An NPN transistor can be made a switch by applying some power to the base.

Is there a limit to how fast I can switch the transistor on and off, or I can switch it on and off as fast as I can turn the signal to the base on and off?

As switching, I mean completely off and completely on. Not in between (if there is even any.)

  • \$\begingroup\$ Yes, there is in between too. It's called the linear region, and it's used for analog audio amps and many other applications. They're not particularly efficient that way, but sometimes it's needed. They tend to respond quicker that way too, if they're not allowed to saturate. \$\endgroup\$
    – AaronD
    Commented Jan 7, 2016 at 3:45
  • \$\begingroup\$ Great Answers on the question, I would like to add that Junction Capacitance also adds to switching time. \$\endgroup\$
    – ammar.cma
    Commented Jan 7, 2016 at 6:46
  • \$\begingroup\$ Completely on/off is typically referred to as Saturation in a transistor. \$\endgroup\$
    – Passerby
    Commented Jan 7, 2016 at 6:53
  • \$\begingroup\$ @Passerby : completely on is saturation, completely off is just off. \$\endgroup\$
    – user16324
    Commented Jan 7, 2016 at 12:40
  • \$\begingroup\$ See this Q&A for part of the story on how to turn a transistor OFF fast. electronics.stackexchange.com/questions/55073/… and a practical application here electronics.stackexchange.com/questions/62271/… \$\endgroup\$
    – user16324
    Commented Jan 7, 2016 at 12:42

3 Answers 3


The major limit to BJT switching time is related to the charge carriers and specifically how long it takes to move carriers into the base, and how long it takes to get them out.

The datasheet will include a few parameters that will give you the theoretical maximum switching frequency*. They are

  1. Delay time (td) - how long it takes to get out of cutoff
  2. Rise time (tr) - how long is the transition from cutoff to saturation
  3. Fall time (tf) - how long is the transition from saturation to cutoff
  4. Storage time (ts) - how long to get out of saturation

Using the datasheet (these parameters are usually listed), you can figure out how fast a transistor can switch between the two states.

$$ f_{max} = \frac{1}{t_d + t_r + t_f + t_s} $$

* This is what transistor can theoretically do, but there are tricks that can be done to improve the switching speed. Also, if you are switching a square wave, then in order to maintain a nice square waveform, the actual switching frequency will be much less.

  • 1
    \$\begingroup\$ How does this relate to the Gain-Bandwidth Product, or the Frequency Transition value listed in the Digi-Key Product Selector for a 2N3904, both listed as 300 MHz while the f\$_{max}\$ calculated with your formula above gives 3.125 MHz based on the Switching Characteristics listed in the datasheet? \$\endgroup\$
    – tcrosley
    Commented Jan 7, 2016 at 5:50
  • 3
    \$\begingroup\$ @tcrosley The 300MHz listed in the datasheet is for small signal. \$\endgroup\$
    – efox29
    Commented Jan 7, 2016 at 5:59
  • \$\begingroup\$ Is the Frequency-Transition number in the product selector the same number, just called different? \$\endgroup\$
    – tcrosley
    Commented Jan 7, 2016 at 11:46
  • 1
    \$\begingroup\$ The transition frequency (\$f_T\$) of a transistor is a characterization of its small-signal linear gain, which is very different from how it operates in a saturated switching mode. This answer is talking about the latter, not the former, since that's what the question is specifically about. \$\endgroup\$
    – Dave Tweed
    Commented Jan 7, 2016 at 17:08

When you try to switch a (bipolar -- NPN or PNP) transistor fast, there are some effects that limit the rate it can respond.

1) The physical base of the transistor extends over some (small) area, and the base lead is only connected to one portion of this. There is some internal resistance between the connection location and the most remote portions of the base. When you switch fast, the time for the charges at the remote portions to be extracted is significant and limits how fast it can turn off. This isn't as severe for turning on because you could get by (for a few ns) with just the local portion turning on.

2) There is a significant capacitance between the collector and base. When turning off, the collector voltage rises (if it didn't there would be no need to turn the transistor off). This couples via the base-collector capacitance and tends to counter the turn-off or -on voltage on the base, thereby making it harder to turn on or off.

3) It actually takes some time for the carriers to transition the base and collector regions.

However, bipolar transistors can be very fast -- they can be switched in picoseconds inside ICs. Discrete devices can't be switched as fast because of greater parasitic capacitances, inductance of the wires that attach to the device, and because the fastest devices have breakdowns of just 1 or 2 V, and so are really only useful inside specialized circuits in ICs.

  • \$\begingroup\$ Not really a problem outside of GHz ICs, but that internal resistance also means that some of the charge gets converted to heat, so if you try and switch a transistor too frequently, that can cause heat buildup. (This being one of the major constraints on CPU speeds) \$\endgroup\$ Commented Aug 17, 2020 at 12:24
  • \$\begingroup\$ That's not the specific reason CPUs heat up. The logic signals are toggling up and down, and they have capacitance -- this takes power to drive (½*C*V^2), and is independent of the resistance of the switch. A second source of power dissipation is the leakage current in the logic switches -- even when a MOS transistor is 'off', it leaks some current, and this non-negligible current in billions of transistors isa significant current. \$\endgroup\$
    – jp314
    Commented Feb 10, 2021 at 20:12

Yes, there is a limit to how fast you can switch the transistor on and off. However, it is dependent on which transistor you use, as well as the characteristics of the driving signal, the power supply, and the load.


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