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

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In an ADC system with a 1 MHz clock and a ramp voltage from 0V to 1.25V over 125 ms, calculations show that for an input voltage (Vi) of 0.9V, 900 clock pulses are counted, and for 0.75V, 750 pulses are counted. The confusion arises regarding why the division of time (90 ms and 75 ms) by the clock period (1 μs) results in 900 and 750 instead of 90000 and 75000. The correct interpretation is that the clock period of 1 μs means that 1 ms equals 1000 clock cycles, thus leading to the lower pulse counts. The discussion clarifies the calculations and confirms the results align with the expected outputs for the given input voltages. Understanding the relationship between time and clock cycles is crucial for accurate ADC pulse counting.
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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]
 

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