Question

Compute the time constant, τ = RC, of the three circuits (i.e., 3 different values of capacitor) and compare them to the half-period of the input voltage. What do you
conclude from the above two simulations?

Answer 1

Circuit 1: 100nF * 1k Ohm = 0.0001= tau
Circuit 2: 1uF * 1k Ohm = 0.001 = tau
Circuit 3: 10uF * 1k Ohm = 0.01 = tau

Answer 2

@ganeshie8 Hey, can you help me out here? really lost.

Answer 3

Hey!

Answer 4

@Curry whats the exact question ?

Answer 5

@ganeshie8 any guidance here?

Answer 6

Aren't you running the input voltage at 500MHz ?

They are asking you to set it to just 500Hz right ?

Answer 7

@Curry

Answer 8

yep

Answer 9

wait, how are you able to tell it's 500 MHz?

Answer 10

nvm, frequency looks good !

Answer 11

Your amplitude looks wrong though, you must be using -2.5V to +2.5V for part a, right ?

Answer 12

Also, they are asking you to plot output waveforms for only two periods

Answer 13

The reason i'm starting from past 5ms is because i'm allowing for steady state to be reached.

Answer 14

No, you're using -5V to 5V for part 1

This is wrong

Answer 15

They told me during lab to use -5, 5. despite it saying -2.5 to 2.5

Answer 16

Ohk

Answer 17

let me change it to -5, 5 then

Answer 18

Does this look good for part1, 100nF ?

Answer 19

Yes

Answer 20

part1, 1uF

Answer 21

basically your plots look good for part 1

Answer 22

do you want me check part 2 too ?

Answer 23

Your input frequence is 500 Hz
the corresponding half period is 1ms

therefore, express RC constant also in milliseconds

Answer 24

Circuit 1: 100nF * 1k Ohm = 0.0001= tau = \(0.1 mS\)
Circuit 2: 1uF * 1k Ohm = 0.001 = tau = \(1 mS\)
Circuit 3: 10uF * 1k Ohm = 0.01 = tau = \(10mS\)

Answer 25

compare those RC constants with the half period : \(1mS\)

Answer 26

Notice that when the capacitance is huge, like 10uF, the time constant is 10mS.

10mS is way larger than the half period 1mS.

Therefore the capacitor cannot fully charge/discharge in the given half period of input voltage. So the voltage across resistor never becomes 0.

Take a good look at the waveform for 10uF and convince yourself..

Answer 27

right, ok

Answer 28

btw what simulator are you using ?

Answer 29

PSpice

Answer 30

Okay nice

Answer 31

so the analysis is, for very large/small capacitances, like 10uF and 100nF, the time period deviates from the half period. Therefore, the capacitor doesn't have adequate time to charge/discharge.

Answer 32

Btw, are you an EE major by any chance? You seem to know so much about circuits.

Answer 33

Rephrase like this :


so the analysis is, for very large capacitances, like 10uF, the "RC time constant" is way larger than the half period of input voltage. Therefore, the capacitor doesn't have adequate time to charge/discharge.

Answer 34

and same for really small capacitors too right?

Answer 35

for really small capacitances, the "RC time constant" is very small compared to half period of input voltage. so the capacitor has enough time to fully charge/discharge

Answer 36

whats the RC time constant when C = 100nF ?

Answer 37

0.1mS

Answer 38

whats the half period of input voltage ?

Answer 39

0.1, wait, nvm, i just understood the other graphs. ok ok, i see now.

Answer 40

what do you mean by 0.1 ?

Answer 41

which makes sense also, cause if it took way less time, it would obviously charge well within that time.

Answer 42

whats the period of input voltage ?

Answer 43

1millisecond

Answer 44

1mS is halfperiod, right ?

Answer 45

mhmm

Answer 46

f = 500Hz

T = ?

Answer 47

is T the period?

Answer 48

Yes

Answer 49

frequency = 500Hz

period = ?

Answer 50

0.002

Answer 51

express it in milliseconds

Answer 52

2 mS