Does Drain Current Increase if Doping Increases?

[rmachzovchev]

TL;DR Summary

If I increase the doping of a transistor, does current / m^2 also increase before device failure?

I'm an EE with only a surface knowledge of solid state. I know this forum is mostly for students but I don't know where else to find a lot of physicists. Also, please forgive me if this is a dumb question.

For a circuit I want to build, I need a transistor that can conduct > 10,000 A, but does not need to tolerate a large drain-source voltage when turned off.

I already know that I can gang a lot of normal* transistors, or make 1 transistor with huge cross sectional area. For reasons I will not bore you with, those options are not on the table.

*normal here means commercially available, typically < 500 A continuous drain current. The sort of thing that DigiKey has lots of.

Here is my question:

All other things being equal, if the doping in the transistor goes up, the available charge carriers should also go up. Therefore, drain-source current should also increase. Is this true?

 

[Lord Crc]

Since you say this is for a project, surely you can discuss this with whomever you'll use for producing these transistors. Clearly cost is not an issue, or you would go for multiple off-the-shelf transistors.

At high current levels the package can quickly be a limiting factor. I'm assuming the manufacturer has some limitations to what kind of packages they support, so that might be an issue as well.

 

[DaveE]

discuss this with whomever you'll use for producing these transistors

Yes, this! No one here (probably) actually knows anything about this extreme application. Talk to the guys that make the 500A transistors. Back in my day, this was the realm of thyristors, check them out too.

 

[Tom.G]

I need a transistor that can conduct > 10,000 A

I already know that I can gang a lot of normal* transistors, or make 1 transistor with huge cross sectional area. For reasons I will not bore you with, those options are not on the table.

Uhmm, I think you need a different table.

 

[jsgruszynski]

Summary:: If I increase the doping of a transistor, does current / m^2 also increase before device failure?

I'm an EE with only a surface knowledge of solid state. I know this forum is mostly for students but I don't know where else to find a lot of physicists. Also, please forgive me if this is a dumb question.

For a circuit I want to build, I need a transistor that can conduct > 10,000 A, but does not need to tolerate a large drain-source voltage when turned off.

I already know that I can gang a lot of normal* transistors, or make 1 transistor with huge cross sectional area. For reasons I will not bore you with, those options are not on the table.

*normal here means commercially available, typically < 500 A continuous drain current. The sort of thing that DigiKey has lots of.

Here is my question:

All other things being equal, if the doping in the transistor goes up, the available charge carriers should also go up. Therefore, drain-source current should also increase. Is this true?

There's no such simple generalization. Multiple locations within a MOSFET have specific and carefully chosen doping levels. There isn't one.

For example, drain-source breakdown voltage depends on the breakdown of the drain-channel diode. Increasing the doping of the channel as well as shrinking the channel for speed reduces this breakdown voltage. Often for non-power devices that are shrink, doping is intentionally reduced by LDD and similar drain engineering which slows the device but prevents breakdown as you shrink (which is combined with reducing logic levels from 5V for TTL to 3.3, 3.0, 2.5, 2.0 1.1 - all to prevent breakdown in addition to using LDD-like low doping.

There are also ultimate limits on the maximum PN junction breakdown voltage simply because you are using silicon (typically ~700V). This is why/how GaN and SiC MOSFETs are gaining so much popularity for switching: their breakdown due to intrinsic semiconductor band gap is far higher than silicon every will be. This also increases their switching speeds.

In general, when you pick the transistor for a switch, you must base the choice on published specs created by manufacturing people who can keep all these details juggled in their heads. And you may have to consider multiple fundamental transistor types ranging from Silicon MOSFETs, GaN or SiC MOSFETs, IGBTs and Bipolars. Each has a trade-off that depends on the specific application requirements - there is never a "one size fits all".

For 10,000 amps I'd be looking at IGBTs and Bipolar in massive parallel circuit configurations. And don't forget about ballast resistors.

https://electronics.stackexchange.com/questions/77045/transistors-in-parallel

 

[artis]

@rmachzovchev

Summary:: If I increase the doping of a transistor, does current / m^2 also increase before device failure?

I need a transistor that can conduct > 10,000 A

For a single transistor not a chance.
Some of the most powerful thyristors even now can only go up to a couple thousand amps like this one from ABB goes up to 4000A
https://search.abb.com/library/Down...LanguageCode=en&DocumentPartId=&Action=Launch

Also remember these thyristors are for low switching/frequency.
So what is your intended switching frequency for the imagined switching device? Increasing the frequency means skin effects will render less and less of the actual conducting material useful which will further decrease the transistor amperage capability. This is the reason why RF transistors have very low amperage ratings compared to lower frequency MOSFETS and IGBT's for example.

Anyway that current rating is I'd say impossible, we need to know the frequency and specific use and maybe then you can use an array of parallel devices or something.
10k Amps is quite frankly the capability of thick copper wire and a generator called "Faraday disc".
A single MOSFET even if it could be made to carry that amount of current would probably have the gate capacitance as high as that of a electrolytic capacitor and it's switching speed would be very low.