I am just beginning to learn basic electronics and I have come across the "Meter Check of Transistor" topic. I am confused on how to identify the collector & emitter leads. I have no problem identifying the base lead. The topic simply said this is the collector and this is the emitter as per some meter voltage readings. The value were 0.655v for the E-B and 0.621v for the C-B and I assume that these are voltage drop values. I am not sure how it came up of finding out the emitter and collector leads. There is a bit of disparity in the voltage values and what I am thinking is that because the emitter is heavily doped (I believe improved conductivity. . i could be wrong), then voltage drop in the E-B junction should be lower (which I think is 0.621 and not 0.655). With this my guess is the Emitter is lead 2 and collector is lead 1. PLease enlighten. .. thanks a lot

 

Edit: Ok, never mind. Look here: http://howtoblogpost.blogspot.com/2011/05/how-to-identify-terminals.html





Step 1.
Is it NPN or is it PNP?

Step 2.
What is the package of the transistor?

 

Edit: Ok, never mind. Look here: http://howtoblogpost.blogspot.com/2011/05/how-to-identify-terminals.html





Step 1.
Is it NPN or is it PNP?

Step 2.
What is the package of the transistor?

PNP. . .I am just reading the topic "Meter Check of Transistor" in "All About Circuits" . . the example transistor has no data at all . .and I am trying to learn how to do a meter check. . here is the link http://www.allaboutcircuits.com/vol_3/chpt_4/3.html

 

I am just beginning to learn basic electronics and I have come across the "Meter Check of Transistor" topic. I am confused on how to identify the collector & emitter leads. I have no problem identifying the base lead. The topic simply said this is the collector and this is the emitter as per some meter voltage readings. The value were 0.655v for the E-B and 0.621v for the C-B and I assume that these are voltage drop values. I am not sure how it came up of finding out the emitter and collector leads. There is a bit of disparity in the voltage values and what I am thinking is that because the emitter is heavily doped (I believe improved conductivity. . i could be wrong), then voltage drop in the E-B junction should be lower (which I think is 0.621 and not 0.655). With this my guess is the Emitter is lead 2 and collector is lead 1. PLease enlighten. .. thanks a lot

Congratulations! You've found the base!

Now take the meter leads & place them on the other two leads.
Next comes the digusting bit!

Spit on your finger & place it between the base & each of the other legs of the transistor,in turn.

If it is an NPN,you will see a reading on the meter when you connect the base to the +ve lead (via your finger),& if it is a PNP,it will appear when you connect the base to the -ve lead.

For simplicity,let's assume an NPN for the following:
Write down the meter reading.

If you now reverse the meter leads & do the trick again,you will find that the meter reading you get connecting the base to the +ve lead reads either much lower or much higher than the previous reading--- the lower of the two readings is the collector.

What you are doing is biasing the transistor "on",so it conducts,& you get a meter reading.
Transistors will work backwards,but not very well,so you get a higher current,hence a lower meter reading when you find the collector.

This test has been around since the 1960s,so it is amazing it isn't better known.
It works very well with small signal BJTs,& still reasonablty well with power devices.

 

Congratulations! You've found the base!

Now take the meter leads & place them on the other two leads.
Next comes the digusting bit!

Spit on your finger & place it between the base & each of the other legs of the transistor,in turn.

If it is an NPN,you will see a reading on the meter when you connect the base to the +ve lead (via your finger),& if it is a PNP,it will appear when you connect the base to the -ve lead.

For simplicity,let's assume an NPN for the following:
Write down the meter reading.

If you now reverse the meter leads & do the trick again,you will find that the meter reading you get connecting the base to the +ve lead reads either much lower or much higher than the previous reading--- the lower of the two readings is the collector.

What you are doing is biasing the transistor "on",so it conducts,& you get a meter reading.
Transistors will work backwards,but not very well,so you get a higher meter reading when you find the collector.

This test has been around since the 1960s,so it is amazing it isn't better known.
It works very well with small signal BJTs,& still reasonablty well with power devices.

great. .. thank you so much