I have some 2N4401 NPN transistors that I'm trying to use with a 12V supply. I saw some strange behavior at cutoff and reproduced it in this simplified circuit. (The base is not connected, in case the schematic is unclear.)

Circuit schematic

If I'm reading the 2N4401 datasheet right, the transistor should be well-behaved in this circuit as long as I keep \$ V_{CE} < 40 \mathrm{V} \$, \$ I_C < 600 \mathrm{mA} \$, and \$ P < 625 \mathrm{mW} \$. Therefore, I expect \$ I_C \$ to be reasonably close to zero, since there's no base current and the transistor is in cutoff.

However, that's not what I saw when I measured \$ V_{CE} \$ for a range of \$ V_Y \$ values from 0V to 32V (the limits of my bench power supply). Here's a graph I plotted from my measurements with three different values of \$ R_Y \$. (I calculated \$ I_C \$ from \$ V_Y \$ and \$ V_{CE} \$.)

Graph of I_C vs V_CE

As you can see, at around \$ V_{CE} = 9 \mathrm{V} \$, \$ I_C \$ suddenly goes up and hooks around to the left, which isn't at all what I expected from a transistor in cutoff. I expected (and want) \$ I_C \$ to be close to 0 for the entire range.

I never went near the absolute limits, so I don't think the transistor is damaged. My graph looks suspiciously similar to graphs I've found of collector-emitter breakdown, but the datasheet claims \$ V_{(BR)CEO} > 40 \mathrm{V} \$, and 9V is way below that. Why am I seeing this behavior?

Edit: I suspect I read the pinout wrong and swapped the collector and emitter when I tested this circuit. Even in that case, I'm still curious whether this behavior has a name.

  • 1
    \$\begingroup\$ In your circuit the base is floating. In the datasheet, the collector cutoff current is measured with Vbe = - 0.4 V. You could retry under those conditions. The other explanation could be that you had emitter and collector swapped like Spehro says. Your expectation that the collector cutoff current should be low, that expectation is correct. \$\endgroup\$
    – user57037
    Commented Mar 29, 2022 at 3:14
  • \$\begingroup\$ Great question, great plot. \$\endgroup\$ Commented Mar 29, 2022 at 15:10

2 Answers 2


Sounds like you have swapped emitter and collector.

You are seeing E-B breakdown. The negative resistance characteristic can actually be used to make an relaxation oscillator. It does apparently damage the transistor however, by reducing the forward beta permanently (some recovery is possible with annealing but that's not going to be possible in a plastic-packaged device).

  • 7
    \$\begingroup\$ Yep, I definitely put the transistor in upside-down! Oops :) \$\endgroup\$
    – ashtneoi
    Commented Mar 29, 2022 at 4:49

Even when you align the transistor properly (as explained in Spehro's answer): if you leave the base floating you will likely get a leakage current that is higher than normal.

This is because any leakage of the collector-base junction will then go on to flow from base to emitter, turning the transistor slightly on. Effectively, this amplifies leakage by the forward current gain of the transistor. Instead of maybe ~10 pA, you could see something high like ~1 nA.

To measure real cut-off leakage, tie the base to the emitter.


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