About my previous question: CMOS tri-state buffer internal structure

I have just one more question: is the value that "activates" the transistor related to the value of the Output? Because if I choose the input "high" and the control "high" I will activate the top transistor in "low" but the output value should b e "high". So it's not related, right?


  • \$\begingroup\$ Can you make this a self-contained question, including the circuit diagram? \$\endgroup\$ – Wouter van Ooijen Mar 25 '16 at 15:35
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    \$\begingroup\$ You need to study how PMOS transistors work. Common source configuration is inverting for either PMOS or NMOS. \$\endgroup\$ – The Photon Mar 25 '16 at 15:35
  • \$\begingroup\$ is the value that "activates" the transistor related to the value of the Output? NO, it is not, the transistors are either in a conductive state or not, depending on the gate-source voltage. Given enough gate-source voltage a conductive channel is formed between drain and source. This channel is formed regardless of the drain (output) voltage. \$\endgroup\$ – Bimpelrekkie Mar 25 '16 at 15:48
  • \$\begingroup\$ Thinking of transistors as "activated" is a shortcut that will mislead you until you learn how they actually work. There is no royal road to learning, as the saying goes. \$\endgroup\$ – pjc50 Mar 25 '16 at 16:22

Very roughly, the action of a MOSFET is controlled by the voltage between the gate and the source terminal. More detailed schematic symbols for MOSFETs make these two terminals look different:

enter image description here

In your circuit, the source is the one connected to the positive power supply node. The action is controlled by how far below the source is the gate voltage.


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