How Many Clock Pulses Are Counted for Different Input Voltages in an ADC System?

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Discussion Overview

The discussion revolves around the calculation of clock pulses counted in an Analog-To-Digital Converter (ADC) system for different input voltages (0.9 V and 0.75 V). The context includes the use of a 1 MHz clock generator and a ramp voltage that increases from 0V to 1.25 V over 125 ms. Participants are examining the mathematical reasoning behind the pulse count results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a solution for the number of clock pulses counted for input voltages of 0.9 V and 0.75 V, calculating 900 and 750 pulses respectively.
  • Another participant questions why the calculation yields 900 instead of 90000, suggesting a potential misunderstanding in the division of time units.
  • Some participants agree on the clock period being 1 μs for a 1 MHz clock, but there is disagreement on the interpretation of the results.
  • One participant mentions obtaining a different result of 112 bpm, indicating a possible alternative calculation or context.
  • There are informal exchanges that divert from the main technical discussion, indicating a casual tone among participants.

Areas of Agreement / Disagreement

Participants express differing views on the calculation of clock pulses, with some agreeing on the results while others contest the reasoning behind the division leading to 900 instead of 90000. The discussion remains unresolved regarding the correct interpretation of the calculations.

Contextual Notes

Participants have not clarified the assumptions behind their calculations, and there is ambiguity regarding the interpretation of time units in the context of clock pulse counting.

mym786
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Homework Statement



The Analog-To-Digital Converter “ADC” system shown in Fig. 1 uses 1 MHZ clock generator and a ramp voltage that increases from 0V to 1.25 V in a time 125 ms.
Determine the number of clock pulses counted into the register
when Vi=0.9 V
and when it is 0.75 V


Homework Equations





The Attempt at a Solution




VR = 1.25 v TR =125 ms
Vi=0.9 V
t1= tr/Vr * Vi = 125ms *0.9 V / 1.25 V = 90 ms
T= 1/f = 1/ 1 Mz = 1μs
N = t1 / T = 90 ms / 1 μs =900
Vi=0.75 V
t1= tr/Vr * Vi = 125 ms *0.75 V / 1.25 V = 75 ms
N = t1 / T = 75 ms / 1 μs =750*




This is the solution of the problem by my prof. My question why N = 90ms/1μs = 900 and NOT 90000 ? The same goes for 750.

Thank you very much.
 

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mym786 said:

Homework Statement



The Analog-To-Digital Converter “ADC” system shown in Fig. 1 uses 1 MHZ clock generator and a ramp voltage that increases from 0V to 1.25 V in a time 125 ms.
Determine the number of clock pulses counted into the register
when Vi=0.9 V
and when it is 0.75 V


Homework Equations





The Attempt at a Solution




VR = 1.25 v TR =125 ms
Vi=0.9 V
t1= tr/Vr * Vi = 125ms *0.9 V / 1.25 V = 90 ms
T= 1/f = 1/ 1 Mz = 1μs
N = t1 / T = 90 ms / 1 μs =900
Vi=0.75 V
t1= tr/Vr * Vi = 125 ms *0.75 V / 1.25 V = 75 ms
N = t1 / T = 75 ms / 1 μs =750*




This is the solution of the problem by my prof. My question why N = 90ms/1μs = 900 and NOT 90000 ? The same goes for 750.

Thank you very much.

I get the same answers as you. A 1MHz clock has a 1us period.
 
berkeman said:
I get the same answers as you. A 1MHz clock has a 1us period.

The number of pulses I get is 112 bpm, but that's not important now. The division really should give 90000 as a result, not 900.
 
Char. Limit said:
The number of pulses I get is 112 bpm, but that's not important now. The division really should give 90000 as a result, not 900.

You just finish working out? My pulse was near that at the end of a 200m IM at noon :wink:

Oops, [/hijack]