There are plenty of quotes on the internet like this one:

"There is nothing like 'analog transistor' and 'digital transistor'. A transistor is a transistor! Simple. One can make either analog circuits or digital circuits out it by making the transistor operate in either linear (active) region or between on-and-off (saturation-and-cutoff) regions respectively. "

Can transistors be changed from operating in active vs saturation-and-cutoff region, after they have been manufactured, and sold. Or, is this mode of operation something that has to be built into the transistor, by balancing N and P regions and whatnot. (If the latter, it sure does seem to me it must be reasonable to refer to things as "active" or "saturation-and-cutoff" transistors, or, with a synonym, analog and digital. But it seems to upset many, so I ask if these modes can be switched between easily after transistors reach market. )

  • \$\begingroup\$ You can set the operating region with base current (BJTs) / gate voltage (FETs). Regarding manufacturing process, some transistor are more suitable for switching (saturation point) some for amplification (active region)(good linearity). \$\endgroup\$
    – user208862
    Nov 17, 2021 at 3:17
  • \$\begingroup\$ The mode of operation is set by the circuit around the transistor, not anything to do with the manufacturing process. That's what the quote you quoted is trying to say: the circuitry determines how the transistor acts. \$\endgroup\$
    – Hearth
    Nov 17, 2021 at 5:06
  • \$\begingroup\$ @Michael P., the operating region is NOT set with base current. Rather, it is set by base voltage (both, the B-E and the B-C junction must be forward biased). In this case, the large base current is an INDICATION for saturation only (not its cause). \$\endgroup\$
    – LvW
    Nov 17, 2021 at 8:39
  • \$\begingroup\$ Neither. Every transistor (be it bipolar or FET) can operate either way. Take an active region transistor amp and overdrive its input hard enough : it now alternates between saturated and off. That's all there is to it. \$\endgroup\$ Nov 17, 2021 at 14:45
  • \$\begingroup\$ @Hearth I understand what the quote means, if transistors in general, incl. in a analog components like diff amp and op amp, reach saturation at 0.7 volt. And something to step down voltage or current is needed to use inputs above 0.7v. But that seems kind of inefficient to me. Not how I would expect it would be organized, so I asked. \$\endgroup\$
    – Doge
    Nov 17, 2021 at 19:12

2 Answers 2


That quote is not wrong, but that doesn't mean there aren't transistors specifically designed for a particular application, switching or linear. It is true that any transistor can be operated in any of the three regions, saturation, cut-off or active/linear, but some transistors are better for switching applications than others.

In a switching application your concern is mainly with the time it takes for a transistor to transit from saturation, through the active region, into cut-off, or vice versa. All transistors can make that transition, but some transistors do it more quickly.

In a linear application, such as an audio amplifier, you never really drive transistors into saturation, or completely into cut-off, and this transition time is of little consequence. You may be more concerned with the linearity of the transistor in its active region, where it spends 100% of its time.

Transistors can't recover instantly from saturation, because it takes time for charge carriers in the base region to vacate, and return the transistor to linear operation, or cut-off. The amount of time it takes for such recovery is a function of junction geometry and doping, and where recovery time is critical you could choose a transistor with better recovery time than others, but you could also mitigate recovery delay by simply designing the circuit in such a way that the transistor is never driven deeply into saturation.

Similarly, if the base-emitter junction is reverse biased, and the transistor is deeply into cut-off, it takes time to repopulate the base-emitter depletion region with charge, and restore it to an active-region state. Again, the solution could be to use a transistor designed to recover from this condition quickly, or simply ensure that the base-emitter junction potential difference is never permitted to become negative, or even approach zero.

So, while there are transistors designed with fast recovery in mind, for switching applications, all of them have the same saturation/active/cut-off states, and they all suffer from the same recovery issues.

In all cases it is a transistor's periphery, those components responsible for biasing and source/load impedance, which are defining whether the transistor is a switch or a linear device, not the transistor itself.

Here are some articles showing the transistor in two different configurations, with wildy differing bahviour, hopefully helping to show why it's not the transistor itself which is deciding its role, so much as the components around it:

Common-emitter (allaboutcircuits.com)

Common-emitter (Wikipedia)

Common-collector (allaboutcircuits.com)

Common-collector (Wikipedia)

These principles also apply to field-effect transistors (where they would be called "common drain" or "common source"). In common-collector configurations, the transistor is always operating in its active/linear region, but the common emitter approach is more suited to switching.

  • \$\begingroup\$ My main concern is, in a diff amp that can take inputs of many volt, and I see no circuitry in schematic that reduces voltage, and digital circuits usually apply 0.7 volt to transistor base to reach saturation, it seems they must use transistors designed and manufactured for different input voltage ranges. Would you say that isn’t the case? \$\endgroup\$
    – Doge
    Nov 17, 2021 at 19:17
  • \$\begingroup\$ @Doge It's not the voltage referenced to ground that matters, that can be anything it wants to be. It's the voltage between the base and the emitter that matters. You'll notice that a diff amp never has the emitter grounded; there's a current source there instead. This adjusts the emitter voltage so that Vbe stays in its normal range of 0.6~0.7 volts. \$\endgroup\$
    – Hearth
    Nov 17, 2021 at 20:41
  • \$\begingroup\$ @Hearth if 1) transistors and response curve is the same and 2) only circuitry differs and 3) what you wrote there is true, I really want to understand it. Because if I understand diff amp, it opens up a lot (basis of so much else. ) But I don't understand what you mean. \$\endgroup\$
    – Doge
    Nov 17, 2021 at 21:02
  • \$\begingroup\$ @Hearth with not grounded, you mean there is resistor in between it and ground? Like they show here, forum.allaboutcircuits.com/threads/…. I can get that, but don't get why that makes arbitrary max volt input at base (say, 0.7x10 = 7 volt) act as 0.7 volt. \$\endgroup\$
    – Doge
    Nov 17, 2021 at 21:07
  • 1
    \$\begingroup\$ @Hearth I think I will find that I have it answered with that. But have to think about it a while. Thank you a lot, it would have been really hard for me to figure that out on my own. And understanding the diff pair seemed like a good stepping stone or exercise. \$\endgroup\$
    – Doge
    Nov 18, 2021 at 1:43

Using a transistor in an analog or digital mode is an application of the transistor. It can fulfill either application to varying degrees depending on its architecture. Read something basic such as this.

enter image description here

The architecture of transistor will set the performance of the device for any given application. For example RF transistors are quite different to small signal and power transistors as they are optimized for RF applications. This is why there are many types of transistor, they are purpose designed. For example transistors that are used in applications requiring low VCE(sat) are designed to provide the lowest possible VCE(sat). Here's an OnSemi power device, the NSS40500UW3T2G, they don't explain the architecture of the transistor, but clearly the VCE(sat) is extremely low at currents up to 4A.

  • \$\begingroup\$ I understand transistors can fullfil both depending on their architecture. You should be able to read that in from the question. What it asks is if the optimization for different modes is something that is done after fabrication, or during. And in the latter case, it must be possible to understand that calling a transistor ”digital-optimized”or ”analog optimized” or optimized for whatever, is not unreasonable at all. \$\endgroup\$
    – Doge
    Nov 17, 2021 at 19:00
  • \$\begingroup\$ @Doge, that's exactly what I said. Transistors are optimized for particular applications, but in general they all have flexible operation in any of modes you describe. \$\endgroup\$ Nov 18, 2021 at 1:07
  • \$\begingroup\$ That they can all work in either saturation-and-cutoff or active mode, is self-evident. Bu that they would do this over different voltage ranges, and this is optimized at fabrication, was more what I was asking about. Most answers or comments seem to imply "no, transistors used in analogue and digital circuits have the same voltage range" and if that is true then ok. It seemed contradictory to me, because it seemed efficient to be able to use gate with more than 0.7 volt saturation, but I am a beginner so might be true. \$\endgroup\$
    – Doge
    Nov 18, 2021 at 1:19

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