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operational amplifier circuits etc.

Discussion in 'Mathematics and Physics' started by PG1995, Apr 19, 2014.

  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Looks like the OP290 has better input offset spec, a little better output swing at 5v, but just very low gain bandwidth.
    So for low frequency stuff it would be better with the better specs.
     
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  2. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    Which parameter(s) from the datasheet led you to say this that OP290 has a little better output voltage swing? Could you please let me know?

    I don't really know how to read an op-amp's datasheet so could you please help me with these queries? Thank you.

    https://drive.google.com/file/d/0B_XrsbDdR9NEZmtPMUNGaENobkk/edit?usp=sharing (LM158/LM258/LM358/LM2904)
    https://drive.google.com/file/d/0B_XrsbDdR9NETGpkQW9hOWU3QXc/edit?usp=sharing (OP290)
     

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  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    When i look at the output high for the LM358 it is normally taken to be Vcc-1.5v, but the data sheet there shows it could be as high as Vcc-1.2v which with a 5v supply means the max output will be 3.8v with RL=infinite.
    For the OP290, it shows worse case 4.0v with typical 4.2v so that's 0.2 to 0.4v higher and that is with a 10k load.

    Q&A:
    1. Either the chip has a built in internal booster circuit (ha ha ha) or else it is a typo
    2. Yes, Vout max=26v with V+ of 30v
    3. Yes.
    4. Right. See if you can find a Vout graph in that case on the data sheet (or a Vout delta graph).
     
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  4. dave

    Dave New Member

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  5. steveB

    steveB Well-Known Member Most Helpful Member

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    Hahaha Mr. Al,

    Typo for sure on #1. Cut and paste gets us all sooner or later. :)
     
  6. PG1995

    PG1995 Active Member

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    Thank you.

    In this post you can see that Vout=4.67005V for -2.3A. For example, if we are using LM358 with 5V supply then 4.67005V falls out of LM358's output range. But LM358 can be used with supply voltage of 5V to 30V therefore using supply voltage such as 12V will extend the output voltage range and hence 4.67005V can easily be output. Please correct me if I'm wrong.

    In this circuit you can see that three capacitors are being used. Let's focus on 1000pF one. In the given circuit the output voltage, i.e. Vout, won't go above 5V even in worst case scenario therefore ideally "1000pF, 5V" capacitor would do. A "1000pF, 5V" capacitor simply means that the voltage of capacitor can be risen to 5V without any damage to the cap, and it takes 1000pC to raise the voltage by 1V. So, using a "1000pF, 20V" capacitor instead of "1000pF, 5V" wouldn't cause any problem in operation of the circuit, right? Thank you.

    Regards
    PG
     
  7. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes if the output should be able to get to a full 4.7 volts in normal operation then it would not work with a 5v supply unless the op amp had a near rail to rail output. The LM358 can only get up to about 3.5v, so that would certainly fall short.
    With a 12v supply we should easily be able to get up to 10v output with say 10ma max output current. That would be a lot better.
    Another thing we did not look at yet either was the input range. The input range may be also limited depending on the op amp. You can get this info from the data sheet too.

    Yes using a 1nf cap rated for 20v is going to be better than using a 1nf cap rated for 5v in almost every application except one where the space is limited and then we might be able to find a 5v cap that is smaller than a 20v cap.
    I do have reserves about using a 5v cap in a 5v application however. I would rather use something higher like 7v or 10v for some extra safety margin. Here we might catch a break though if the output could really only produce 3.5v, as we might find that a 5v cap works ok. We'd have to check the possible range of the voltages on both leads of the cap carefully and that would include power suppl turn on and turn off modes.
    If the power supply was raised to +12v, we'd have to consider the fact that the output may get up as high as 11v if there was some sort of temporary analog error either during turn on or turn off or during normal operation. Then a 10v min rating might be a good idea.


    Steve:
    Yes, ha ha, they forgot to turn on *voltage regulation* (compare to *spelling correction*) in their *text editor* <chuckle>
     
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  8. steveB

    steveB Well-Known Member Most Helpful Member

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    PG,

    I agree with Mr Al about using higher voltage caps. My rule of thumb is to use twice the rated voltage. To appreciate why this is important, take a look a the the MIL-HDBK-217.

    http://snebulos.mit.edu/projects/reference/MIL-STD/MIL-HDBK-217F-Notice2.pdf

    Although reliability depends on which capacitor type you care dealing with, a factor of 2 safety margin in voltage is a good general rule that dramatically improves reliability of capacitors in operation. Granted you are not trying to make a reliable product, but if a cap fails during your demonstration, you won't be a happy camper.
     
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  9. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Yep, 2x. I built an amp from scratch and two 100 uF electrolytic were rated at 50V which was the approximate supply voltage. I had pieces of cap everywhere when two exploded. Later, I put MOV's on the main filter caps which were 4 x 9600 uF and I THINK 85 V surge. My slow/turn-on and protection circuit was getting fried.

    Exploding caps on purpose, was fun in another era.
     
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  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Wow, that takes me way back. One class i was in we used to connect various components like resistors to the two leads of a variac and hang the component (on the long leads) out the window. Then we'd switch on the variac and give the class right below us a thrill as the component blew up right outside their window <chuckle>. They loved it :)
     
  11. PG1995

    PG1995 Active Member

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    Hi

    The datasheet of LM358 says:

    I believe it means that in case of single supply operation the supply voltage should be between 3V and 32V for normal operation. Is this correct?

    In case of dual supplies it simply means that minimum supply voltage should be +1.5V and -1.5V and maximum should be +16V and -16V. Is this correct? Thank you.

    Regards
    PG
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes that sound right to me. The usual supplies would be plus and minus 15v though not plus and minus 16v, but they are indicating that it could take plus and minus 16v, and if you look down farther in the data sheet they say that is an "absolute maximum" so it's not really a good idea to run it that high.
     
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  13. PG1995

    PG1995 Active Member

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    Hi

    I was using this set up to apply dual polarity voltage to an op-amp but the regulator highlighted in red smoked and blew away. What was wrong with the set up? Please let me know. Thanks.

    By the way, I asked about the similar set up in post #18 and in post #19 JimB okayed it.

    Regards
    PG

    Update: There is obviously something wrong with the set up of regulator in red highlight because I just checked another regulator in similar configuration and it became so hot with in just a fraction of second.
     

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  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yeah, there's no negative supply there anywhere so you cant get a voltage that is actually LESS than ground. You would need a 7905 regulator and a negative supply.

    If you want to try it with two 7805's then you have to set the second one up such that its ground terminal connects to the OUTPUT of the first 7805, and that same output becomes the new ground for the op amp. That gives you plus and minus 5v where ground of the 18v supply looks like -5v to the op amp, and the output of the second 7805 looks like +5v to the op amp, and the common connection of the two looks like ground to the op amp.
    You may also have to load the first 7805 a little more with a resistor like 100 ohms so that it's pass transistor is always conducting even if the upper (second) 7805 supplies it with a little bias current through the output terminal of the first. If any other new ground connections can produce a current INTO the output of the first 7805, the 100 ohm resistor may have to be reduced in value somewhat.
     
    Last edited: Aug 12, 2014
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  15. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    yep, you either need two isolated 5 V supplies or a stacked isolated supply as MrAl said.
    You may have to supply a minimum load of 5-10 mA.

    So, you have the required 3V; e.g. 3+3+5+5 or a minimum of 16V.
    The Common point would not be ground in this case, but it's like stacking two 9V batteries to get a +-9V supply.

    Save the magic smoke. It might come in handy.
     
    Last edited: Aug 12, 2014
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  16. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    I have never used 79xx series. Why I would need a negative supply. I believe that all I need is a 7905 and a power supply. For example, If I use a +18V power supply to power a 7905 then I can get -5V out of it, can't it?

    Is this what you are suggesting? Please give it a careful look because I don't want to blow out another part.

    Where should the ground of Reg #1 be connected to?

    Regards
    PG
     

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  17. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Your attachments can't be found. Reminder, check the power dissipation (18-5) * 0.1a) IS 1.2 w.

    I added a few words in bold. make sure there is no unintentional connection to earth ground. Many time the - is connected to ground.

    See my post #94
     
  18. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    With two separate ISOLATED 18V supplies, not with one. Make sure the outputs have no connection to the AC ground pin.
     
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  19. PG1995

    PG1995 Active Member

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    Thank you.

    I'm sorry but I don't really get it. Could you please edit my diagram to show what you are really saying? You can edit in MS paint.

    I have got only 18.5V power supply and it's basically an old laptop charger. I have 7805 and 7905 regulators. Can't it make it work using a single supply and the given regulators? I can't get another supply. Thanks.

    Regards
    PG
     
  20. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Me and drawing is kinda tough now. Scanner isn't fixed either.

    Take two 7805 regulators and put them in series. Vin(1) to 18+; Gnd(1) to Vin(2); Gnd(2) to 18-

    Take +5 from Vout(1) and -5 from Vout(2). Make the common point the junction of Gnd(1) and Vin(2).
     
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  21. JimB

    JimB Super Moderator Most Helpful Member

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    You got me very worried there PG! :eek:
    There is actually a BIG difference between the circuit in post#18 and what you have used here.
    In post #18 the circut implies that there are two separate and isolated 9volt supplies. If there are, the circuit will work.

    Connecting the circuit of post #18 to a single supply will result in the release of magic smoke.

    JimB
     
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