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 Author Topic: operational amplifier simulation  (Read 18132 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
alfalion
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 « Embed this message on: February 24, 2011, 12:03:00 am » posted from:Baia Mare,Maramures,Romania

Hi there. I'm new to this forum. I like electronics very much and i was wondering if a simulation of operational amplifiers can be made. Something like changing the input voltage or some resistor and see what you get at the output using EJS.
Some of the most important opamp categories can be found here
So is it possible?
Thank you.
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: February 24, 2011, 11:31:28 pm » posted from:Taipei,T\'ai-pei,Taiwan

I hope the following simulation is what you need!
You can use sliders to change R1,R2, Vin.
Click invert to find the circuit type similar to your case!

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
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• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
 ejs_operationalAmplifier.jpg (18.1 KB, 673x392 - viewed 624 times.) Logged
alfalion
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 « Embed this message Reply #2 on: February 25, 2011, 05:00:24 pm » posted from:Baia Mare,Maramures,Romania

Yes, this is what i was wondering about. Thanx.
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Tminus
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 « Embed this message Reply #3 on: June 08, 2011, 04:52:02 pm » posted from:Satu Mare,Satu Mare,Romania

Hello Mr. Fu-Kwun Hwang, can you explain or comment the variables (what they do) used for this simulation? I want to better understand this simulation. Thankyou very much.
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Fu-Kwun Hwang
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 « Embed this message Reply #4 on: June 08, 2011, 05:52:58 pm » posted from:Taipei,T'ai-pei,Taiwan

The key point is: the same current flow through both resistors $R_1,R_2$.

For non-invert case:
$V_{out}=I\,(R_1+R_2), \qquad V_{in}=I\, R_1$
so $V_{out}=\frac{R_1+R_2}{R_1} V_{in}=(1+\frac{R_2}{R_1}) V_{in}$

For invert case:

$V_{out}= -I\, R_2, \qquad V_{in}=I\, R_1$
so $V_{out}=-\frac{R_2}{R_1}\, V_{in}$
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srisak31
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 « Embed this message Reply #5 on: October 14, 2015, 03:18:15 pm » posted from:BANG BUA THONG,NONTHABURI,THAILAND

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chamDz
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 « Embed this message Reply #6 on: October 26, 2015, 05:00:24 pm » posted from:,,Sri Lanka

i have some concerns regarding this
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Wisdom consists of the anticipation of consequences. ..."Norman Cousins(1913-1990, American author)"