PrathikP

modified 6 years ago

MOSFET Body Diode Question

176

02:09:14

A while ago, I had built some DC-DC converters on my breadboard. I originally intended to use the IN5822 power diode, but unfortunately it didn't fit in my breadboard (not that the breadboard is a good place to build a DC-DC converter) because of which I ended up using UF4007 diodes in parallel, which have thinner leads. I could have soldered thinner wires to the 5822, but I'm a lazy goose and like simple solutions! Today, the idea of using a MOSFET's body diode struck me. MOSFETs fit in my board, which makes it possible to use. By shorting the gate and source terminals, i would be able to use it as a high current diode, but then I suddenly wondered whether using (or abusing?) a MOSFET in such is way is a good idea or not. Can the MOSFET possibly sustain any kind of damage if I use it this way? Let me know in the comments. 

Incase anyone noticed that the MOSFET I'm abusing has a gate threshold voltage of 0.5V instead of 3V, which is the Vto of the IRF540N, it is because of the fact that, for some twisted reason, the cut-in voltage of the body diode of an EC MOSFET is equal to Vto! I measured the cut-in voltage of the IRF540N body diode and found that it is around 500mV; and according to the datasheet, the diode drop voltage at 16A is 1.2V. 

MOSFET used: IRF540N http://everycircuit.com/circuit/5282567524450304

published 6 years ago

kiani

6 years ago

Could you show plz. How did you measure the cut in diode voltage to be 500mV.

kiani

6 years ago

Also does EC mosfet has body diode included in its model...! Therr is no body!

PrathikP

6 years ago

I used a multimeter. That's how.

PrathikP

6 years ago

I didn't do it on EC, if you're under that impression. I measured the cut-in voltage of a real IRF540N

PrathikP

6 years ago

Because of the 500mV cutin, i sent Vto=0.5V for the FET who's bodydiode I'm using. The other fet is set to 3v Vto

PrathikP

6 years ago

Oh hang on i think youre asking me how i found out that Vto = cutin in EC. That i found out by simply passing 1mA of current through the body diode.

kiani

6 years ago

Right, i see.. I am not sure if EC model has a body diode at all,,. but yoir circuit is working... Or if yhete is a sort of body diode is included how it is modeled and how to be taken to account,, only EC knows snd they are not telling us..

PrathikP

6 years ago

As far as i know, the definition of cutin voltage is "the minimum voltage required to turn on a diode". Some sources say that the cut-in voltage is equal to the barrier voltage of the diode. I don't like these definitions because there is no mention of how much current must flow through the diode when the voltage measurement is made. So i use the definition "the minimum voltage required to send 1mA through the diode" since most multimeters use that value of current and any smaller current is too small. Any higher is too much. I think 1mA is the perfect current to make that measurement

kiani

6 years ago

What do you mean by turn "turn on a diode". The dide datasheet specifies Vf, {ie. Forward voltage} and forward current, {If}... As a diode has a natural equivalent resistance {1n4007, approx. Is 1 ohms. (not sure)}. so if a diode is forward biased with a voltage higher than the junction voltage {0.6! V},, only (I forward}, floes, no more no less. . I guess that is the min current the diode needs.

PrathikP

6 years ago

"turn on of a diode" is vary vague, which is my point. I've seen these definitions on the internet. That's why i added the 1mA thing

kiani

6 years ago

http://www.datasheetcatalog.com/datasheets_pdf/1/N/4/0/1N4005.shtml

kiani

6 years ago

Just checked some data sheet, the forward characteristic shows, a forward voltage of 1v {for 1n4005! } at 10mA... They don't mention min. Current required to setup any particular forward voltage, which seems to vary with forward currents... Its is interesting how much current do we really need to say a diode is turned on.!? As forward voltage changes with forward current (in this case from 1 to 1.5 v).. For 10mA to 10A.

PrathikP

6 years ago

That's why i use the above mentioned definition. Multimeters use around 1mA of current. (Datasheet of my meters say 1mA+-0.6). So by using that definition, multimeters aren't affected in any way

PrathikP

6 years ago

The uf4007 has a drop of 0.7V at 10mA. The uf and 1n are same except for recovery time. So it can't be 1v for the 1n4005. I think

hurz

6 years ago

did i get it right "diodes in parallel" ???

PrathikP

6 years ago

Yup!

hurz

6 years ago

Bad idea if not from the same die and productline and in same package. Answay, even the it might need some extra balancing resistors to not burn one of them quickly.

PrathikP

6 years ago

Hmm I'm aware of that. Like i said, it was a quick and simple solution. Not the best, i know. I think I'll try using the mosfet's body diode next time.

PrathikP

6 years ago

Or I'll just solder some thinner wires to the 5822 lol

PrathikP

6 years ago

I used 2 uf4007s in parallel. I wouldn't put 2 completely different diodes in parallel of course

hurz

6 years ago

without reading it all, there is no body diode in everycircuit mosfets

hurz

6 years ago

put two uf4007s in real in parallel and it is just one diode for you working, the other one is laszy. This will destroy sooner or later the working diode and the lazy one must take over.

PrathikP

6 years ago

Hmm yes

hurz

6 years ago

Consequence, if you put two diodes in parallel to get the power of two diodes, you will be disappointed, cause you get only the power of ONE diode, till its destroyed and then the second diode takes over till its also destroyed. Two diodes in parallel is always a bad idea, since you put some resistors at each diode in series to algin there tollerrances.

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