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#1
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Plese, can someone help me to understand from the internal strcture of 741 opamp that how possitive input give positive output and how negative input give inverted output, especislly after the differential amplifier section. The internal structure is on http://en.wikipedia.org/wiki/Op-amp!! and plese try to make it discriptive than mathamatical , I know this is too much but plese help me !!!!!!!!
Last edited by chrischristian; 02-24-2008 at 12:17 PM. |
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#2
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Your description is slightly off the mark. The inputs do not affect the output independently as you have implied. The absolute magnitude of the inputs is irrelevant as long as they remain within the "Common Mode Range". This parameter is given on datasheets and typically is contained to a range which is smaller than the power supply rails. There are amplifiers whose common mode range includes one or both of the supply rails.
The critcal point, about all the opamps you will ever see, besides all the ones that you won't, is that the output depends ONLY on the difference between the two inputs. Example #1 Av = 100,000, V+=12 V, V- = -12V, CMR = {-10.5,..10.5} IN+ = 8.000001 VDC IN- = 8.000000 VDC Vo = (8.000001 - 8.0)*100,000 = 100 mV Example #2 Same as Example #1 IN+ = 6.000002 IN- = 5.999998 Vo = (6.000002 - 5.999998)*100,000 = 400 mV Example #3 Same as Example #1 IN+ = 8.000000 IN- = 8.000001 Vo = (8.000000 - 8.000001) * 100,000 = -100 mV Example #4 Same as Example#1 IN+ = -8.000000 IN- = -8.000001 Vo = (-8.000000 - (-8.000001)) * 100,000 = +100 mV !! It should be clear from these examples that the differential input stage acts only on the difference between the two inputs and not their absolute magnitude.
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We never have time to do it right, But we always have time to do it over. |
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#3
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Thank you for your time , but my question was different, please read the attachment !!!
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#4
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You're right, I don't know how to explain it to you. Until you stop considering the inputs one at a time I don't think anyone will be able to get through to you. That's just your tough luck I guess.
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We never have time to do it right, But we always have time to do it over. |
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#5
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I think I am right in this - op amp internal circuitry is often extremely complex since transistors are easy to create and other componenets more awkward - hence a glut of transistors.
T8 (in conjunction with other components) forms a constant current source. Output to the second stage is taken from the collector of T6. A positive input at the inverting input increases current through T2 & T4 raising the voltage at T6 collector. Since this current flows through T8 (constant current source) the current through T1 & T3 decreases (not that we are bothered at this stage). A positive input at the non-inverting input increases current in T1 & T3 and due to the constant current source (T8) the curent in T2 & T4 decreases lowering the voltage at T6 collector ! Steve. |
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#6
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Here is a good tutorial on op amps:
http://cktse.eie.polyu.edu.hk/eie304/Op-amp.pdf The 741 op amp is analyzed, beginning on page 24.
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General info: If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. |
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#7
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Does this help?
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#8
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Hey! Every one out there ! thank you very much for your time and effort now I feel relieved.
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| Tags |
| 741, circuit, integrated |
Related Site Pages
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| Video Lecture | Op Amps Characteristics (Part 2) - Internal Circuitry - Op Amps and Op Amp Circuits | |||
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