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Author Topic: RC differentiation circuit  (Read 31047 times)
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sridhar10chitta
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on: December 13, 2005, 10:39:46 am »

I experience a lot of difficulty in explaining the waveforms of an RC differentiation circuit.
Consider a battery of voltage V connected through a switch to a capacitor C and one end A of a resistor R. The other free end B of the resistor R is connected to the negative terminal of the battery.
When the switch is closed, a sharp rise in voltage (A with respect to B) is observed across the resistor.

Most of the text books I referred to state that since "the voltage across the capacitor cannot change instantaneously provided the current remains finite" one observes the sharp rise in voltage across the resistor.
Refer "Pulse, Digital and Switching Waveforms" by Jacob Millman and Herbert Taub.

When students ask me for an explanation of what is happening across the capacitor involving potentials and charges, I experience difficulties.
Can someone provide a reasonable explanation using potentials, electric fields and charge movement or one at a more microscopic level ?

Thanks

Sridhar Chitta

Assoc. Prof
Dept of Instrumentation and Control Engg
Vignan Institute of Technology and Science
Village Deshmukhi Nalgonda Distt
Pin code: 508284
near Hyderabad Andhra Pradesh
India
website: www.vignanits.ac.in
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Sridhar Chitta
Head of Deptt
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Hyderabad 508 284 India
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Reply #1 on: December 13, 2005, 11:31:34 am » posted from:Taipei,T'ai-pei,Taiwan

The capacitor C is defined as the charge per voltage applied. C=Q/V

Before you turn on the switch, if there is no charge on the capacitor, the voltage Vc=0;
When you turn on the circuit, the current starts to flow into the capacitor (which in turn, increase the voltage of capacitor Vc=C*Q).

You can check out our RC circuit applet
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sridhar10chitta
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Reply #2 on: December 13, 2005, 04:47:37 pm »

Dear Prof Hwang

your reply 1 and my reply 1.
Thank you for the reply and the link to the applet.

They are very illuminating.

The point where I find difficulty in my understanding and where I need to explain to the students is how does the potential of the capacitor attain the same voltage (i.e V) in the circuit that I had described in my question.

The plate connected to the capacitor may attain V volts since it is connected directly to the battery positive thro' switch at the immediate instance when the switch is closed.
But what about the other plate i.e terminal A of the resistor since there is no apparent connection to the battery and since the resistor and the negative terminal of the battery is coming in between.

For this I need to blend probably electrostatics and electric circuit behaviour it seems.
They have and including me have background about Physics by Resnick and Halliday about potential and electric fields but the fine links needed to explain the phenomenon is not explained clearly in Resnick text book
also.


Thanks
Sridhar
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Sridhar Chitta
Head of Deptt
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Hyderabad 508 284 India
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Reply #3 on: December 14, 2005, 09:45:12 am » posted from:Taipei,T'ai-pei,Taiwan

From the definition: If there is a charge Q on the capacitor (capacitance is C) plate,
the voltage between the capacitor is Vc=Q/C;
So Vc is the voltage diffenence between capacitor plates.


If there is no charge on the capacitor, then the voltage difference is 0.
If the voltage source Vo is connected directed to one side of the capacitor, and the other side of the capacitor is connected to a resistor R and back to another end of the voltage source.

The voltage on both side of the capacitor are all Vo (the difference is 0).
And the voltage across the resistor is also Vo.
This will cause the current I(t=0)=Vo/R to flow, which in turn will charge the capacitor.
After a very small time interval dt, the charge on the capacitor will be Qc(dt)=(Vo/R)*dt.
And the voltage difference between capacitor will be Qc(dt)/C=Vo*dt/(R*C).
Then the voltage between the resistor will be I(dt)=(Vo- Vo*dt/(R*C))/R
So the current will become smaller when more charge are being accumulated between capacitor plates.
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sridhar10chitta
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Reply #4 on: December 14, 2005, 12:02:15 pm »

my reply2

The concept about the voltage not being able to change quickly is appreciated by the students and by me as well.
Where I get stuck is when I tell them that since the potential is V on one plate then the other plate instantaneously assumes a potential V because the voltage across the capacitor cannot change.
I referred to a few text books and also some internet sites and have put together an understanding of the action and am mailing to your email id since I do not know how to send via the forum's mail method....

Please give me your comments.
Thanks
Sridhar
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Sridhar Chitta
Head of Deptt
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Hyderabad 508 284 India
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Reply #5 on: December 14, 2005, 12:24:29 pm » posted from:Taipei,T'ai-pei,Taiwan

I just created another RC circuit simulation. Click the following URL if you want to download it or find out how it is created with EJS. RC circuit
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Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
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sridhar10chitta
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Reply #6 on: December 15, 2005, 04:12:39 pm »

Dear Prof Hwang

At last I managed to see the applet....our systems are slow and one of the Java plugins was not available....Thanks.

I will modify the doc file that I had sent to you via mail....based on the applet.
There will not be any field between the plates initially at t = 0+.

I will also add a couple of points reg emf source operation like battery so that the role of the source is clear to the students.
"The important thing to understand is that electrons are motivated to and/or from the cell's electrodes via ionic reactions between the electrode molecules and the electrolyte molecules. " from http://www.ibiblio.org/obp/electricCircuits/DC/DC_11.html.

I will add some points regarding the potential changes from the problems at
http://qemp.deas.harvard.edu:8182/students/lectures/specificlecture/?lectureID=3728.
especially question No. 9 in the above site.

If there are any suggestions you can make on the electric field E2 I will be happy to include it, else I will delete it.

Regards
Sridhar
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Sridhar Chitta
Head of Deptt
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Hyderabad 508 284 India
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Reply #7 on: January 12, 2010, 02:48:42 pm » posted from:Quezon,Nueva Ecija,Philippines

As what i have exepected in this forum, it is really full of educational topic.
Keep it up guys.


“Love has no other desire but to fulfill itself. To melt and be like a running brook that sings its melody to the night. To wake at dawn with a winged heart and give thanks for another day of loving.” -*-

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Abner001
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Reply #8 on: September 20, 2010, 05:37:05 pm » posted from:Islamabad,Islamabad,Pakistan

help me . . . !!! a square wave frequency 1KHz and amplitude 5v is applied to a differentiator circuit using a .5 micro farad capacitor and variable resistor.
what is the maximum allowable value for the resistor in order for the circuit to be a good differentiator?
what would the circuit look like?
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Fu-Kwun Hwang
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Reply #9 on: September 20, 2010, 07:45:45 pm » posted from:,,Taiwan

RC time comstant should be much smaller than period of the square wave.
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Abner001
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Reply #10 on: September 21, 2010, 05:33:12 pm » posted from:Multan,Punjab,Pakistan

Thanks Admin... ...
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