Here's an interesting problem. Someone hands you a part. It you don't know what it is, but it's has three leads. He tell they tell you it's a B J T. Checking the diode characteristics tells you it's an NPN. So which lead is the collector and which lead is the emitter?

I used a speech to text converter on this, so I was busy cleaning it up if you happen to see it in between state.

 

So which lead is the collector and which lead is the emitter?

Мeasure forward voltages V_B-X and V_B-Y.
Smaller voltage is V_B-C.

 

A 3 lead component (with two junctions) can be a dual diode also.

 

Мeasure forward voltages V_B-X and V_B-Y.
Smaller voltage is V_B-C.

That method is not very reliable. The two forward voltages will probably be pretty close together and minor differences in currents can swap the relative magnitudes. Even if the currents are the same, it is very possibly not definite.

I just took a 2n3904 and a BC337 and made some measurement that show this.

2n3904:
Current = 599 µA: Vbe = 697 mV, Vbc = 691 mV
Current = 5.85 mA: Vbe = 784 mV, Vbc = 776 mV

In both cases, the Vbc was smaller, but only my 8 mV

BC337:
Current = 598 µA: Vbe = 695 mV, Vbc = 696 mV
Current = 5.85 mA: Vbe = 776 mV, Vbc = 776 mV

Here, Vbc was smaller than Vbe by only 1 mV at low current and they were the same at the higher current.

A far better test is to just measure the h_FE. Since nearly all BJTs are highly asymmetric, the forward h_FE will usually be something like one or two orders of magnitude greater than the reverse, so you don't even need much of a setup to distinguish them.

For my little test, I used 4 AA batteries (V = 6.67 V) and put 100 kΩ resistor in series with the base, which should give me about 60 µA of base current. Assuming an h_FE of 300 (picking in the high side deliberately), my collector current would be 18 mA, so call it 20 mA, and I want that to put the transistor near saturation, so I need to drop about 6 V, which calls for 300 Ω. I'll use a 330 Ω collector resistor.

Whichever orientation produces the greater drop across the collector resistor is the forward orientation.

2n3904:
FWD: 4.45 V; ß ≈ 225
REV: 73 mV; ß ≈ 3.7

BC337:
FWD: 4.53 V; ß ≈ 229
REV: 991 mV; ß ≈ 50

I was a bit surprised by the high reverse h_FE for the BC337. But maybe it's not as asymmetrical since it's CB reverse breakdown voltage isn't quite as high. I'd need to test several more, and at some other currents, to start drawing any conclusions, though.

 

The way I would do it, But hey! That’s only me.
Take a 10 volt supply current limited to a very low current, say 2 mA.
Reverse bias each PN junction. The one which clamps the voltage to around 6.5 to 7 volt, that would be the BE junction. The BC junction would not clamp the voltage at all.
.

 

The way I would do it, But hey! That’s only me.
Take a 10 volt supply current limited to a very low current, say 2 mA.
Reverse bias each PN junction. The one which clamps the voltage to around 6.5 to 7 volt, that would be the BE junction. The BC junction would not clamp the voltage at all.
.

I'd recommend using something quite a bit smaller, such as 10 µA to 100 µA (which is what many manufacturers of small-signal transistors use for emitter-base breakdown voltage tests). A current of 2 mA, especially is left for very long, possibly even just a few seconds, can cause permanent h_FE degradation.

 

Very true, specially for small signal transistors.

 

with small base current compare gain. orientation with higher gain is correct. i used to test them like that using analog meter 50 years ago, using finger skin as base resistor (barely touching...).