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According to the datasheet, a 2N2222 can handle VCE(max)VCE(max) up to 40 V40 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 V15 V as VCC, but after that, the transistor gets turned on, and it starts burning at 22 V22 V. I bought my transistor from markets in Shenzhen. The full label is
I just connected the RBRB to the ground to ensure that the transistor is off.
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asked Nov 3 at 9:20
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- 2You need to provide more details like PCB layout, full schematic, photographs, power supply type and surge resilience, anti-static precautions in place and used etc.. – Andy aka Nov 3 at 9:49
- 1I appreciate your comment. I just assembled them on the breadboard, and the power supply is a variable linear 0-30 volt 3A power supply(MEGATEK-MP-3003). – John Jin Nov 3 at 10:19
- 3Immediately fail, or fail after a specific period or number of operations? Is that a switch on the transistor base? – HandyHowie Nov 3 at 11:09
- 1I appreciate all your comments. My transistor starts burning after turning on the power supply. 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 bought my transistor from markets in Shenzhen. The full label is “2N2222A.”. I just connected the Rb to the ground to ensure that the transistor is off. – John Jin Nov 4 at 4:06
- 1Does the label list a manufacturer? – mkeith Nov 5 at 17:12
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The only possible answer is: you don’t have that circuit.
So one of the following must be true:
- Your transistor isn’t actually 2N2222,
- Your circuit is not wired as you think,
- Your voltage isn’t 22 V,
- You have something (unsuitable) connected at output,
- 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, Vout 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.
Another possibility is that your transistor has a different pinout than you’re expecting, such as the P-variant:
Which would mean you actually have the following, which has VEB of 22 V, where the limit is 6 V (see datasheet portion below).
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.
answered Nov 3 at 10:30
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- 5That evil PN2222 was what came to my mind as well. – Janka Nov 3 at 13:50
- 1This is why I always physically test the pinout even when I think I know which one it is. – Ian Bland Nov 3 at 17:52
- 1@IanBland Good suggestion, I added a section on that. – jonathanjo Nov 3 at 18:56
- 2Dear Jonathanjo, thank you very much for your answer. I will recheck the transistor to assure myself about the circuit behavior. I think that my transistor is a counterfeit part, but I will check your answer step by step and let you know. – John Jin Nov 4 at 4:11
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.
answered Nov 3 at 13:46
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- If I’m not mistaken, reverse browndown minimum is given as 6V. Do you know if the transistor is likely to fail with CE open? BE open? – jonathanjo Nov 4 at 7:40
- 1This wouldn’t happen with BE open, since that’s the junction conducting and heating in this scenario. It could still happen with C disconnected altogether. The lower the reverse breakdown voltage, the less power is dissipated per amp, and the less likely damage is, but I think 6V is atypical. Most devices will break down at around 9V, as far as I know, but I haven’t measured any to know what the distribution is. – Simon Fitch Nov 4 at 12:46
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:
- The transistor is not fake and actually meets its specifications.
- The transistor actually has the pinout you think it does. Check it with the diode function on a multimeter!
- 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:
- 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. 9×4=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.
answered Nov 4 at 20:52
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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.)
answered Nov 4 at 20:00
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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 V700 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.
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answered Nov 3 at 20:21
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- 3Dear David, Thank you very much for your explanation. My load is a simple resistor, as I mentioned in the main question. – John Jin Nov 4 at 4:21
- 3The OP showed the circuit in the question. There is no reason the transistor should be turning on at all since the base is pulled low with a 1 k resistor. – mkeith Nov 5 at 17:16
continued from comments above
/ Update 2022nov16hkt1313
I am also going to test the SOT23 version MMBT2222A. I got it also from ShenZhen (same as the OP). This 16 year old shop has 3 Crowns reputation, so should be most reliable.
The Fairchild MMBT2222A datasheet says that for pulse or low duty cycle operation, the absolute maximum value would be different from the specification, so I will be using a PWM signal with different frequency and duty cycle to verify that. I will be using the manual/UART controlled XY-PWM signal generator as the test signal.
I will also try the Fairchild SOT23 MMBT2222A.
Now I am thinking of monitoring the temperature rise of the 2N2222 in “wrong” pinout/configuration, and increasing the Vce voltage from 5V to 24V, and see if the 2N2222 burns out, as reported by the OP.
I am going to test the Malaysia STMicroelectronics 2N2222A below.
answered Nov 15 at 2:47
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- 3Can you answer the question? Do the experiments. Take notes (locally, not here.) Post a summary of the results with a short description of how you got the results. You’ve got a long bunch of “I’m going to” and nothing actually done. You’re writing an answer to a question, not a thesis for a PhD exam. – JRE Nov 15 at 8:18
- Good cooking takes time. If you are made to wait, it is to serve you better, and to please you. – Menu of Restaurant Antoine, New Orleans academia.edu/43166378/The_Mythical_Man_Month – tlfong01 Nov 16 at 10:37
- 2Yes, but if you are in a restaurant and the chef came out and told you “We are going to get a kettle and a big propane burner to make a crawfish boil. Then we will see where we can get a 100 pound sack of crawfish.” then you would be rather disappointed because you won’t be getting your crawfish for lunch today. Do it. Get it all together. Post it in one clean, complete shot instead of dribbling it out over days. – JRE 2 days ago
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