In the last few days I spent working with NPN transistors and measuring their characteristics. My aim was measure VBE and IE when transistor will turn on (transistor as switch). For all tests VCE = 3.3V.

I started with S9013. I got data like these:

VBE=0.5V ; IE = 0.007mA
VBE=0.6V ; IE = 0.05mA
VBE=0.8V ; IE = 100mA

(there is a jump from 0.6V to 0.8V when transistor is turned on)

When I tried to measure the same for BC546B, I got:

VBE = 0.5V ; IE = 0.004mA
VBE = 0.6V ; IE = 0.17mA

around VBE=0.68V, the transistor turned off and VBE started to jump around from negative to positive values.

According to a datasheet VBE(SAT)=0.7V; VBE(ON)=0.55-0.7V.

Could someone explain me what happened? The transistor was new. I wouldn't expect that it's broken.

Update. Answer for questions in the comments: There isn't any resistor to limit current via led diode. I considered resistance of diode as limiting factor for current. Transistor case wasn't hot. The current was too small during measuring.

Schema how I tested it

  • \$\begingroup\$ AFAIK, transistor datasheets are averages/ideal, not 100% exact. As most stats are given as Min/Typical/Max, there is variation on individual parts/batches. \$\endgroup\$
    – Passerby
    Oct 29, 2013 at 2:24
  • \$\begingroup\$ "VBE started to jump around from negative to positive values." --- When I do this kind of test I use a common-emitter setup (driving a LED via a resistor) which can never have negative VBE. What circuit are you using? \$\endgroup\$ Oct 29, 2013 at 10:22
  • \$\begingroup\$ How were you limiting the collector current? What was the current limit on the 3.3V supply? Was the transistor hot when it "turned off"? \$\endgroup\$
    – user16324
    Oct 29, 2013 at 10:23
  • 3
    \$\begingroup\$ "I considered resistance of diode as limiting factor for current" - you should not think of diodes as having resistance. \$\endgroup\$
    – John U
    Oct 29, 2013 at 11:56
  • \$\begingroup\$ Can you take some pictures of your setting? Are the wires long? \$\endgroup\$
    – johnfound
    Oct 29, 2013 at 12:16

3 Answers 3


From your question, you've only shown 2 data points for your BC546B transistor. If you look at the datasheet, here are the electrical characteristics, specifically the base-emitter on voltage

enter image description here

So the minimum is 0.55, and the maximum is 0.7 to 0.77. The manufacturer's datasheet guarantees your transistor will be somewhere within that range.

Right now, you've only shown us 0.5V and 0.6V for Vbe.

I'd also recommend measuring the base current, Ib.


I think you should replace this diode by a resistor whose value will let the transistor have a high current (near maximum supported current by the transistor). This way you can protect your transistor form over current (damage) and you still can observe what you want. You will still notice in what point it will jump to this maximum current by changing Vbe.

Since the diode is a non ohmic element, I think is harder to see what you want. You do not have a V x I linear relation. Note that the second transistor works with higher currents than the first one. This way the voltage accross de diode will also vary in a different way.


As Passerby mentioned, Data sheets are never ideal. The only information that I generally extract from them is maximum ratings. I have come across data sheets for GaAs RF Fets where the scattering parameter plot was completely wrong (Becomes capacitive with frequency).

If you intend to use it as a switch, why measure its knee point? I imagine it would be driven with a 3V or 5V MCU source or soem kind of "on/off" arrangement on the base. If this is the case, then the knee point is irrelevant as you ideally should saturate it.

In General, I try to use FETs as switches. Their high gate impedance is ideal for minimal drive requirements and with a low Qg they can be very fast and delvier significant current to a load.


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