enter image description here In the B part of this question , i am not getting why the base current is zero. Everything seems to be fine but still not getting ...

  • \$\begingroup\$ Draw arrows on it to show the path you think the base current would follow and it should become obvious. \$\endgroup\$
    – Transistor
    Jul 6 '19 at 15:15
  • \$\begingroup\$ What is Vbe? Then Ic? \$\endgroup\$ Jul 6 '19 at 15:19
  • \$\begingroup\$ Replace the arrow in Q1 with a diode and ask yourself if any current should flow through that diode. \$\endgroup\$ Jul 6 '19 at 15:39
  • \$\begingroup\$ @HarrySvensson but it has 3 isolated pins so Rb pulls up Vbe to Re to Vcc thus Vbe=0 \$\endgroup\$ Jul 6 '19 at 16:03
  • 1
    \$\begingroup\$ The PNP transistor will only be open if the base current "see" a path for a current to flow to GND ( negative terminal of the supply). \$\endgroup\$
    – G36
    Jul 6 '19 at 18:15

The only possible path to the base current would be via the reversed collector-base junction:

enter image description here

No base current, no voltage in the base resistor = base-emitter voltage is zero, transistor is open.

If you create an alternative path to get current flowing through the base resistor, enough to polarize the base-emitter junction, you can turn the transistor on:

enter image description here


When we are talking about transistor characteristics there are three different operating regions. When we required a switching operation from a transistor, we use cutoff region and saturation region of a transistor. When there is no base emitter current(base is connected to positive in this case) it is operating in the cutoff region and collector current is approximately zero at this point. I can see that in figure(b).

When the base emitter current is at its maximum the transistor operates in the saturation region and maximum collector current can be obtained. Resistor settings in Figure (c) shows that the transistor operates in saturation region.

When the transistors are used in signal amplification the operating point will be selected between above two regions.enter image description here

That is the normal operating point of the transistor as shown in the image. Image (a) is invisible in your question. I guess the image (a) shows the circuit to obtain this operation. Therefore the above three figures to represent those three operating points of a transistor.


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