I've stumbled onto something that I don't understand.
I've wired up a phase shift oscillator using a BC337 transistor (datasheet here). According to the datasheet, \$V_{BE(on)}=1.2\ V\$
However, using my multimeter (and after doing the theoretical calculations), I have found that my \$V_{BE}\$ is about 0.6V, which should apparently force the transistor into cutoff.
Here is my circuit:
simulate this circuit – Schematic created using CircuitLab
The oscillator is functioning, producing a slightly distorted sine wave of about 1.5 V peak-to-peak and at about 560 Hz. I measured \$I_E\$ to be around 7 mA, but couldn't measure \$I_B\$ without my multimeter ruining the output of the circuit.
As far as I know, this output should indicate that the transistor is in active mode, but if \$V_{BE}\$ is that small, shouldn't it be in cutoff?
Best Answer
The value in the datasheet is a maximum value, so any value lower than 1.2 V is within the specified limits.
Furthermore, it is specified with 300 mA collector current, where your circuit is requires only about 1 mA collector current, so it's not suprising that you measure Vbe well below the 1.2 V value.
In fact, there's a typical curve provided:
That shows that around 0.6 V is expected when the collector current is this low.