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

Thread Starter

Beanz1114

Joined Nov 13, 2014
18
Hello good people of AAC,
Of late I have seen rectifier circuits in IC design that use MOSFET connected diodes (connecting Gate to Drain) instead of PN junction ones while claiming that the former has lower voltage drop compared to the latter. No justification was really given. I scoured the depths of the World Wide Web in a quest to find the reason behind this but, alas, my search has been in vain. I understand that by connecting the gate of the MOSFET to the drain, the MOS will only operate either in cutoff or saturation. The current in saturation is given by the oh-so famous square law that we all know by heart. However, this doesn't really explain how it compares to the EXPONENTIAL forward current of the PN junction diode in terms of voltage drop does it? Wouldn't a PN junction have a lower voltage drop than MOSFET since it has exponential forward bias current?
 

AnalogKid

Joined Aug 1, 2013
10,971
Agree. If you want a FET to saturate with a drain-source voltage of 0.5 V for example, less than the forward voltage of a typical rectifier diode, and the gate is tied to the drain, then Vgs is 0.5 V, not enough to even begin enhancement. So the FET never would reach this operating point.

ak
 

Thread Starter

Beanz1114

Joined Nov 13, 2014
18
Are you sure that diode connected MOSFET will have a lower voltage drop than "normal" diode?. I don't see how this can be possible.
To get lower voltage drop you need to use a MOSFET as switch where the gate is connected to the control voltage.
https://en.wikipedia.org/wiki/Active_rectification
http://www.mouser.com/ds/2/277/MP6914_r1.0-526360.pdf
Thanks for your input! Actually I'm not that sure myself.
However, from the Wikipedia link you just gave it does say that "Replacing a diode with an actively controlled switching element such as a MOSFET is the heart of active rectification. MOSFETs have a constant very low resistance when conducting, known as on-resistance (RDS(on)). They can be made with an on-resistance as low as 10 mΩ or even lower. The voltage drop across the transistor is then much lower, meaning a reduction in power loss and a gain in efficiency. However, Ohm's Law governs the voltage drop across the MOSFET, meaning that at high currents, the drop can exceed that of a diode."

Am I right in assuming from the Wiki page that at low currents, the voltage drop of a MOS diode is lower than a PN diode as per Ohm's Law?
 

Thread Starter

Beanz1114

Joined Nov 13, 2014
18
Agree. If you want a FET to saturate with a drain-source voltage of 0.5 V for example, less than the forward voltage of a typical rectifier diode, and the gate is tied to the drain, then Vgs is 0.5 V, not enough to even begin enhancement. So the FET never would reach this operating point.

ak
Thank you for your input AnalogKid! So you agree that PN junction diodes have lower voltage drop than MOS connected diodes? The Wiki link that Jony130 gave suggests otherwise, I don't know what to think here lol. I might be missing something here. Why wouldn't Vgs = 0.5V be enough to begin enhancement?
 

AnalogKid

Joined Aug 1, 2013
10,971
So you agree
NO, I do not agree, because NO, that is not what I said.

I'm saying that your description of a "MOS connected diode" does not work so the question is moot. The Wiki quote in post #4 is correct as far as it goes, but it says nothing about how to drive a gate to achieve active rectification. Your description of how to do that is wrong. Yes, there is such a thing as a diode-connected bipolar transistor, connecting the base to the collector. A similar arrangement with a MOSFET

will
not
work.

ak
 
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hp1729

Joined Nov 23, 2015
2,304
Hello good people of AAC,
Of late I have seen rectifier circuits in IC design that use MOSFET connected diodes (connecting Gate to Drain) instead of PN junction ones while claiming that the former has lower voltage drop compared to the latter. No justification was really given. I scoured the depths of the World Wide Web in a quest to find the reason behind this but, alas, my search has been in vain. I understand that by connecting the gate of the MOSFET to the drain, the MOS will only operate either in cutoff or saturation. The current in saturation is given by the oh-so famous square law that we all know by heart. However, this doesn't really explain how it compares to the EXPONENTIAL forward current of the PN junction diode in terms of voltage drop does it? Wouldn't a PN junction have a lower voltage drop than MOSFET since it has exponential forward bias current?
Have you ever played with a synchronous rectifier design? The diode portion of the MOSFET conducts first. When the MOSFET turns on voltage across the MOSFET decreases even more. My point is the MOSFET has a lower resistance than a diode, but I guess that may not hold true for a diode connected MOSFET per the topic. Or does it?
 
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dannyf

Joined Sep 13, 2015
2,197
Wouldn't a PN junction have a lower voltage drop than MOSFET since it has exponential forward bias current?
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.
 

Roderick Young

Joined Feb 22, 2015
408
In IC design, sometimes a FET was used in with gate tied to drain to make a resistor. Used to be called a Pinch Resistor decades ago, I don't know if that's still the case. I'd never seen that configuration used as a diode. As others have mentioned, there is a synchronous rectifier configuration. That's when drain and source of the FET are used in place of, or more commonly, in parallel with, a PN diode. There is additional circuitry to sense when the FET should be on, and that circuitry drives the gate of the FET.
 

grahamed

Joined Jul 23, 2012
100
I agree that connecting Gate to Drain will not turn a FET on. Problem is that if you do try to use a FET as a diode by simply turning it on you create a switch not a diode - you must turn it on synchronously with the waveform.

I have used the configuration called a "perfect diode" involving a FET and a (preferably matched) pair of BJTs and a couple of resistors. Since many FETs exhibit RDS of a few 10s of milliohms it can be truly almost perfect.

When I get chance I'll post the circuit but I guess it's there on the net.
 
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hp1729

Joined Nov 23, 2015
2,304
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.
Diodes increase in forward voltage at higher currents, too.
 

crutschow

Joined Mar 14, 2008
34,201
Yes, a MOSFET used as a diode can have a much lower ON voltage than a junction diode, if the gate is driven by the proper circuitry to generate a synchronous diode. Such a circuit connects the gate to source when blocking voltage in the normal direction and connects to the gate to turn it ON when conducting in the reverse direction (since a fully on MOSFET conducts equally well in both directions).
This is shown in grahamed's posted circuit. (Note the added diode across the MOSFET is not needed as that is in parallel with the MOSFET substrate diode.)

Here is such a bridge rectifier circuit with a dedicated LT4320 driver.
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.
Since MOSFETs can have ON resistances in the low milliohm region, the MOSFET drop can be much less than a junction diode's up to very high currents.
It should be apparent though, that this MOSFET bridge will not have any advantage over diodes if the peak voltage being rectified does not go significantly above the MOSFET's threshold voltage to fully turn it on.
So for a low voltage bridge you want low threshold voltage (logic-level type) MOSFETs.
Note that, with a typical bridge capacitor output filter, the diodes only conduct near the peak voltage level, thus the substrate diode in the MOSFETs will not actually conduct any current before the MOSFET turns on.
 
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ian field

Joined Oct 27, 2012
6,536
Hello good people of AAC,
Of late I have seen rectifier circuits in IC design that use MOSFET connected diodes (connecting Gate to Drain) instead of PN junction ones while claiming that the former has lower voltage drop compared to the latter. No justification was really given. I scoured the depths of the World Wide Web in a quest to find the reason behind this but, alas, my search has been in vain. I understand that by connecting the gate of the MOSFET to the drain, the MOS will only operate either in cutoff or saturation. The current in saturation is given by the oh-so famous square law that we all know by heart. However, this doesn't really explain how it compares to the EXPONENTIAL forward current of the PN junction diode in terms of voltage drop does it? Wouldn't a PN junction have a lower voltage drop than MOSFET since it has exponential forward bias current?
MOSFETs are often used as voltage polarity protection when low diode Vf is needed.

You have to connect drain and source so that the body diode would be normally conducting, or it would defeat the point of the exercise - apparently the channel works either way round, so as long as you bias the gate correctly, the channel will conduct better than the diode Vf.

This was all explained in EPE magazine some years back, but the article could be difficult to track down.
 

crutschow

Joined Mar 14, 2008
34,201
MOSFETs are often used as voltage polarity protection when low diode Vf is needed.

You have to connect drain and source so that the body diode would be normally conducting, or it would defeat the point of the exercise - apparently the channel works either way round, so as long as you bias the gate correctly, the channel will conduct better than the diode Vf.

This was all explained in EPE magazine some years back, but the article could be difficult to track down.
Here's my article with simulation results on that.
 
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