2N2222 burning out in the lab at 22 V

Question

According to the datasheet, a 2N2222 can handle \$V_{CE(max)}\$ up to \$40\ V\$. However, in the lab, I've seen different 2N2222s fail (get turned on) and burn with the below circuit. I need to know why this happens.

My transistor starts burning after turning on the power supply. As I checked the transistor, it works fine up to \$15\ V\$ as VCC, but after that, the transistor gets turned on, and it starts burning at \$22\ V\$. I bought my transistor from markets in Shenzhen. The full label is 2N2222A.

I just connected the \$R_{B}\$ to the ground to ensure that the transistor is off.

Answer 1

The only possible answer is: you don't have that circuit.

So one of the following must be true:

1. Your transistor isn't actually 2N2222,
2. Your circuit is not wired as you think,
3. Your voltage isn't 22 V,
4. You have something (unsuitable) connected at output,
5. Faulty test equipment is leading you completely astray.

Obviously the easiest things to check are the wiring, the resistor values, and the voltage. Comments have made the suggestion that you have a counterfeit part -- certainly a possibility. Also of course if the test equipment is faulty, V_out could be something else. If you're getting surprising results: always check the test equipment. In this particular case, given that the transistor is burning up, it's pretty evident something is very wrong.

The first thing to check is the pinout, both from datasheet for the exact part number, and by confirming on the actual device.

Variant Pinout

Another possibility is that your transistor has a different pinout than you're expecting, such as the P-variant:

^{From Wikipedia}

Which would mean you actually have the following, which has V_EB of 22 V, where the limit is 6 V (see datasheet portion below).

^{simulate this circuit – Schematic created using CircuitLab}

Identifying Pins

The classic method for identifying the pins of an out-of-circuit NPN transistor is to measure the voltage drop with the diode-mode of a multimeter. You expect approximately 0.6 V measuring from B to C and from B to E, and OL in all the other pairings. The one with the slightly larger drop is the emitter. The 2N3904 I just measured gave 0.664 V and 0.684 V. (Sorry, no 2N2222 to hand.)

I recommend you try this on your parts which are burning up and see if you can confirm the pinout.

Datasheet Fragment

^{From datasheet}

Answer 2

One possible explanation: the 2N2222 is most famous for having two different pinouts, as shown in this Wikipedia article.

As a result, it's often connected with emitter and collector reversed, and in such a configuration the base emitter junction can easily become reverse biased beyond its reverse breakdown voltage (which is only 10V or so).

That junction conducts well, and quickly overheats. The transistor is toast before you even realise you bought one of the upside-down versions.

Answer 3

According to the datasheet, a 2N2222 can handle VCE(max) up to 40 V. However, in the lab, I've seen different 2N2222s fail (get turned on) and burn with this circuit

the power supply is a variable linear 0-30 volt 3A power supply (MEGATEK-MP-3003)

My transistor starts burning after turning on the power supply

One possibility then is that the power supply is pushing way more than 40V into that transistor as it is turned on.

Use a digital oscilloscope to capture the transient that occurs then the power supply is turned on using the power switch. Some poorly designed supplies wildly overshoot the output voltage when the mains power is turned on.

This will be the culprit, assuming that:

1. The transistor is not fake and actually meets its specifications.

2. The transistor actually has the pinout you think it does. Check it with the diode function on a multimeter!

3. Verify that the emitter and collector are where you think they should be by measuring the current gain when the transistor is on. Use a the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

4. The circuit, wired as shown in the question, is checked with voltage measurements with a multimeter when supplying say 12V to the circuit. Base to emitter should be 0V. Collector to emitter should be 12V. The voltage across the collector resistor should be 0.000V.

To completely eliminate the power supply as the potential source of trouble, use 9V batteries in series as a power source instead. You can use crocodile jumpers to connect them in series. 9x4=36V - the transistor should handle that without a problem.

I have assembled your circuit using known-good 2N2222 transistors and nothing happens when the power is turned on: the transistor remains off.

Answer 4

As I checked the transistor, it works fine up to 15 as VCC, but after that, the transistor gets turned on, and it starts burning at 22V.

I made simulations and some measurements with a real 2N2222 transistor.

Then let's see what happens at 15 V VCC voltage. If we do not connect the collector, it will be an 8.8 V Zener diode. By connecting the collector, it becomes a reverse-biased transistor, but it can handle the 25 mA collector current indefinitely (250 mW). It's getting a little warm.

At a supply voltage of 24 V, the Zener current increases to 12.5 mA, but this does not destroy the transistor. It works continuously and gets a little warm. When we connect the collector here, the transistor still works, only the higher collector current heats up the transistor very quickly. It will last a few seconds until we can measure currents. If it is turned off in time, the transistor remains functional. After about 10 seconds it overheats and breaks down (700 mW).

The actual measured values were: UB = 980 mV, IC = 86 mA.

I sometimes use reverse biased base-emitter junction as a low capacitance 9 V Zener if I need one. (For the simulation, I modified the transistor, connected a diode and a Zener between the base and emitter.)

Answer 5

Nobody has asked you what kind of load you are switching, although since it isn’t shown, it is another aspect of the diagram not representing the actual circuit you are operating.

If you are switching an inductive load, when you turn off the transistor, the current in the inductor cannot immediately change. Since the switch is now high impedance, the current which the inductor is trying to maintain can create a very high voltage at the collector of the transistor, leading to breakdown and eventually “burning” the transistor. For example if the load is the coil of a small signaling relay, it’s possible for the collector to see \$700\ V\$ or more at the peak of the inductive “kickback”

The solution is to place a snubber diode from the collector to V+ oriented with the cathode at the collector, so that any voltage higher than V+ present on the collector will be conducted to the V+ supply. This will take the energy stored in the inductor and return it to your power rail instead of dissipating it through the transistor.

If your load is resistive or capacitive, this would not apply.