741 Op amp Inverting and Noninverting Inputs

Discussion in 'Analog & Mixed-Signal Design' started by Electrozapper, Sep 21, 2016.

  1. Electrozapper

    Thread Starter New Member

    Dec 9, 2015
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    Okay, to some this question may seem silly; however, I have been in a full study of the "guts" of the 741 op amp. At one point in my study I (so to speak) tore into the inner workings of the 555. Now, I am into the 741. Now, I understand most of the circuitry. I can understand the current mirror parts, the amplifier sections (gain stage), the differential stage, and such. BUT, my question is about the inputs themselves. You see, before I looked into the schematic of this little "beastie," I had in my mind a possible "schematic" of what the inputs might be, based upon their descriptions: Inverting and Non-inverting inputs. That is to say, for the non-inverting inputs I had IC formed BJTs configured as either common collector or common base since that would not invert the input signals. Okay, well and good so far. In fact, the "guts" do reveal a common collector (emitter follower) configuration. All is well....so far. The world is beautiful and makes sense. Then, I was tossed a curve ball that I totally fumbled. The invert input I wild-eyed speculated would be the common-emitter configuration....after all, it does invert the output signal. HOWEVER, no dice. I still had the (as you all know) the common collector/emitter follower staring at me. Okay, by now you see my reason for confusion. How can this emitter follower be the inverter input?!? As I have gone through the "innerds" of this monster (the 741 op amp, that is), what have I missed? Is there a section I missed dedicated to the output of the non-invert pin input? Is there another section dedicated to the output of the invert pin input? I didn't see any. In fact, I did trace an invert section (common emitter if memory serves) in the (I think it was....I don't have the schematic in front of me)...but I think it was in the Gain Stage. Might have been in the Bias Generator....but, I think the common emitter inverting occurred in the Gain Stage. Anywho, can someone explain or refer to materials that better explain to me how a common-collector amplifier configuration is going to invert the invert pin's input? AND, there is always an "and" or a "but" isn't there?....moreover, can someone explain how that common-emitter configuration in the Gain Stage (?) isn't going to invert the non-invert pin's input signal?
     
  2. Bordodynov

    Active Member

    May 20, 2015
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    Here you have to think about

    741Opamp_Inverting_and_Noninverting.png
     
  3. AlbertHall

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    Jun 4, 2014
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    So Q14/Q20 buffer (common collector - non-inverting) the output from Q15/Q17
    Q15/Q17 give amplification of the signal on Q15 base - common emitter, inverting
    Where does the Q15 base signal come from considering the inverting input? (The easier bit)
    Where does the Q15 base signal come from considering the non-inverting input? (The harder bit, look up long-tailed pair)
    upload_2016-9-21_14-14-0.png
     
  4. dl324

    Distinguished Member

    Mar 30, 2015
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    Try studying the differential amplifier in the input stage.

    And try using paragraphs to express your thoughts in a more organized manner.
     
  5. AlbertHall

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    Jun 4, 2014
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  6. hp1729

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    Nov 23, 2015
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    Inverting and non-inverting does not refer to the design of the circuit so much as the relationship of input signals to the op amp. A positive going signal on the non-inverting input will move the output higher. A positive signal on the inverting input will move the output down.
    The input signal should be looked at as relative to each other, not with their relationship to ground.
     
  7. #12

    Expert

    Nov 30, 2010
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    Other people have said some of this, but, "inverting input" doesn't mean the input stage inverts. It means the sum of all stages from the inverting input pin to the output pin equals "invert". Look to the differential pair as the heart of all op-amps and comparators. The shared emitter resistance is what forces this: Whatever one transistor does, the other one must do the opposite.
     
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  8. Electrozapper

    Thread Starter New Member

    Dec 9, 2015
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    My thanks for all the helpful responses. Hmmmm, okay, a parallel question suddenly comes to my mind. If one places an input signal at the Non-Inverting Pin, does the Output Pin Source Current? Secondly, if one places an input signal at the Inverting Pin, does the Output Pin Sink Current? I realize these questions may seem odd but I have been relooking at the schematic and want to verify something...something I think I figured out. Anyway, thanks for all the assistance, gang. Really appreciate it.
     
  9. AlbertHall

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    Jun 4, 2014
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    Assuming the inputs and output voltages are within their appropriate ranges:
    If the voltage on the inverting input is lower than the voltage on the non-inverting input then the output will sink current.
    If the voltage on the inverting input is higher than the voltage on the non-inverting input then the output will source current.
     
  10. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    Such a mass of text with no paragraphs. Hard to read.

    Do you speak that way or are too anxious?
     
  11. Electrozapper

    Thread Starter New Member

    Dec 9, 2015
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    Once more, many thanks for the helpful responses. Okay, I realize that I should have listed this at the first. It is an oversight I must now correct. It is where I am getting the basis for my inquiries on the 741 op amp. Just yahoo or google it: http://shop.emscdn.com/KitInstrux/741/741_datasheet_revB.pdf

    As you can see, the circuit (741 IC op amp) that I am studying is from a company known as Evil Mad Scientist, and the product is called "XL741." It is, as you notice, a discrete version of the IC chip that would otherwise be a 007 mysterious "black box." To be honest, it appears that the schematic used by this firm closely resembles that published by Texas Instruments. At least, that is what my studies have revealed so far. I have seen "other" schematic versions of this BJT device from other companies.

    My sincere apologies for not providing that data at the start. If this added data changes any of your prior responses, please let me know. Again, thanks for your much needed assistance. I shall, in a manner of speaking, "chew the fat" on what has been provided and see if there are any other questions that I must advance.

    Thank you.
     
  12. Electrozapper

    Thread Starter New Member

    Dec 9, 2015
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    Okay, I have a question on AlbertHall's response of Wednesday, 6:20am (Sept 21, 2016). Your schematic is a replica of the LM741 by Texas Instruments. For clarify I shall quote the section in reference:

    So Q14/Q20 buffer (common collector - non-inverting) the output from Q15/Q17
    Q15/Q17 give amplification of the signal on Q15 base - common emitter, inverting
    Where does the Q15 base signal come from considering the inverting input? (The easier bit)
    Where does the Q15 base signal come from considering the non-inverting input? (The harder bit, look up long-tailed pair)

    Alright, uhhhh, Q15/Q17 is actually (I think) Q16/Q17 (The Darlington Pair). Uhhh, let me see, Q15 goes with Q14 as a current limit (if memory serves from prior studies) when the 741 Sources Current. I think. Anyway, I shall use your included schematic for this.

    Now, you asked of me, "Where does the Q16 base signal come from considering the inverting input? (The easier bit)" I shall try to trace that answer and you (so to speak) "spot" me on this one. If I get this right, the signal comes from the following (please, correct me if I am wrong): A proper forward bias is applied to the Inverting Input Pin so as to turn that side's NPN Transistor of the differential amplifier on. Thus, Q2 and Q4 are on. Now, this action provides an open current path (conventional or electron) for Q16/Q17 (Darlington Configuration) and R11 to Ground. Amplification of the signal occurs and the Signal is also inverted since the Darlington Pair Configuration is of the Common-Emitter Amplifier type. The Output Pin "Sinks" Current (that is, current flowing out of the 741 op amp (using electron current flow theory). If I understand this little puppy, the Vbe Generator Section (R7, R8, C1, and the unlabeled transistor) forward bias Q20b. Am I close?

    Now, to the second and more difficult part of your statement: "Where does the Q16 base signal come from considering the non-inverting input? (The harder bit, look up long-tailed pair)." I looked up the long-tailed pair and, thus far, it did not shed any light that I didn't already have. So, let me take a stab at this one and you, once more (and thanks for your gracious help) correct me if I miss the mark. I am speculating (wildly, at that) that when the right side of the differential amplifier (the side we looked at a moment ago) is (so to speak) "Off," that the left side of the differential amplifier is "On" and conducting current. That would be Q1 and Q3. So Q1b is forward biased "ON" with an input signal. Since Q1/Q3 are "ON," that means Q2/Q4 are off and provide no current path for the Darlington Pair (Q16/Q17, etc.). Thus, the Darlington Pair are also "Off" as are the Q2/Q4 side of the differential amplifier. Okay, alright then, that means (I hope and think) that Q14 and R9 "Source Current" (using electron current flow theory, that is, current flows into the 741 op amp). I believe that the Vbe Generator (that section with R7, R8, C1, and an unlabeled transistor in your given schematic) biases Q14b on.

    Okay, sir, am I even in the ball field? Or, has my team not yet arrived at the ball park? Thanks for your patience and your help. As you can tell, however, I am approaching the circuit (some electronics tech friends of mine have told me) from a "design" standpoint. I mean, someone had to sit down and plot out each path, each action, each step and phase so as to get the outcome they desired. It is a methodical process, it would seem. Anyway, again, thanks.
     
    Last edited: Sep 23, 2016
  13. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    First, the schematic I posted has two Q15s :rolleyes: hence the confusion there.

    Yes, that is all OK, except for the on and off parts. While the amp is working in its linear region nothing is on or off. Where you say on, it means conducting a little more and off means conducting a little less.
     
  14. #12

    Expert

    Nov 30, 2010
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    Yes. Once you have the differential pair figured out, everything else is setting current flow limits and adding gain until the output stage has enough voltage and current to drive a load.

    Look at the line from the Q4 and Q6 collectors over to the push-pull output stage.
    Q8,9,10,11,12,and 13 are just setting current flows. The entire output stage is Q13 to Q20 and it can seemingly stand alone.
    Q15 and Q22 are current limiters and C1 is the speed limiter.

    This is how I started...figuring out the circuits inside chips. I still do it to figure out which way the current is going to flow at Q1 and Q2 when I calculate the input resistors. The main thing you can't transfer to discrete designs (one transistor at a time) is the current mirrors. They depend on matched transistors inside the single silicon chip. After you get outside that monolythic world with equal temperatures on the matched transistors, current mirrors are much more difficult to match in the real world. The failure to match and track each other during temperature changes make discrete current mirrors lose a lot of their convenience. You lose so much gain in the emitter resistors required to get a fair match that there are better ways to design most circuits than trying to make current mirrors.

    However, you can buy matched pairs! It's possible, but the price of matched pairs usually shoots down the design compared to buying an op-amp with all that matching done internally for a rather low price. That's why op-amp chips are so wonderful. You can get nearly anything optimized in an op-amp, depending on the part number. Low current, excellent input voltage match on the differential pair, high speed, good output power, low voltage operation, single supply operation...you name it, somebody makes it...and mostly cheap.
     
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  15. Electrozapper

    Thread Starter New Member

    Dec 9, 2015
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    I send out many, many thanks to you, AlbertHall. AND, thanks for the clarity on the On/Off states. I admit, I had wondered if "On/Off" was an imprecise terminology. Thanks so much for the clarify.
     
  16. Electrozapper

    Thread Starter New Member

    Dec 9, 2015
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    #12, I ALSO want to express my deep appreciation and thanks for your educational input. Both You and AlbertHall have gone a long way to further my electronics educational progress. Again, many, many thanks to you! I find such help a valuable resource.
     
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  17. OBW0549

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    Mar 2, 2015
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    I miss the old days when semiconductor manufacturers' data sheets-- especially those by Linear Technology and National Semiconductor-- included a transistor-level schematic of the chip's innards. There was an enormous amount that could be learned by studying those diagrams and figuring out how they worked.
     
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  18. dl324

    Distinguished Member

    Mar 30, 2015
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    I learned a fair amount from the 1979 Signetics Analog Applications Manual. It had information about how the devices were constructed and pointed out some of the parasitic devices that were formed. It also talked about opamp design (current mirrors, level shifters, transistor beta, compensation, thermal considerations, current limiting, etc) in detail.
     
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  19. #12

    Expert

    Nov 30, 2010
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    I still have the 1988 version of that book.:)

    Wisdom from old guys: Never throw away a data book!
    I had to look up stuff in the 1977 Audio book from National Semiconductor for this site!
     
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  20. dl324

    Distinguished Member

    Mar 30, 2015
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    I agree.

    When I left my last job where I actually worked with components, my Boss told me to take any of the databooks I wanted from my bench. I took a couple dozen, which I still have, but should have taken all of them. I was too young to know better.
     
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