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...).

 

Do people actually test transistors now days ?
Esspecialy when SMT , they seem to just get lost !
Another dying art .

 

Well….
People are still testing vacuum tubes!
And there a few very sophisticated ones out there, that plot several curves and calculate Gm, cutoff, etc.

 

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.

Hi,

[1] Look at the part number, look up the data sheet. This is preferred unless you can't see the part number on the transistor package.

[2] Number the leads 1 and 2 (and the base B). Bias it as if #1 was the emitter and #2 to a 10k resistor to +5v, check if it 'turns on'. If not, bias it as if #2 was the emitter, check if it turns on. Don't use voltages over 5v because if you hook it up backwards, one of the junctions could become damaged and the current gain can go way down. A lot of emitter base junctions only take up to 5v reverse bias. Since you know it is NPN, you know what polarities to use so you have some advantage which you don't have if you don't know if it is NPN or PNP. I am assuming a small transistor, so if it is a power transistor you have to test with somewhat higher current levels.

[3] If it is a small transistor, stick the leads into a meter that has an NPN test socket. Switch leads until you see a decent current gain. You can probably tell if it is NPN or PNP that way too. Some meters have an NPN socket and a PNP socket.

 

The beginning premise was no data sheet, it's an unknown part, how do you tell which is the collector and which is the emitter? The only thing that was specified is that it is a BJT. So feed 1 µa to the base and presumed emitter and measure the gain?

 

The beginning premise was no data sheet, it's an unknown part, how do you tell which is the collector and which is the emitter? The only thing that was specified is that it is a BJT. So feed 1 µa to the base and presumed emitter and measure the gain?

If your measuring the gain, I'd probably use more current, say 10 µA into the base (if it's an NPN, out of if it's PNP). To prevent breaking down the base-emitter junction, keep the supply voltage below 5 V (like perhaps 3.3 V). That would let you use a base resistor of around 270 kΩ. Use a collector resistor that will saturate the transistor if the gain is more than about 100, which would work out to about 2.7 kΩ. Then, after identifying the base, try it in both directions. One should saturation and the other will probably fall well short. The orientation that saturates is the correct one.

If you are doing it by detecting breakdown, then keep the current in the 1 µA to 10 µA range and measure the reverse voltage. Be sure to use a voltmeter with a high input impedance (the 10 MΩ of most DMM's should be sufficient). For this, you need to use a higher voltage supply since you are intentionally causing reverse breakdown.

 

The beginning premise was no data sheet, it's an unknown part, how do you tell which is the collector and which is the emitter? The only thing that was specified is that it is a BJT. So feed 1 µa to the base and presumed emitter and measure the gain?

Hi,

Here are two simple circuits that shows how you can test this. This is a basic "switch" test to see how well the transistor works as a simple switch.
The one on the left has the right connections, the one on the right has emitter and collector reversed.
Note that in the circuit on the left we call the upper terminal the collector, and in the circuit on the right we also call the upper terminal the collector even though it is the physical emitter. That's because we are assuming it is the collector when we go to test it. After the test, we should have the terminals properly labeled.
Also note the supply voltage is 5v so that we do not risk reverse biasing the base emitter with a higher voltage which can damage the transistor. You can use 4.0v for example, but don't use 5.5v for example.

You can quickly see that the one on the left has collector voltage less than 200mv, while the one on the right has 'collector' (which is the real emitter) voltage almost equal to the supply voltage of 5v. The one on the left acts like a transistor switch, the one on the right does not do anything significant.
You can also see the current levels are much less in the circuit on the right and that's the one that is connected backwards. In a real device in a real circuit you may see current levels higher than that, but they will always be less than those in the left circuit which is connected correctly. The spice models may not show the true operation when the emitter and collector are swapped, but when the transistor is not connected properly the 'collector' (physically the real emitter) current will still always be lower than when it is connected properly.

Once you see the switch behavior, you can go on to change the base resistor in order to check the gain if you want to do that too. If you want to use a potentiometer for the 100k resistor, you should probably also connect a 10k resistor in series with it so the base current cannot go too high when the pot shaft is turned.

The main reason we see a difference when we reverse the terminals is because the transistor doping is asymmetrical. The emitter is more heavily doped than the collector.

 

[...] can cause permanent h_FE degradation.

That's what I came here to say. I saw in a video that there's a random noise generator that uses reverse Vbe voltage and it's very easy to cause this degradation.

 

The beginning premise was no data sheet, it's an unknown part, how do you tell which is the collector and which is the emitter? The only thing that was specified is that it is a BJT. So feed 1 µa to the base and presumed emitter and measure the gain?

That's why I have this. The clips are very easy to test components that doesn't fit.

 

i used to test them like that using analog meter 50 years ago

technology is awesome, it is remarkable what few parts can do nowadays.
it also makes things so convenient so that fewer and fewer actually know what is behind it.
the old method still works, heck even without AVO meter, one can do that by fashioning galvanometer using nothing but some fine wire and a battery.
needle, piece of cork and glass of water are optional.

 

View attachment 371889
That's why I have this. The clips are very easy to test components that doesn't fit.

Where did you get that?

 

Where did you get that?

Lol, AliExpress is full of stuff like that.
Link here. Also, it measures the ESR of components. It shows the current the LED reached at what voltage, it's just a point, but it's better than multimeter (don't know about precision).
Also, you might want to check out this. It's a voltage breakdown tester. It goes to the current you set, and measure the voltage, if you allowed to reach to that point. You can easily discover the maximum VCE voltage without burning the transistor down. But beware, because it's high voltage and 5 mA! It makes a high pitch noise (switching) when you press the button to test.

 

technology is awesome, it is remarkable what few parts can do nowadays.
it also makes things so convenient so that fewer and fewer actually know what is behind it.
the old method still works, heck even without AVO meter, one can do that by fashioning galvanometer using nothing but some fine wire and a battery.
needle, piece of cork and glass of water are optional.

Instead of the cork and glass of water, I stuck the needle through a piece of rubber band and suspended the rubber band in the middle of the coil.