Determine input voltage given resistances

  • Thread starter Thread starter orangeincup
  • Start date Start date
  • Tags Tags
    Input Voltage
AI Thread Summary
The discussion focuses on calculating input voltage based on given resistances in a CMOS gate scenario, assuming infinite input resistance. For R1=1K and R2=infinity, the input voltage is determined to be 5V, as the infinite resistance acts like an open circuit. For R1=1K and R2=10K, the voltage divider rule yields an output voltage of approximately 4.545V. Participants clarify the application of the voltage divider equation, emphasizing that the input voltage is developed across R2, not R1. The conversation concludes with confirmation of the calculations and understanding of the concepts involved.
orangeincup
Messages
121
Reaction score
0

Homework Statement


[/B]
Calculate the expected input voltage for each set of resistances below. For the CMOS gate, assume the input resist of the gate is infinite.

Homework Equations


i=v/R
Voltage divider rule v*R1/(R1+R2)=v1

The Attempt at a Solution


1)For R1=1K and R2=infinity, would V simply be 0 since the resistance is infinity? I'm mostly confused with how the CMOS gate and the infinity resistance will effect the voltage.

2)For R1=1k and R2=10K
v1=5V*1K/(1K+10K)= 0.45V?

I assume for all the rest, the values with infinity I'd calculate as in step 1, and the values with two specific numbers I'd calculate as I would in step 2?
 

Attachments

  • cmos.png
    cmos.png
    5.8 KB · Views: 432
Physics news on Phys.org
Check your voltage divider equation: The input voltage is developed across R2, not R1.

orangeincup said:
1)For R1=1K and R2=infinity, would V simply be 0 since the resistance is infinity? I'm mostly confused with how the CMOS gate and the infinity resistance will effect the voltage.
Redraw the input circuit without R2 (an infinite resistance is equivalent to an open circuit):

Fig1.png

What's the input voltage?
 
  • Like
Likes orangeincup
Oops,

1)The input voltage would just be 5V then wouldn't it?

2) v1=5V*10K/(1K+10K)=4.545V?
 
orangeincup said:
Oops,

1)The input voltage would just be 5V then wouldn't it?

2) v1=5V*10K/(1K+10K)=4.545V?

Looking good!
 
  • Like
Likes orangeincup
Back
Top