Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Phase Shift between input and output wave form of a RC circuit

Status
Not open for further replies.

skj

New Member
I had connected a 3.3k resistor to a 0.33µ Capacitor,other end of the capacitor is connected to common,now a 50Hz sinewave signal is connected to open end of resistor and common,and output is monitored between capacitor plus resistor junction and common.Waveforms of input and output were observed on a calibrated dual channel CRO,a phase shift of 3.34mS was observed I want to varify this result by calculation, can any member guide for a formula for this.​
 
I had connected a 3.3k resistor to a 0.33µ Capacitor,other end of the capacitor is connected to common,now a 50Hz sinewave signal is connected to open end of resistor and common,and output is monitored between capacitor plus resistor junction and common.Waveforms of input and output were observed on a calibrated dual channel CRO,a phase shift of 3.34mS was observed I want to varify this result by calculation, can any member guide for a formula for this.​

hi
Look here.
**broken link removed**

Calculate the Zcap for the 0.33uF at 50Hz and find the phase angle
 
Φ=tan^-1(X_C/R) will give you the phase angle in degrees. Do you know how to figure for X_C (Capacitive Reactance)?

If not, it is 1/2(pi)(f)(C) where pi is 3.1415.., f is the frequency in Hertz and C is capacitance in Farads.
 
Hello there,

Sorry to have to correct, but the phase angle (often called "Theta") is equal to:
TH=-atan(w*R*C)
which is also equal to:
TH=-atan(2*pi*F*R*C)
which in this case is equal to:
TH=-atan(2*pi*50*3300*0.33e-6)
or:
TH=-atan(0.1089*pi)
or:
TH=-atan(0.34211943997593)
which is equal to:
TH=-0.3296370988258 rads
The negative sign means the output lags the input in time

This is the phase shift in rads however, not seconds or milliseconds. To get that we have to divide by 2*pi and multiply by the period of 1/50, and we get:
-0.001049267474092 seconds, or -1.049267474092 milliseconds. The negative sign just means the output lags the input in time

Since this does not confirm the real life measurement, that means either the frequency is not 50Hz or either the cap or resistor (or both) is (are) not the stated value(s). Perhaps the cap and resistor and also the frequency could be measured.
 
Last edited:
Φ=tan^-1(X_C/R) will give you the phase angle in degrees. Do you know how to figure for X_C (Capacitive Reactance)?
If not, it is 1/2(pi)(f)(C) where pi is 3.1415.., f is the frequency in Hertz and C is capacitance in Farads.

that might be one way of figuring the phase angle, but the way I posted is straight from a text book.

Then that text book is incorrect. It happens more often then we might think. :) The book may have been talking about something else however (see below).

If you want to assume you already know the lead or lag, then we could do a similar thing:
TH=tan^-1(R/X_C)

but you'll note that your original post had this:
TH=tan^-1(X_C/R)
which is not correct for the response indicated by the OP. What it IS correct for is calculating the phase of the current through the whole circuit so perhaps the book was talking about that rather than the voltage across the cap. For the OP here though we need the other formula because he wants the output taken from across the cap...

I had connected a 3.3k resistor to a 0.33µ Capacitor,other end of the capacitor is connected to common,now a 50Hz sinewave signal is connected to open end of resistor and common,and output is monitored between capacitor plus resistor junction and common.Waveforms of input and output were observed on a calibrated dual channel CRO,a phase shift of 3.34mS was observed I want to varify this result by calculation, can any member guide for a formula for this.​
 
Last edited:
Which to me is telling me he observed a phase angle shift from the input (straight 50Hz line signal), not that he was asking for voltage across the cap. He wanted to mathematically prove the phase angle shift. I provided that. Where does he speak of taking voltage across the capacitor?
 
Hi again,

From the OP's first post:

"and output is monitored between capacitor plus resistor junction and common"

That sounds to me like he is looking across the capacitor. If he wanted to monitor the current through the circuit he would have had to look across the resistor and divide the voltage by the resistance to get the current, or use a current probe.

It's not a bad idea to know both however so it's not a total waste. Better to know more than less :)
 
Hi
All ! Thanks for your valuable comments on my query I am sorry but there was some mistake in my first post, the value of resistor actually was 33k and some additional resistors are also coming in picture ,for clarity I am attaching part of actual circuit in attached pdf file.I have also mentioned there how CRO was connected

phase shift of 3.34mS was measured on a calibrated YOKOGAWA's digital CRO using curser measurement facility,

Please post your comments after corrections reported by me.
Thanks.

skj
 

Attachments

  • RC Phase Shift.pdf
    9 KB · Views: 274
Hello again,

Oh ok that's better. I was going to guess that the resistor was 5 times larger, but with the resistor 10 times larger and some output resistance too we get the same effect almost.

The phase shift now is:
3.33657ms (with the output lagging the input)

which agrees very well with your measurement of 3.34ms.

You might also want to know that the amplitude at A3 is 0.253284 times the input at A1, and the junction between the 24k and 12k resistors will be 0.089394 times the input at A1 without the op amp inverting input connected to the circuit (with the op amp connected this last amplitude could change dramatically so perhaps it goes to the non inverting input).
 
Last edited:
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top