# Resistance of mosfet's internal diode

I am an IoT solutions developer and not an engineer, so please excuse my vocabulary if not upto the mark.

I have used Mosfets such as FDD6637 and IRF9310 in my circuits for power switching and they work great for my use cases because of their low Rds(on).

From the datasheet, there is an internal diode and I am showing the characteristics in the attached image.

My questions:

• What would be the internal resistance of this diode? Will it be the same as Rds(on)? I can't seem to find it in the datasheet.
• Do you know of any P-channel mosfet that does not have this diode but has similar Rds(on) (10 to 15 mOhms at -4.5v?). I do not want reverse current to flow in a particular case, as I need my switching to work in only one direction.

EDIT: I have shown my problem in the rough diagrams below (sorry if the symbols are not accurate).

Works but battery powers load through internal diode (apparently):

Even though above circuit works, the question is about the internal resistance of diode and the drop across the mosfet when main power is off.

Following does not work because main voltage shows up at battery connector:

In the above circuit, I expect the mosfet not to conduct when there is +9v. But it is conducting because of the internal diode. This circuit would take advantage of the low Rds(on), so it is desired.

• Vocab? You mean vocabulary? ;-) – winny Oct 3 at 9:57
• "•What would be the internal resistance of this diode?" It's rarely stated. You need to look at a I-V-curve for the diode, if supplied. – winny Oct 3 at 9:58
• Diodes don't really have a resistance - the I/V graph looks completely different. – pjc50 Oct 3 at 9:58
• For practical purposes ALL MOSFETs that you will encounter will have the reverse diode - it's caused by the structure of the device on a silicon substrate and is unavoidable without very special approaches. If you tell us what you are trying to actually achieve we can help provide a good solution. | A method to get low voltage drop and mono directional switched current flow is to use two identical FETs in series with one polarity reversed. Join s-s and g-g. One D is the new S and one D is the new D. Drive gg negative wrt to ss to turn on as usual. This works because MOSFETS are 2 quadrant ... – Russell McMahon Oct 3 at 11:15
• ... and Vds polarity is not important as long as Vgs polarity is "correct". but THIS SOLUTION IS NOT USUALLY NEEDED. what IS YOUR ACTUAL APPLICATION? – Russell McMahon Oct 3 at 11:16

I answer to your questions below, starting from the first one on the body drain diode.

• What would be the internal resistance of this diode? Will it be the same as Rds(on)? I can't seem to find it in the datasheet.

The I-V characteristic of a diode is not linear, so it is almost never modeled as a single resistor: this is the reason why, in the datasheet of the MOSFETS you use, such parameter is not even mentioned. The only exception to this situation I'm aware of is the characterization of high power diodes like this: in the datasheet of those power devices, the so called slope resistance $$\\mathrm{r_T}\$$ is specified. In such devices, the magnitude of this model resistance is usually of few milliohms: however, in MOSFETs you cannot expect any precise relation between the (unspecified) $$\\mathrm{r_T}\$$ and the (very well characterized) $$\R_{DS_\mathrm{ON}}\$$ since these two resistances do not model the same physical phenomena even if the semiconductor structure is the very same one.

• Do you know of any P-channel mosfet that does not have this diode but has similar Rds(on) (10 to 15 mOhms at -4.5v?). I do not want reverse current to flow in a particular case, as I need my switching to work in only one direction.

Power MOSFETs where the body drain diode is not present have been produced in the past, but they are not easily available and surely do not have the low $$\R_{DS_\mathrm{ON}}\$$ you need in your application. I suggest another approach which may be a more viable alternative:

simulate this circuit – Schematic created using CircuitLab

The $$\p\$$-channel MOSFETs shown are used in the so called anti series connection: when the common $$\V_{GS}\$$ is $$\>0\$$, the MOSFETs are OFF and the two body drain diodes are connected back to back, so they are not conducting. When $$\V_{GS}\ll V_\mathrm{th}\$$, both the two MOSFETs are ON and the battery feed the load through a $$\2R_{DS_\mathrm{ON}}\$$ resistance: by choosing properly the devices, you solve the problem at the cost of an additional MOSFET.

• The D690S is perhaps sold only to Train producers and those who need megawatts of DC power made from the AC power grid, @arun: for them, spending \$129 is as easy as for us is spending a dollar. ;) – Daniele Tampieri Oct 3 at 14:22
• That circuit will not work properly because the FETs will stay turned on when 9V is applied. The OPs circuit partially gets around this by using a small value for R1 which reduces Gate voltage, but this causes high FET voltage drop in battery mode. The proper solution is an 'ideal diode' circuit electronics.stackexchange.com/questions/223935/… – Bruce Abbott Oct 4 at 3:20
• @BruceAbbott correct. I just substituted the $p$-MOS switch in the above circuit by other two antiseries connected switches: some work may be needed in order to make this circuit properly and a wired (ideal, made as shown in your link) diode resistor OR circuit may be the right choice in this application. – Daniele Tampieri Oct 4 at 5:53
• @BruceAbbott In my actual circuit, I have this 1n5822 to prevent the situation you have mentioned.. I have updated the question.. It does have a huge drop, but its ok on the main power as anyway input is a bit higher than 9v and efficiency is not that important on main power. The solution you have pointed to is above my head. – arun Oct 4 at 6:30
• With the diode it works, and as a bonus you don't need the extra FET! Since the primary battery can't stand any charge current I would make R1 a bit lower (eg. 220 Ohms) to ensure Gate voltage is well below the threshold with higher input voltage. – Bruce Abbott Oct 4 at 8:38

For your purposes, there is no such thing as "a" diode resistance.

When you plot current vs voltage for a resistor, you get a straight line, and the resistance is simply the slope of the line. Well, actually that's conductance, and resistance is the reciprocal of conductance, but I hope you get the point. Resistance does not change with voltage or current.

Diodes are different. If you plot current vs voltage, you get an exponential curve, and it should be obvious that the slope of this curve depends on exactly where you pick to measure - and it varies enormously over quite a small voltage range.

So instead of resistance, you can talk about incremental or local resistance, which is specific to a single voltage or current, and is found by measuring the slope of the V-I curve at the specified level. It will not be accurate over a wide range, but can be useful when analyzing small changes. This is commonly done with zener diodes, for instance.

Or, you can model the V-I relationship as a simple resistor as long as you know the current you'll be operating at. In your case, if you want accurate numbers you're going to have to measure them yourself, or you can look at the data sheets for Vds vs Is. Divide the two and you'll have your number for that current. As a quick example, your figure shows Vos as -1.2 volts for Is of -2.5 amps, which means that, at 2.5 amps, it behaves like a 0.5 ohm resistor. Well, 0.48 if you want to get picky.

The thing is, this will change with current and with temperature. Any curve you find in a data sheet will be "typical", and will often show those numbers for minimum, maximum and normal (room) temperatures.

And finally, just to make your life more difficult, for very large currents the bulk resistance of the diode will come to dominate the response, and the exponential curve will become a close approximation to linear.

You can measure that, if you like, but you'll be getting close to whatever dissipation limits the FET has, and don't blame me if you Let The Magic Smoke Out.

• "is found by measuring the slope of the V-I curve at the specified level". Thank you.. I will check it out.. – arun Oct 3 at 14:19