3

I am trying to do this experiment, but I can't read the current Ib. Note that the multimeter I use in this experiment is very sensitive. I don't know much about BJT transistors, so it might be an invalid transistor connection.

enter image description here

enter image description here

enter image description here

JRE
  • 71,321
  • 10
  • 107
  • 188
bleckemr
  • 33
  • 4
  • 2
    Do you know the "internal" impedance of your ammeter? In place of measuring "current" Ib, measure the voltage on Rb with an electronic voltmeter (of which you know the input impedance, do correction if needed against Rb). – Antonio51 May 12 '22 at 16:05
  • 1

    so it might be an invalid transistor connection. So, you are right ! +1.

     – Antonio51 May 12 '22 at 20:45

2 Answers2

4

Just looking at the schematic, the collector current will be a bit less than 2mA before correcting for drop across Rb. If that transistor is a 2N2222A hFE should be between 50 and 300 so base current would be around 7 to 40uA, before correction. So 20uA isn't really out of line.

Here's a simulation, result is 9uA Ib.

schematic

simulate this circuit – Schematic created using CircuitLab

Spehro Pefhany
  • 397,265
  • 22
  • 337
  • 893
  • But our lab assistant said that it must be 5 microamps, and my friends found correctly except me. I guess internal resistance of ammeter is negligible. – bleckemr May 12 '22 at 20:31
3

For info about internal resistance, https://www.eeeguide.com/series-resistance-microammeter/

Just verify if it is easier ... measure directly ib, or v(Rb)/100k?

enter image description here

If you reversed the BJT, then you are "right"!

enter image description here

Antonio51
  • 13,128
  • 1
  • 7
  • 22
  • Our teacher said it must be 5 microamps and everyone except me obtained it. – bleckemr May 12 '22 at 20:32
  • 1
    Logical. See my answer ... – Antonio51 May 12 '22 at 20:42
  • 1
    +1 Good catch @Antonio51 – devnull May 12 '22 at 20:46
  • 1
    Thanks! I actually tried it flip vertically but I was trying to be quick so there should be other mistakes. Again thank you for your clear explanation. – bleckemr May 12 '22 at 20:57
  • 1
    It is the "only" simple means (from an external view) to distinguish emitter and collector, so you have learned something important ! ... It is the reason why on the multimeters, you can swap all "pins" to detect the "true" pinout of the BJT ... Thanks ... @devnull – Antonio51 May 12 '22 at 21:05