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741 differential amplifier question

Discussion in 'General Electronics Discussion' started by jackorocko, Apr 28, 2010.

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  1. jackorocko

    jackorocko

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    Apr 4, 2010
    I was under the assumption that when hooked up as a differential amplifier the 741 op-amp will output the difference between the inputs. If that assumption is correct then what am I doing wrong in my simulation to cause a -2.5V on the 741 output?
     

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  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Remember that the 741 has gain greater than unity.

    the output will be the difference in the input voltages multiplied by the gain (for small differences in voltage).
     
  3. jackorocko

    jackorocko

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    Ok. so taking away the negative feedback resistor gives me the desired output of 0V. So then I have to ask myself, why is the negative feedback resistor network even needed? From the looks of it, I would say the 2.5 volts I am seeing on the output is coming directly from the voltage dividing network through the feedback resistor. Isn't that how the current is flowing?

    I know I am making this harder then it needs to be but, I must be missing something simple.

    for reference, I been reading this http://www.electronics-tutorials.ws/opamp/opamp_5.html
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Taking away that feedback resistor will increase the gain to a very large value and cause the output to clamp itself to one supply rail or the other. With your circuit it chooses the negative rail. Since you are measuring the voltage with respect to that rail, you see 0V.

    For this type of circuit you really need a double-ended power supply (say +/- 9V).

    I'm surprised the tutorials site doesn't mention that. They don't mention it explicitly (even on page 1) but you need to connect 2 9V batteries (say) in series. The +ve end is the +9V, the -ve end is -9V and the common connection in the middle is 0V

    The circuit you have is still not correct. You need to build it more like the example you showed me.
     
  5. jackorocko

    jackorocko

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    Apr 4, 2010
    Yeah, I sort of assumed I needed a negative and positive supply voltage.

    Maybe this attachment will make things clearer. I used 9V for the wheatstone and input voltage and used 5 +/- for the 741 itself. As you can see I added a graph to show what the circuit is doing when the variable resistor is being changed over a 20k ohm range. Between 12k and 14k ohms the voltage swings rapidly from one rail to the other and the output voltage of the 741 is either -/+ the input voltage. The arrows on the negative feedback resistor in my simulation shows current flow. This current flow across the neg. feedback resistor is what makes me believe that the neg. feedback circuit is whats supplying the output voltage.

    This is where I am confused, because the simulation does not seem to be working as I would understand it to operate.
     

    Attached Files:

  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    That simulation shows *exactly* what I would expect.

    What do you expect?
     
  7. jackorocko

    jackorocko

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    Apr 4, 2010
    Well lets start off with a couple things first.

    1. I would expect the output voltage to be the gain of the amplifier only when the wheatstone is balanced. I would assume that is when the variable resistor was set to equal 10k ohms. The graph doesn't show that at all.

    2. I also expected to have a higher output voltage spanning a larger range of resistances. The simulation shows that I have a voltage swing of 6V's across a 2k ohm range. That too me doesn't make sense since the output should be the voltage difference times the gain (A) of the amplifier. I would expect the voltage output to infinite in the ideal situation with the ideal power source.
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The 741 has a very high gain. The 10M resistor providing negative feedback is not going to reduce it much.

    The output will be the difference of the input voltages multiplied by the gain (probably several thousand)

    Your point (1) makes no sense to me. Your point (2) seems to suggest that you are expecting the output to vary more than it does?

    Firstly, the output voltage cannot exceed the supply rails, secondly the 741 cannot even drive the output completely to the supply rails.

    The specs say the output can swing within 2 or 3 volts of the supply rails, and that for decent CMRR the inputs shouln't get too close to the supply rails either.

    Depending on how the simulation of the 741 is defined, this may well be the reason the output is shown to only go as far as it does toward the supply rail. Equally, since you are driving the inputs beyond where the common mode input voltage range, odd things may happen -- I have no idea what the simulation might do to handle that.

    Having said all of that, the output of the simulation looks like what I'd expect. I take your point that the crossover point should happen when the bridge is balanced, but your lack of input resistors may well be the cause of that. I'm afraid my experience in breaking the rules isn't that good.
     
  9. jackorocko

    jackorocko

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    Apr 4, 2010
    to clarify point (1). If the voltages on the negative and positive inputs of the 741 are equal then the output to me would have to be zero. But you said in your first post
    But anything multiplied by zero should be zero.

    http://en.wikipedia.org/wiki/Differential_amplifier

    The wiki tells me the formula for figuring this all out. But in that formula I am still confused on how to figure Ad and Ac. Any help?
     
  10. jackorocko

    jackorocko

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    Apr 4, 2010
    I am not sure what you mean by this? I want to do things by the book, the way it should be done. I am experimenting to try and learn as much as I can on my own. So I am not looking for a shortcut, if my design is flawed or the material I am reading is wrong. So be it, just point me in the right direction please.
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    1) I believe your input voltages are going too close to the +ve supply rail

    2) you are not building the comparator the way it's shown in the demo circuit. The demo circuit has input resistors for the 741.

    The reference to my inexperience with breaking the rules is that I can't easily predict what the effects of your changing the circuit are. I have no idea if the same results would be seen in practice, or whether it's an artefact caused by your simulation software.
     
  12. jackorocko

    jackorocko

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    Apr 4, 2010
    Well, here is what I ended up with.

    I chose to make the V output of R2, R3 to be closer to R1, R4 when R4 has a lower resistance. I am assuming R4 is a variable resistance. This keeps the negative voltage drop across the transistor at about -1V. Wouldn't a negative voltage (>-6) drop across the emitter-base junction fry my transistor. Ground = 0 Base = -6 or am I misunderstanding what b-e reverse max voltage?

    This circuit is proving more useful then my last. I ran into the problem that over a distance the resistance would change due to my light source being weaker. I think this circuit allows me more flexibility then the last given the same scenario I wanted to accomplish. Would you agree with that statement?
     

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    Last edited: May 2, 2010
  13. jackorocko

    jackorocko

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    guess that's a no? lol
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The BE junction is only reverse biased by 1 volt according to your graph. That won't hurt it.

    The graph also shows your variable resistance changing very rapidly, i.e. not a smooth transistion, however the output looks to be what I'd expect. Tray making the rate of change slower and see what happens.
     
  15. Zmechanic

    Zmechanic

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    May 6, 2010
    First off, you need to look at the differential amplifier configuration of an op-amp, not just the op-amp itself.


    NOT the article you had before. Yes, the op-amp is a differential amplifier, but its internal gain can and often times is more than a million. You must control it with negative feedback. I won't blurt out the equations for the control loop theory behind it, but your current configuration is causing the op-amp to function with all its gain. Therefore, a 1microvolt (1 millionth of a volt) difference between the + and - terminals could easily translate to 1 volt or more. The thermal noise ALONE from a 10K resistor could generate upwards of 6uV (pk-pk) of noise (@10KHz bw). Plus there is noise from the op-amp itself. Now, you've got 6uV times a million, that's 6V. You get a randomly swinging signal of 6V that you can't even control because it's noise.

    Long story short, you have to have feedback resistors.

    What are you trying to accomplish with that circuit? Are you trying to do strain gauge measurements?
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    ZMechanic, which circuit diagram were you looking at?
     
  17. Zmechanic

    Zmechanic

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    May 6, 2010
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    ZMechanic, that happens to be *exactly* what his last circuit diagram looks like.

    It probably pays to read the last couple of posts rather than just replying to the first post in the thread. :)

    However, his last circuit diagram shows the output controlling a relay. For that I'd actually expect to be looking at higher gain so the op-amp acts more like a comparator. There's little point in driving a relay with a circuit that produces an analogue voltage.
     
  19. jackorocko

    jackorocko

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    Apr 4, 2010
    Well why didn't you say this before? You do make a very good point. I was trying to over complicate the problem. thanks again steve.
     
  20. jackorocko

    jackorocko

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    Apr 4, 2010
    I assume this is what you had in mind?

    The only question I have is why does my transistor not blow up when there is 10 volt potential difference between the BE junction? In the simulation the only way I can make the transistor blow is to hook up a neg supply rail to the Emitter and ground to the base. Guess I am missing something again... :confused:
     

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