diode in motor drive and battery charging applications (especially when using p-channel high side mosfets), and similar uses.

It is not clear to me what you intend this circuit to do. Is one mosfet supposed to be the AC input, and the other the switched output? I would think that the ground between source and drain would prevent anything from happening at all, other than being able to short your upper AC input (?) to ground.

For either mosfet, the body diode will act to clip the AC to the diode forward voltage when the diode becomes forward biased.
 

tmfkam

Senior Member
This looks as though it should work, assuming the AC being switched is isolated from the PicAxe. I searched for 'Mosfet Active Rectifiers' and found this: https://www.maximintegrated.com/en/app-notes/index.mvp/id/652 it may offer some inspiration?

Or try looking how Solid State Relays are constructed? I saw some 16A relays the other day from Anglia that may have been manufactured in this sort of way. They could switch AC or DC and had a centre tap for half wave operation, in an 8 pin Dip package.
 
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Jeremy Harris

Senior Member
My experience has been that the body diode has near identical ratings to the FET itself, but it is not a fast diode at all, and the bigger the FET (in terms of max Ids) the slower the body diode will be, in general. This limits its usefulness in higher frequency (anything over around 1KHz or so) applications quite often.
 

Circuit

Senior Member
Buzby, thank you, yes I was aware of that configuration but my understanding is that the circuit I illustrated has less voltage loss than the bridge rectifier.

Jeremy, thank you - in fact although I show AC in the diagram what I actually want is to be able to switch a DC motor circuit whichever way the motor is turning; therefore AC in the loosest context - I guess that "bipolar switch" would be a better descriptor.

Tmfkam - thank you, your reference was spot on; to quote from it "Therefore, designers interested in squeezing the last percent of efficiency out of their power supply place a Schottky diode in parallel with the synchronous rectifier MOSFET... A Schottky diode in parallel with the silicon body diode turns on at a lower voltage, ensuring that the body diode never conducts". I suspected that was the case. I looked at the Anglia site but I couldn't locate the SSR you referred to - could you give me a reference link?
 

tmfkam

Senior Member
I looked at the Anglia site but I couldn't locate the SSR you referred to - could you give me a reference link?
I'll find it when I'm at work. It was in a marketing email. I was interested until I saw they cost in the region of £10 each...

Here is the link to the Arrow information page: https://www.arrow.com/en/products/g3vm-401cr/omron

Here is a direct link to the datasheet for the G3VM SSR. http://static6.arrow.com/aropdfconversion/4975d13bd9a348b01297f4b0a0c12bb7fb46d5fd/en-g3vm_cr_fr.pdf If you scroll down to page 6 you'll see that the internal structure of the 'relay' switch is made up of two mosfets in a similar layout to your proposal.
 
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Circuit

Senior Member
I was interested until I saw they cost in the region of £10 each...
Thanks again. Yes, the cost of small SSRs was what prompted me into looking at cooking up my own. Initial tests on a breadboard show this seems to work okay.
 

rq3

Senior Member
Here is a direct link to the datasheet for the G3VM SSR. http://static6.arrow.com/aropdfconversion/4975d13bd9a348b01297f4b0a0c12bb7fb46d5fd/en-g3vm_cr_fr.pdf If you scroll down to page 6 you'll see that the internal structure of the 'relay' switch is made up of two mosfets in a similar layout to your proposal.
Similar, but different. The OP circuit had the upper source and lower drain connected to ground. It can't switch anything except shorted current to ground.
 

Circuit

Senior Member
Similar, but different. The OP circuit had the upper source and lower drain connected to ground. It can't switch anything except shorted current to ground.
Ah, I don't believe that I have... Please look again at my schematic, I think you are misreading it. Both sources are connected to the digital ground in my schematic - note that the lower MOSFET image is inverted correctly; both drains are connected as the bipolar switch. As it is, the circuit is now soldered up permanently and is working perfectly well as expected and as illustrated in my schematic; the PICAXE is able to control a 12 volt +/- feed to a DC motor and switch it on/off whichever way the motor is turning.
 

rq3

Senior Member
Ah, I don't believe that I have... Please look again at my schematic, I think you are misreading it. Both sources are connected to the digital ground in my schematic - note that the lower MOSFET image is inverted correctly; both drains are connected as the bipolar switch. As it is, the circuit is now soldered up permanently and is working perfectly well as expected and as illustrated in my schematic; the PICAXE is able to control a 12 volt +/- feed to a DC motor and switch it on/off whichever way the motor is turning.



This is your original schematic. As I said before, you didn't make it clear what's input, what's output, what's reference (ground). As drawn, without comments, it cannot do what you state. If what you have built works, then carry on!
 

rq3

Senior Member
Ah, I don't believe that I have... Please look again at my schematic, I think you are misreading it. Both sources are connected to the digital ground in my schematic - note that the lower MOSFET image is inverted correctly; both drains are connected as the bipolar switch. As it is, the circuit is now soldered up permanently and is working perfectly well as expected and as illustrated in my schematic; the PICAXE is able to control a 12 volt +/- feed to a DC motor and switch it on/off whichever way the motor is turning.
Aha, now I get it. You are not switching AC, you are switching the SAME DC supply in opposite polarity. Very different. Yes, you have successfully built a DC half-bridge, assuming (shame on me) that the motor load is where I think it is. However, you are not switching AC, you are just switching the polarity of a DC signal to a load.

When asking for input in engineering, too much information is not enough.
 

Buzby

Senior Member
Aha, now I get it. You are not switching AC, you are switching the SAME DC supply in opposite polarity.
I don't think so. Although the OP stated AC, but is using bi-directional DC, I think the circuit works as expected, assuming the AC has no common connection to the DC.

If you imagine the top AC terminal at +12v relative to the bottom terminal, and the PICAXE output is high, then the upper FET will turn on.
Electrons will flow 'upward' through the top FET, and 'upward' through the body diode of the lower FET, thus completing the AC circuit.

If the AC 'polarity' is reversed the roles are inverted, and current still flows between the top and bottom terminals.

If the PICAXE is low, then neither FET will turn on, but one of them will be conducting though the body diode, but the other will block the current.

Well, that's how I see it.

Cheers,

Buzby

A clever circuit !
 

rq3

Senior Member
assuming the AC has no common connection to the DC.
And in this case, the OP schematic, as presented, has a full AC short to the common DC ground.

Hence the need for full disclosure of schematic, and intended use. This is akin to "I have a rabbit farm, and my rabbits are dying. What should I do?". Many rabbit farmers weigh in with information on preventing death of rabbits. Then on post XYZ the rabbit farmer lets it be known that he also has a fox farm, and the fox have full access to the rabbits.

If the OP presents his full functional schematic, with ALL components in place (motors, etc), I can guarantee that it will neither look nor act like the originally presented schematic.
 
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Circuit

Senior Member
I don't think so. Although the OP stated AC, but is using bi-directional DC, I think the circuit works as expected, assuming the AC has no common connection to the DC.

If you imagine the top AC terminal at +12v relative to the bottom terminal, and the PICAXE output is high, then the upper FET will turn on.
Electrons will flow 'upward' through the top FET, and 'upward' through the body diode of the lower FET, thus completing the AC circuit.

If the AC 'polarity' is reversed the roles are inverted, and current still flows between the top and bottom terminals.

If the PICAXE is low, then neither FET will turn on, but one of them will be conducting though the body diode, but the other will block the current.

Well, that's how I see it.

Cheers,

Buzby

A clever circuit !
Buzby, thank you. Yes, you are absolutely correct on all points and I stand by my schematic and it works as shown -in fact as a component-built SSR.
 

Circuit

Senior Member
And in this case, the OP schematic, as presented, has a full AC short to the common DC ground.

Hence the need for full disclosure of schematic, and intended use. This is akin to "I have a rabbit farm, and my rabbits are dying. What should I do?". Many rabbit farmers weigh in with information on preventing death of rabbits. Then on post XYZ the rabbit farmer lets it be known that he also has a fox farm, and the fox have full access to the rabbits.

If the OP presents his full functional schematic, with ALL components in place (motors, etc), I can guarantee that it will neither look nor act like the originally presented schematic.
Oh dear! I think you are still misreading the schematic. There is no full functional schematic beyond this one - the working device is built as is; two FETs and four wires leading off the board. Two are connected to the PICAXE digital control as labelled and the remaining AC leads act as a simple digitally-controlled switch. Almost anything can be switched, provided it is within the specifications of the MOSFETs. I intend using this device in a number of applications.

The AC switched circuit is clearly not part of the DC digital circuit of the PICAXE - AC does not share a common ground with DC. Obviously, I am not switching the SAME DC supply because that would, as you infer, cause a short and it is not wired as half an H-Bridge but as a AC/DC-switching, bipolar solid-state relay. Whether AC or reversing DC through the switch is irrelevant - only a question of frequency. AC is the usual descriptor where polarity changes regardless of frequency or waveform. It works just as well at 50Hz as it does with a few seconds one way or the other. Similarly, the AC connections can be either input or output; by definition the direction of current is changing.

Buzby has read the whole thing perfectly and his narrative explanation of his understanding is correct in all elements. I do trust that this clarifies the matter.
 
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