How does the voltage drop of a PN diode compare to that of a MOSFET connected diode?

grahamed

Joined Jul 23, 2012
100
#18 well spotted about the diode. I forgot that I added the diode as the body diode did not appear in the model I found.I did not look very hard. In this configuration it doesn't do anything.

I was not aware of the LT4320, but I like it, but not its price. I think I'll stick to discrete component version..
 

crutschow

Joined Mar 14, 2008
27,424
#18 well spotted about the diode. I forgot that I added the diode as the body diode did not appear in the model I found.I did not look very hard. In this configuration it doesn't do anything.

I was not aware of the LT4320, but I like it, but not its price. I think I'll stick to discrete component version..
There's more to generating the proper gate signal than may be apparent.
Your discrete version has a small problem with a capacitor on the output if you are trying to generate a DC voltage.
It can conduct in the reverse direction for a period of time when the MOSFET is ON and the input sine-wave peak drops below the voltage on the capacitor.

Edit: Also the reverse Vbe across the left diode-connected transistor will exceed the rating for most transistors.
It could be replaced by a diode with a higher reverse voltage rating although the value of resistor although R1 needs to be reduced to about 5kΩ for proper operation.

And I think the price of the discrete parts to drive 4 MOSFETs for an ideal bridge rectifier would be comparable to the cost of the LT4320,
 
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anhnha

Joined Apr 19, 2012
881
In operation, the substrate diode conducts in the MOSFET reverse direction (with gate biased ON) until the threshold voltage of the MOSFET is reached, where it starts to turn on. So at voltages above the threshold the forward drop of the MOSFET can become much less than a junction diode, depending upon the MOSFETs ON resistance.
I can't understand why drop voltage decreases if RON is smaller. Could anybody explain?
 

anhnha

Joined Apr 19, 2012
881
If you need that explaining - you may do better at woodwork!
So what do you mean by that?
I see it now. Just feel discouraged by this kind of comment. English is not my native language and many times I couldn't make my self clear. This made me scared as a beginner.
Every hear of Ohm's law: V = I * R?
Of course, I know. What do you mean by RON? Is it a channel resistance or the dynamic resistance of body diode in piecewise linear diode model?
I assumed that RON is ON resistance of the channel and it is in parallel with body diode.
With RON being the channel resistance, by graphical method to solve the voltage across the diode, I can see that Vdrop across will decrease as RON become smaller.
Is this what you meant to say in that post?
Hope you see the problem.

Body diode model.PNG
 
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AnalogKid

Joined Aug 1, 2013
9,290
Yes, it is possible - bjt is the preferred switching device at very high current.
Once a mosfet is in its linear region, it behaves like a resistor (ohmic). So if Ids * Rds < 0.7v, a mosfet is more efficient. However, for very high current, it is likely the other way around.
Don't think so. A power MOSFET with Rdson of 0.002 ohms would have to sink 400 A to reach a Vds of 0.8 V. That part in a 50 A application kills any bipolar transistor's Vcesat. Note that for large bipolar transistors, normal transistor assumptions do not apply. Vcesat is much greater than 0.1 V and Vbe is much greater than 0.7 V. Plus there's that whole gain thing.

ak
 

crutschow

Joined Mar 14, 2008
27,424
..................
Of course, I know. What do you mean by RON? Is it a channel resistance or the dynamic resistance of body diode in piecewise linear diode model?
I assumed that RON is ON resistance of the channel and it is in parallel with body diode.
With RON being the channel resistance, by graphical method to solve the voltage across the diode, I can see that Vdrop across will decrease as RON become smaller.
Is this what you meant to say in that post?
...............
That's it exactly.
Ron is the channel resistance is parallel with the body diode and obviously the lower it is the lower the voltage drop for a given current since all the current is going through the channel as long as this drop is less than the diode forward drop.
 

ian field

Joined Oct 27, 2012
6,539
So what do you mean by that?
I see it now. Just feel discouraged by this kind of comment. English is not my native language and many times I couldn't make my self clear. This made me scared as a beginner.
If that made you scared.......................................................
 
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