Op amp

Discussion in 'General Electronics Chat' started by mad_mat222, Aug 16, 2013.

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  1. mad_mat222

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    Aug 14, 2013
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    Are op amp circuits only capable of puting out an on/off output or are they capable of a variable output ie if controlled by a variable resistor.
     
  2. ScottWang

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    If you treat it as a comparator then it will output a High/Low voltage as a logic signal, but they are not output an ON/OFF.

    If you treat it as an amplifier and using a pot to adjust the input voltage, and then the output voltages will following the input voltages according to the magnification of formular .
     
  3. mad_mat222

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    Aug 14, 2013
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    Do you have a pic of a what you mean? I have a basic understanding of op amps. I spoke to someone who mentioned that they can only be turned on or off. I will be using it in an comparative circuit.
     
  4. ScottWang

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    What you said about the on/off, they are pointed to the two output bjts of op amp, not the output pin of op amp, you can google any datasheet of op amp, and go to take a look the internal stucture of op amp.
     
    Last edited: Aug 16, 2013
  5. mad_mat222

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    Understand that but I was of the understanding that the actually output was constant (able to be varied, but still constant). I also understand that they can be used to invert the voltage ie + to -. Just want to know how to use it to create a the above mentioned varying output.
     
  6. ScottWang

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    You can google invterting op amp.
    Also you can set the gain to match what you need.
     
  7. jegues

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    They can have variable output.

    Wire up a op-amp in a voltage follower configuration, feed in a time varying signal into the input and measure the signal at the output.

    You better believe it will be varying alongside the input.
     
  8. studiot

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    I really suggest you find out more about these things before you rush into an application.

    If you genuinely want a comparator you should get a comparator, which is a special type of chip optimised for the application, not a general purpose op amp.

    If you want a 'comparative circuit' most folks understand this to mean a differential amplifier which is an entirely different type of circuit.

    Again you should find out enough to understand the difference between 'inverting the voltage' from positive to negative and from positive going to negative going.

    An op amp with a purely positive supply cannot generate a negative voltage by itself,

    So what do you actually mean and want to do?
     
    Last edited: Aug 16, 2013
  9. ScottWang

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  10. w2aew

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    You may find my video on the basics of op amp operation helpful:
    http://www.youtube.com/watch?v=K03Rom3Cs28
     
  11. WBahn

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    Unless you told that person that you were pretty sure that they were wrong, these two statements are a bit contradictory.

    The basic idea behind an opamp is that the output is equal to the difference between the two inputs (noninverting input minus inverting input) multiplied by a very large number. In practice, that number is typically between 100,000 and 10 million. For this discussion, let's assume ideal behavior except for a finite gain of exactly 1 million (an ideal opamp has infinite gain)

    Thus, if the difference between the two inputs is 1μV the output might be 1V and if the difference is 20μV it would be 20V. If the voltage at the inverting input is higher than the voltage at the noninverting input, they you have a negative differential voltage (despite the fact that both voltages might be positive). Thus if the difference is -3μV the output would be -3V.

    But since the output of a real opamp can't go outside of the supply rails --another "real" restriction we will impose on our otherwise ideal opamp -- this means that the difference between the two inputs has to be a very small voltage in order for this relationship to apply. For instance, if your opamp is powered by +12V and 0V rails (and has a gain of one million), then for any difference between the inputs greater than just 12μV you will only get 12V at the output, while for any negative difference you would get 0V at the output.

    From this you can immediately see how an opamp could be used, in theory, as a comparator. Apply a fixed voltage of, say, 6V to the inverting input. If the voltage applied to the noninverting input is 6V or less, the output will be 0V, while if the voltage applied is 6.000012V or more, the output will be 12V. For the tiny range of voltages between those limits, known as the "active" or "linear" region of operation, the output will be somewhere inbetween.

    As noted by studiot, this is not a good way to use an opamp. Yes, in theory it works like this and, in practice, it comes pretty close. In fact, opamps DO get used this way in a number of instances and they work ust fine. But generally you either want your circuit to behave like a comparator, in which case you aren't concerned too much about the details of how it behaves when the input differential voltage is between that tiny voltage range, or you want it to behave like an amplifier, in which case that is pretty much all you care about. It turns out that practical opamp circuits can't behave really well in both regions, so some of them are optimized for performance within the linear region (we call those "opamps") and some are optimized for how well they transition from one side of this region to the other (we call those "comparators").

    So how do we use an opamp as an amplifer in practice if the linear region is only a handfull of microvolts wide? We use external circuitry to take our large signals of interest and transform them into signals within this range. We also design the circuits so that the output of the opamp interacts with the inputs so as to stabilize the behavior according to our goals. This process is known as using "negative feedback".

    Let's take a real simple example.

    Connect an input signal to the noninverting input and set it to 2V. Now connect two identical resistors, in series, between the output and ground forming a voltage divider in which the voltage between the two resistors is half the output voltage. Now apply that divided voltage to the invering input.

    Let's say that the output voltage is initally at 0V. We have a differential voltage of 2V, meaning that out opamp would like to make the output voltage 2,000,000V. Of course, it can't do this but it doesn't know that so it heads off in that direction and the output voltage starts to increase. At some point it is at 1V and now the opamp only wants to make the output 1,000,000V, but since that is still greater than what the output is actually at, it keeps increasing the output voltage. At some point the output gets to 3.9999V resulting in the voltage at the invering input being 1.99995V for a differential input voltage of 50μV so the opamp want to put out 50V, which is still greater than it's current ouput so it continues increaseing the voltage. When the output reaches 3.999992V, the voltage at the noninverting input is 1.999996V making the differential input voltage 4μV and the desired output voltage 4V, which is very close to what it actually is, but still not quite; so it continues to increase the voltage.

    However, when the output reaches 3.999992000016V, the resulting differential voltage is exactly what is needed to make the opamp want to put out the same output that it is already putting out, so it doesn't change the output.

    Should the output overshoot and create a slightly higher output voltage than this, the opamp would respond by wanting to output a lower voltage than it is currently outputting and the voltage would come down. For example, let's say the output went to 3.999993V. This would result in an input differential voltage of 3.5μV making the amplifier want to put out 3.5V, thus resulting in the output wanting to move downward.

    The key is that whenever there is an error between what the amplifier is actually putting out and what it should be putting out, that the external network (the voltage divider, in this case) results in the creation of an error signal that is multiplied by the open-loop gain of the amplifier and then subtracted from the current output. This is the essense of "negative feedback".

    Thus, the output will stabilize at a value that is, for all intents and purposes, 4V. But notice that it won't stabilize at exactly 4V, because that would result in a desired output of 0V and the opamp output would head downward. Therefore, we see that we have to have a non-zero error signal in order to drive the output to a value that is very close to the desired value. The amount of the error signal that is needed is simply the final output voltage divided by the open-loop gain of the amplifier and, consequently, the higher the gain the smaller the error.

    If the voltage at the noninverting input were to change to 5V, this process would be repeated and the output would stabilize at a value of 10V (not exactly, but so close that it would be very hard to tell the difference without a lot of effort). Whatever the voltage at the noninverting input, the voltage at the output will qucikly change so that it becomes twice that voltage, and it will generally take less than a microsecond to do so, hence for most purposes it can be considered instantaneous. Thus, with just two resistors, we have taken a device that has a gain of one million and and a linear region of just 12μV and create a circuit that has a gain of 2 and a linear region of 6V -- a much more practical and useful set of numbers.
     
    Last edited: Aug 17, 2013
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  12. LvW

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    I think, the part of the question regarding the "variable resistor" was not yet answered up to now.
    Of course, you can have a variable output (at a constant input signal) using a variable resistance within the feedback chain. For example, this can be achieved using a FET in its linear region (used as a voltage controlled resistor) or other parts like thermistors or LDR`s.
    Remember, the first WIEN oscillator circuit used a filament bulb for realizing an amplitude controlled gain.
     
  13. WBahn

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    While I agree with the good point you are making, I read the original question as implying that the opamp has to be on/off unless something like this is done. Now. what was meant by "controlled by an variable resistor" is unclear. Does putting a pot from output to ground and using the wiper voltage at the output qualify?
     
  14. studiot

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    Until the OP replies to the question in my post#8, you guys may be putting in a lot of abortive effort.

    I am not sure that his 'comparative circuit' is meant to be a comparator.

    From his description and subsequent questions it could easily be a differential amp providing an output proportional to the difference between a reference and some monitored sample as in a voltage regulator.
     
  15. WBahn

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    Oh, I tend to agree.

    But these things do seem to take on a life of their own, don't they? :D
     
  16. studiot

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  17. mad_mat222

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    Aug 14, 2013
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    Hi all and thanks for advice so far. I don't want to use a pot I want to use a thermistor as the temperature sensor. Further searches of goggle are leading me to look at chips such as the LTC1392 which might make a thermistor controlled with a variable output a little easier. These circuits exist eg digital thermometers. There are obviously sensing temperature and then putting out a variable output which is then used to control some sort of display.

    I want to do the same thing but use the output to control another circuit. Whether I have to use op amps or transistor or micro power I don't care, just has to be functional and reliable.

    Thanks in advance. Hopefully this helps tames the thread same and prevents it taking a life of its own:)
     
  18. WBahn

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    Please take a step back and tell us, in nontechnical terms, what you NEED to accomplish. What is your underlying problem that you are trying to solve? For instance, "I need to monitor the temperatures at two points in a tank and turn on a circulation pump whenever the difference exceeds 10°C."

    If possible, try to list your: NEEDS, CONSTRAINTS, WANTS, IDEAS. At that point, we are in a position to understand your questions and to ask appropriate clarifying questions in return.
     
  19. mad_mat222

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    Aug 14, 2013
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    Wanting to measure temperature between 30 - 60 degrees. The output needs to be between 2.5 down to 0 v as it gets hotter. I got to the voltage dividing circuit to linearlise the output and no further. Options seem to be using transistors, a chip or op amp. As long as it works I'm not fussed on what's in the circuit.
     
  20. Rbeckett

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    Scott,
    I appreciate the links to the info on OpAmps, but the question he was asking is a bit more than just reading a DS for data. I believe he is trying to ask if he can use an opamp to create a voltage follower or analog output. I may be wrong, but those Datasheets probably wont help solve his question very well and may confuse him even further. Studiot was onto the answer in his reply I think.

    Bob
     
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