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LM358 strange behavior !

Discussion in 'Electronic Design' started by [email protected], Nov 4, 2012.

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  1. Guest

    Hello,

    Excuse my english, I'm French ...

    The LM 358 is a mono-voltage amplifier capable of approaching its negative supply (there is a NPN transistor in the output stage). I had to build a mono-voltage subtractor (a voltage substractor built with a LM358 powered with 0v-12v supply) and with the consent of the electronic simulator ISIS I built (wired) such a subtractor. I was surprised that it could not entirely do the job. For some subtractions, like 2v - 1.8v for example, the substractor said 0.62v instead of 0.2v. But for some other, it answered the right thing, wich was 0.3v (a voltage lower than 0.62v) sometimes. So it was ableto go under 0.62v !

    Why does a 2v - 1.8v substraction gave 0.62v instead of 0.2v !


    Resistors, all identical, were 2.5k one, so I do not suspect bias currents problem. I tried many other chip, it didn't change anything. I changed the 0v suply into a -12v one and the substractor worked well for every input voltage.


    Does someone have an explanation?

    cdlt.

    Michel.
     
  2. Guest

    Le dimanche 4 novembre 2012 11:44:51 UTC+1, John Fields a écrit :

    Ok. It is the basic soustractor schématic (not the instrumentation one with 3 amplifiers), with unity gain, visible here :

    http://crteknologies.fr/electronique/cours/ao_soustracteur.gif

    In my case, R1=R2=R3=R4=2.5 k, and Us = U2 - U1.

    Michel.
     
  3. Guest

    Your feedback resistors (R1-2) are asking the LM358's
    output to sink too much current. Close to ground, its
    ability is quite limited. See "Output Current Sinking
    Characteristics" on page 7 here:
    http://www.ti.com/lit/ds/symlink/lm358-n.pdf

    The reason for the limitation can be understood easily by
    looking at the LM358 schematic on page 20--the output stage
    is an emitter follower in parallel with a current sink. Only
    12uA is guaranteed at 200mV, Vcc=+15v ("Output Current," pg. 4).

    Cures: use larger resistors, or connect a load resistor
    or a current sink to the output.
     
  4. Jamie

    Jamie Guest

    Try putting a load R on the output, like a 10k or something in that
    range. You maybe in the cross over zone and that op-amp does not have
    much for shoot-through current.


    Jamie
     
  5. John S

    John S Guest

    Oh, good for you! Recommending James Arthur's suggestion as your own is
    sheer genius.
     
  6. John S

    John S Guest

    Oops! In your vernacular, it should have been "Recommending James
    Arthur's sugesstion as your own is shear genus."
     
  7. Jamie

    Jamie Guest

    I don't know what the **** you're talking about, I have no idea who
    James arthur is and I don't give a shit. Maybe you got some infatuation
    with him, I don't, who ever he is.

    The fact that a 324, 358 having those issues being common knowledge
    as it is, leads me to believe that you're just a sit in stooly. Yes,
    one that is so full of shit that I smell it all the way over here.

    Oh, btw, incase you want to expand that fake no hands on knowledge of
    yours, a 358/324 etc can also oscillate if the output is brought down to
    or near zero output with no load on it.

    What a piece of shit you are. You belong with Phil, down under, the
    table, that is.

    Jamie
     
  8. Robert Baer

    Robert Baer Guest

     
  9. The LM358 has an input CM range that is guaranteed to go down to the
    negative rail over the full temperature range. One of the things that
    makes it so useful for an extremely cheap and common op-amp (it's half
    an LM324). So, it's not that.
    The LM358 is guaranteed to drive the output within 20mV of the
    negative rail (5mV typically) over the full temperature range, with a
    <=10K load (critically important, but perhaps not obvious from the
    data sheet) connected to the negative rail. The subtractor has < 10K
    of load but does not have that configuration of loading on the output.
    That's his problem.


    Best regards,
    Spehro Pefhany
     
  10. Guest

    His R1-R2 combination presents 5k from -Vin to output, so he'd need to
    sink 320uA for his example (Vin+ = 2.0v, Vin- = 1.8v, Vout = 200mV).

    R1=R2=100k would go a long way toward fixing that.
     
  11. Which overwhelms the 50uA nominal internal sinking capacity.

    Sure, and make the other two 100K as well to kill most of the offset
    due to bias current.


    I suppose one could do something like this and keep the 2K5 resistors:

    |\|
    -|-\
    | >-----+----
    -|+/ |
    |/| |
    |
    |
    |
    |
    1M8 |
    ___ |/
    +12 -|___|--| 2PC4081R
    |>
    |
    |
    ===
    GND

    LOL.
     
  12. Guest

    LOL!

    I nearly suggested a JFET load, Vgs=0v. Bigger resistors was easier
    though.
     
  13. Fred Bartoli

    Fred Bartoli Guest

    a écrit :
    Hey, a resistor to a negative supply would be nice too :)
     
  14. Guest

    Sure, if you've got one. (a negative supply, that is.)

    Bob Pease did a neat trick--or was it Jim Williams?--using the e-b
    junction of a powered optoisolator as a negative current generator.
    The phototransistor served as a solar cell, to pull an LM358 output
    all the way down. Cute.
     
  15. Fred Bartoli

    Fred Bartoli Guest

    a écrit :
    Was joking of course. If you have a neg supply then just use it for the
    opamp.
    I think it was Bob Pease but I doubt this has enough current for an LM358.
     
  16. Guest

    Missed it--whoosh. I often don't use negative supplies for V- for
    other reasons, like, it might be dirty, or to save power, or reduce
    loading on V-, etc.
    Could be. If an ordinary phototransistor isolator has a current
    transfer ratio of 20%, and the phototransistor has a hfe of ~100, that
    suggests i.b ~= 0.2%. So, 10mA into the IRLED could make 10 or 20uA.
     
  17. Guest

    That's not a hazard, it's a feature--crowbar. LM34, right?

    Grins,
    James
     
  18. John S

    John S Guest

    Exactly! I like Phil because his knowledge exceeds Jamie's (Maynard's)
    by several orders of magnitude and he communicates it clearly. I have
    learned much from him and I suggest Jamie try to do the same.
     
  19. John S

    John S Guest

    Well, no wonder that you never learn anything. He is a prolific poster
    here and provides great information.

    Maybe you got some infatuation
    Infatuation? Is that your practice word of the day?
    I must ask, what is a sit in a stooly? I don't think I've ever done
    anything like that. It must be something with which you are familiar and
    I am not.
    Well, I've already replied that I would rather learn from Phil than see
    your junk.
     
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