When transistor (bipolar transistor in common emitter amplifier) is closed (not conducting) or opened saturated (enough base current, collector current is limited by the resistance of the load), it dissipates the power well within its limits. However switching from opened to closed obviously goes over all intermediate values (it is probably not possible even theoretically to be otherwise, the question is just how fast).

Should I normally care that transistor would stay within its power limit at the intermediate values, if I truly believe that thermal issues are highly unlikely (controlled by digital input that has only two values and changes fast)?

There are no time based limits on the specification sheet. Are there any generic rules about this maybe?

  • 2
    \$\begingroup\$ What do you mean here: closed or opened saturated? We don't talk in terms of water valves in EE, so can you be clear which anology you refer to and are you restricting this question to MOSFETs by using the term "saturated" when tied to one specific condition of open or closedness. Open means open-circuit i.e. passing no current in EE circles. Be clear. And yes, you should calculate the dissipation when switching (usually) and definitely yes generically. \$\endgroup\$ – Andy aka Jan 30 at 18:52
  • 3
    \$\begingroup\$ In general, you can't ignore the power during switching. \$\endgroup\$ – Justin Jan 30 at 18:58
  • \$\begingroup\$ "Open" and "saturated" means in my context that transistor (bipolar transistor in common emitter schematics) has base current sufficiently large, so that further increasing it does not increase the collector current more. In other words, I_c < h21_e * I_b. In the language I have read most of the literature, closed transistor means not conducting, opened - conducting. \$\endgroup\$ – h22 Jan 30 at 18:59
  • \$\begingroup\$ If the transistor is in switching mode, then (usually) those transitions will account for the biggest dissipation. If it's just an electronic switch to turn on or off at various times, as long as the switching is not some low varying ramp, and the switched currents are within limits, you shouldn't worry about it, since typically even if it heats up a bit it has time to cool down until another turn on or off is needed. Of course, if this all sounds too vague and unreliable, is because the input has been equally vague. \$\endgroup\$ – a concerned citizen Jan 30 at 19:29
  • 1
    \$\begingroup\$ @h22 In English, open means not conducting and closed means conducting. For example, an open-circuit means the wiring loop has not been completed (closed, en-closed) so no current flows. \$\endgroup\$ – DKNguyen Jan 31 at 4:42

Decades ago I worked on a satellite project. Lotta good stuff to learn; most of my circuits worked first time, so they let me handle integration of the entire "box" we were building.

One of my circuits was a stripped-down version of the customer-suggested relay-driver schematics; they suggested an AC_coupled positive-feedback 2-transistor circuit. I thought "We don't need that."

Turns out the SLOW switching of 2N2222s at 28 volts and 0.2 amps (for latching relays) was killing the transistors. At 50% point, that's 14 volts and 0.1 amps, or 1.4 watts.

Why is that bad? The die is about 1mm cube; the thermal timeconstant of that is 11.4 milliSeconds (a cubic meter of silicon has 11,400 seconds thermal Tau). The thermal capacity of silicon is 1.6 picoJoules/cubic_micron per degree C. The thermal capacity is thus 1.6 picoJoules/cubic_micron * 1000 * 1000 * 1000 === 1.6 milliJoules per degree C.

The heat rise in a millisecond was 1.4 milliWatts / 1.6 milliJoules/degree C, or about 0.8 degree C.

Why was 2N2222 killed? All the heat was dissipated in the 10 micron thick collector implant, not in the 1,000 micron thick die (maybe only 300 micron, of a typical wafer, but does not matter). Our heat rise was not 0.8 degree C per millisecond, but 30x to 100X higher.

And any current imbalance in the emitter region will lead to thermal runaway.

Cure? We kludged the PCB to use the extra transistors and the positive feedback. The customer did not want to respin the PCB. To teach me a lesson.

I learned my lesson on Safe Operating Area.

That PCB/box is still in orbit today. Kludged, to avoid SOA failure.


The 2N2222 was really 2N2219, a large metal case device to dissipate 0.5 watt.

I made the mistake; I'd striped out the switch-the-device-quickly extra components. Thus the 10 micron thick active region (the collector implant) had to dissipate the 1.4 watts, briefly, and the device developed hot spots due to slight emitter variations.

By reducing the "briefly" from milliseconds to nanoseconds, by positive feedback, the switching event became non-destructive. I needed to learn about SOA.

The 2N22xx family has a thin base and achieves 100+MHz bandwidth at moderate current.

| improve this answer | |
  • \$\begingroup\$ Is using a 2N2222 to switch 28V at 0.2A an unusual use of that transistor? Or was it the conditions it was being used in that exasterbated things? \$\endgroup\$ – DKNguyen Jan 31 at 5:13

While turning on and off, your transistor produces so-called switching losses:

$$P_{SW} = f_{Sw}\cdot(E_{on} + E_{off})$$ $$f_{Sw}=\text{Switching frequency}$$ $$E_{on}=\text{Turn-on energy}$$ $$E_{off}=\text{Turn-off energy}$$

As you can see, the switching losses depend strongly on your switching frequency.

If they're relevant to your application or not depends on the ratio of switching losses to conduction losses.

| improve this answer | |

This figure from "Review of Switching Concepts and Power Semiconductor Devices" (no author) illustrates the power losses when a bipolar device is used as a switch: enter image description here

There are two considerations. One is that the average power, when multipied by the thermal resistance (junction-to-ambient if there is no heat sink, or junction-to-case if the heat sink temperature is known) gives the temperature rise of the device. This rise is added to the ambient temperature, and if the result exceeds the maximum junction temperature, the device will fail. The second is that the transistor remain in the Safe Operating Area (SOA), often plotted as a family of curves (operating temperature or pulse duration) for single non-repetitive pulses on a graph of collector current versus emitter-collector voltage. During the switching time, the transistor's simultaneous voltage and current must remain inside this region.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.