# How many transistors in a ...

• posted

So, I kinda hit a threshold in a recent YouTube video when the announcer just casually threw out "the transistor acts like a switch..."

I'm kind of curious about the world proportion of transistors that operate in linear mode as opposed to being entirely off or entirely on.

So a typical computer would have millions of transistors in the processor and memory and GPU and whatnot that are acting as logic, and maybe a thousand or less that are in power supplies and in the monitor that are in linear mode.

So -- how many transistors are there in a modern switching power supply controller, and how many are on hard? Ditto in LCD screen controllers, and in hard drives, etc.?

I'm figuring that at any given moment the proportion of operating linear- mode transistors to operating "switch" mode transistors is probably on the order of 10^(-4) or 10^(-5) -- but that's just a guess.

```--
Tim Wescott
Wescott Design Services ```
• posted

Probably most transistors in the world are in DRAM and flash chips. I don't know whether to call them hard switchers or not. Probably not.

A big LCD TV or monitor will only have 10 million or so poly transistors, small potatoes compared to a multi-gigabyte flash stick.

There are an estimated 1e19 transistors made per year.

```--
John Larkin         Highland Technology, Inc
picosecond timing   precision measurement  ```
• posted

Karl Kruszelnicki says "7000 for every grain of rice."

• posted

Do you count transistors in a switching power supply as linear or digital? How many transistors do you count for power transistors? My understanding is they are made by building a bunch of small transistors on one chip wired in parallel. Do you count them as one or many?

```--

Rick C```
• posted

I wonder how many electrons flow through those 1e19 transistors?

```--

Rick C```
• posted

You might want to rephrase "on hard" as "operating in a finite number of discrete states" (vs. "a continuum")

• posted

Plotting the load line vs. time, for any digital transistor, will still resolve a continuum. There will be high density towards the VDD/VSS ends of the plot, but it will not be wholly exclusive of the middle.

What, then, is the population-density-grouping threshold between "analog" and "digital"? 51%? (50% would be linear: half and half, so it can't be less than that by definition.) 90%? 99%? 99.99%?

(To be fair: 90 or 95% is probably the practical threshold. It's a reductio ab adsurdum sort of argument, but it's one with a convenient end-run. Such is the difference between abstract math and real engineering.)

Tim

```--
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Contract Design ```
• posted

I see the issue as "design intent". Is the device intended to be operated in a set of discrete states or along a continuum. Class A and D amplifiers have obviously different goals for their means of operation -- despite the fact that the "end result" (Vout) is intended to be similar.

• posted

It's only a continuum if you don't look closely enough. Current is discrete electrons after all. So it's sort of the ultimate digital!

```--

Rick C```
• posted

Electrons are discrete, sure, but it doesn't follow that current is so too.

Jeroen Belleman

• posted

Current is only "discrete electrons" if you stop them from interacting.

Even in valves you get space charge effects if you put two electrons close enough together. Saurbrey found a photo-multiplier non-linearity that popped up when you had two photo-electrons traveling from the photocathode to first dynode at the same time

```--
Bill Sloman, Sydney```
• posted

Eh? How does that work? Current is charge flowing past a point over time. If the charge is discrete, how is the current not?

```--

Rick C```
• posted

How shall I put this? You can detect the presence of charge only by the field surrounding it. The field can have any value, depending on the distance to the charge. A change of field is also a current, Maxwell's displacement current.

In other words, in your picture above, you'd detect a current by merely pushing the electron a little closer to the point at which you measure.

It's only when charges have to hop over some sort of barrier that their discrete nature becomes apparent.

Jeroen Belleman

• posted

So the field will increase and decrease as electrons move past. That sounds pretty digital to me.

```--

Rick C```
• posted

It isn't.

```--
Bill Sloman, Sydney```
• posted

That's like saying the tides are digital since there is only one moon :)

```--

John Devereux```
• posted

That would be true for a beam of electrons in a vacuum, yes, but for a current in a piece of wire, the conduction electrons will space themselves such that it's impossible to tell them apart. Read up on "shot noise".

Jeroen Belleman

• posted

Even in that case, you measure a continuous current during the time of flight of the electrons.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted

I don't think so, or these Schottky and stochastic cooling pick-ups we have in our accelerators wouldn't work. Yet work they do.

OK, these particles are moving at a fair fraction of velocity c.

Jeroen Belleman

• posted

The field is nonuniform, but the displacement current between end points is continuous. I built a laser-produced-plasma experiment many years ago that had two conducting grids with a kilovolt or so between them.

Current started flowing as soon as the plasma formed, not when it reached the grids.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant ```

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.