Why do we use a CMOS for inverting a circuit when the PMOS already achieves that?











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The output in a PMOS is as follows:

i/p o/p

0 1

1 0

Why can't I just use this instead of using a CMOS for inverting logic?

(Please explain in simple terms as I am a beginner in this topic and subject)










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  • 2




    FWIW, what OP describes is not a characteristic of PMOS transistors, but of common source/common emitter stages.
    – Vladimir Cravero
    Nov 17 at 15:01















up vote
8
down vote

favorite
1












The output in a PMOS is as follows:

i/p o/p

0 1

1 0

Why can't I just use this instead of using a CMOS for inverting logic?

(Please explain in simple terms as I am a beginner in this topic and subject)










share|improve this question


















  • 2




    FWIW, what OP describes is not a characteristic of PMOS transistors, but of common source/common emitter stages.
    – Vladimir Cravero
    Nov 17 at 15:01













up vote
8
down vote

favorite
1









up vote
8
down vote

favorite
1






1





The output in a PMOS is as follows:

i/p o/p

0 1

1 0

Why can't I just use this instead of using a CMOS for inverting logic?

(Please explain in simple terms as I am a beginner in this topic and subject)










share|improve this question













The output in a PMOS is as follows:

i/p o/p

0 1

1 0

Why can't I just use this instead of using a CMOS for inverting logic?

(Please explain in simple terms as I am a beginner in this topic and subject)







mosfet digital-logic cmos nmos pmos






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asked Nov 17 at 14:10









Harshit Pandey

454




454








  • 2




    FWIW, what OP describes is not a characteristic of PMOS transistors, but of common source/common emitter stages.
    – Vladimir Cravero
    Nov 17 at 15:01














  • 2




    FWIW, what OP describes is not a characteristic of PMOS transistors, but of common source/common emitter stages.
    – Vladimir Cravero
    Nov 17 at 15:01








2




2




FWIW, what OP describes is not a characteristic of PMOS transistors, but of common source/common emitter stages.
– Vladimir Cravero
Nov 17 at 15:01




FWIW, what OP describes is not a characteristic of PMOS transistors, but of common source/common emitter stages.
– Vladimir Cravero
Nov 17 at 15:01










2 Answers
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up vote
25
down vote



accepted










In a word: Efficiency.





You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high.



When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not logic low.



Your actual truth table is:



I/P    O/P

0 1
1 Z


You can overcome this inability to drive low, by using a resistor to pull the output low when the transistor is off. However to be able to strongly drive low, you need a low value resistor.



enter image description here



This resistor is always across the output, which means that when you turn the PMOS on to drive high, a large current will flow from the PMOS through the resistor to ground. This uses lots of energy. If you have billions of switches, you can see that the power consumption will be very high.



The better approach is to replace this resistor with an NMOS transistor. This is called CMOS. By using a NMOS device, you can think of it as being able turn off the resistor when the output is driven high (PMOS is on).



Using the NMOS you can also get a strong logic low because when switched on, the NMOS is effectively a short.



CMOS therefore by using complementary transistors, has very low static power dissipation - when an output is being held either high or low, almost no power is consumed.






share|improve this answer






























    up vote
    7
    down vote













    CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching.



    From here, be the circuit below for a simple inverter:





    schematic





    simulate this circuit – Schematic created using CircuitLab



    When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a short-circuit. The opposite for when IN = 1: the NMOS is a short-circuit and the PMOS is a open-circuit. It's either Vdd (5V) or ground at the output which is being driven "strongly".



    As a result you have lower power dissipation.






    share|improve this answer





















      Your Answer





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      2 Answers
      2






      active

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      2 Answers
      2






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      up vote
      25
      down vote



      accepted










      In a word: Efficiency.





      You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high.



      When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not logic low.



      Your actual truth table is:



      I/P    O/P

      0 1
      1 Z


      You can overcome this inability to drive low, by using a resistor to pull the output low when the transistor is off. However to be able to strongly drive low, you need a low value resistor.



      enter image description here



      This resistor is always across the output, which means that when you turn the PMOS on to drive high, a large current will flow from the PMOS through the resistor to ground. This uses lots of energy. If you have billions of switches, you can see that the power consumption will be very high.



      The better approach is to replace this resistor with an NMOS transistor. This is called CMOS. By using a NMOS device, you can think of it as being able turn off the resistor when the output is driven high (PMOS is on).



      Using the NMOS you can also get a strong logic low because when switched on, the NMOS is effectively a short.



      CMOS therefore by using complementary transistors, has very low static power dissipation - when an output is being held either high or low, almost no power is consumed.






      share|improve this answer



























        up vote
        25
        down vote



        accepted










        In a word: Efficiency.





        You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high.



        When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not logic low.



        Your actual truth table is:



        I/P    O/P

        0 1
        1 Z


        You can overcome this inability to drive low, by using a resistor to pull the output low when the transistor is off. However to be able to strongly drive low, you need a low value resistor.



        enter image description here



        This resistor is always across the output, which means that when you turn the PMOS on to drive high, a large current will flow from the PMOS through the resistor to ground. This uses lots of energy. If you have billions of switches, you can see that the power consumption will be very high.



        The better approach is to replace this resistor with an NMOS transistor. This is called CMOS. By using a NMOS device, you can think of it as being able turn off the resistor when the output is driven high (PMOS is on).



        Using the NMOS you can also get a strong logic low because when switched on, the NMOS is effectively a short.



        CMOS therefore by using complementary transistors, has very low static power dissipation - when an output is being held either high or low, almost no power is consumed.






        share|improve this answer

























          up vote
          25
          down vote



          accepted







          up vote
          25
          down vote



          accepted






          In a word: Efficiency.





          You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high.



          When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not logic low.



          Your actual truth table is:



          I/P    O/P

          0 1
          1 Z


          You can overcome this inability to drive low, by using a resistor to pull the output low when the transistor is off. However to be able to strongly drive low, you need a low value resistor.



          enter image description here



          This resistor is always across the output, which means that when you turn the PMOS on to drive high, a large current will flow from the PMOS through the resistor to ground. This uses lots of energy. If you have billions of switches, you can see that the power consumption will be very high.



          The better approach is to replace this resistor with an NMOS transistor. This is called CMOS. By using a NMOS device, you can think of it as being able turn off the resistor when the output is driven high (PMOS is on).



          Using the NMOS you can also get a strong logic low because when switched on, the NMOS is effectively a short.



          CMOS therefore by using complementary transistors, has very low static power dissipation - when an output is being held either high or low, almost no power is consumed.






          share|improve this answer














          In a word: Efficiency.





          You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high.



          When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not logic low.



          Your actual truth table is:



          I/P    O/P

          0 1
          1 Z


          You can overcome this inability to drive low, by using a resistor to pull the output low when the transistor is off. However to be able to strongly drive low, you need a low value resistor.



          enter image description here



          This resistor is always across the output, which means that when you turn the PMOS on to drive high, a large current will flow from the PMOS through the resistor to ground. This uses lots of energy. If you have billions of switches, you can see that the power consumption will be very high.



          The better approach is to replace this resistor with an NMOS transistor. This is called CMOS. By using a NMOS device, you can think of it as being able turn off the resistor when the output is driven high (PMOS is on).



          Using the NMOS you can also get a strong logic low because when switched on, the NMOS is effectively a short.



          CMOS therefore by using complementary transistors, has very low static power dissipation - when an output is being held either high or low, almost no power is consumed.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered Nov 17 at 14:23









          Tom Carpenter

          37.5k267114




          37.5k267114
























              up vote
              7
              down vote













              CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching.



              From here, be the circuit below for a simple inverter:





              schematic





              simulate this circuit – Schematic created using CircuitLab



              When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a short-circuit. The opposite for when IN = 1: the NMOS is a short-circuit and the PMOS is a open-circuit. It's either Vdd (5V) or ground at the output which is being driven "strongly".



              As a result you have lower power dissipation.






              share|improve this answer

























                up vote
                7
                down vote













                CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching.



                From here, be the circuit below for a simple inverter:





                schematic





                simulate this circuit – Schematic created using CircuitLab



                When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a short-circuit. The opposite for when IN = 1: the NMOS is a short-circuit and the PMOS is a open-circuit. It's either Vdd (5V) or ground at the output which is being driven "strongly".



                As a result you have lower power dissipation.






                share|improve this answer























                  up vote
                  7
                  down vote










                  up vote
                  7
                  down vote









                  CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching.



                  From here, be the circuit below for a simple inverter:





                  schematic





                  simulate this circuit – Schematic created using CircuitLab



                  When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a short-circuit. The opposite for when IN = 1: the NMOS is a short-circuit and the PMOS is a open-circuit. It's either Vdd (5V) or ground at the output which is being driven "strongly".



                  As a result you have lower power dissipation.






                  share|improve this answer












                  CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching.



                  From here, be the circuit below for a simple inverter:





                  schematic





                  simulate this circuit – Schematic created using CircuitLab



                  When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a short-circuit. The opposite for when IN = 1: the NMOS is a short-circuit and the PMOS is a open-circuit. It's either Vdd (5V) or ground at the output which is being driven "strongly".



                  As a result you have lower power dissipation.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 17 at 14:23









                  Renan

                  4,28222144




                  4,28222144






























                       

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