op amp supply voltage

Discussion in 'Homework Help' started by notoriusjt2, Oct 31, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    [​IMG]

    for the op amp to work correctly the -V supply on the op amp should be -9.0V right? since the +V is equal to +9V
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    One way to determine the answer is to get the data sheet for an ICL7660.
     
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,783
    943
    Isn't there a misdrawn component there?
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,647
    2,346
    Hello,

    I see no problems with the circuit.
    Look at the two positions of the 7660, to see how the capacitors are switched.

    Bertus
     
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,783
    943
    I'm switching between a datasheet and the circuit on this page now and looking at the capacitor connected to pins 3 and 5 of the chip I believe there is a misdrawn component on this circuit. Specifically the cap is backwards.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    See the datasheet for the charge pump IC, or, draw the 2 states of the internal switches as separate circuits. Pay attention to what charges are on the capacitor with what polarity. Compare the two circuits (one for switches in state 1, one for switches in state 2). What happens?
     
  7. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,783
    943
    http://www.intersil.com/data/fn/fn3072.pdf

    circuit as drawn on page 6 of the pdf is what I have referenced as being different from this one. Is it a different circuit, that still gives a negative voltage? One of these two has a cap backwards. Who wants to build both and find out which is wrong? :)
     
  8. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    on page 6 of the PDF, it states

    VOUT = 2VIN

    so Vout = +18V?
     
  9. bertus

    Administrator

    Apr 5, 2008
    15,647
    2,346
    Hello,

    When you look at the circuit, the capacitors are now in series.
    In the other position they are parallel.
    So what will be the negative voltage?

    Bertus
     
  10. DonQ

    Active Member

    May 6, 2009
    320
    11
    Non-polarized caps (no + - signs). The curves on the caps, when used correctly, show which winding of a rolled cap is on the outside. On low-voltage, non-polarized caps, it is not an issue.

    But this is a very non-standard way of using a 7660 (see data-sheet. Really, look closely). I won't elaborate yet because this is homework, but you need to not just believe the schematic, or that the designer knew what he was doing (as is much too often the case).

    Does it work this way? Maybe.
    Should it ever be used this way in a real circuit? No!

    It will require knowing how the 7660 works, but somewhere in the world, someone will find this in their notes from when they were in school, and design this into something important.

    Not on my watch!
     
  11. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    so when the switches are in the other position, the capacitors will be in series. i see that, but i am getting lost trying to follow that out...

    the power comes in the AC source, then flows through the 150k resistor, then down to terminal 3 or 5. im confused
     
  12. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    The signal/AC Source is not connected to the charge pump, the lines simply cross. There isn't a "\bullet" (standard) or "\circ" (in this schematic) to indicate a connection. The signal only goes to the non-inverting input of the amp.

    Follow the path of +9v, Ground, and the -V output.
     
  13. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    9V goes to the V+ of the op amp and terminal #8 of the IC. terminal #8 on the IC is the Vin for the IC. that is going to be there no matter which way the switch is flipped.

    Ground goes to terminal #5 on the IC which is also the Vout

    the -V input goes to terminal #3 on the IC, which based on the IC datasheet should be a Ground.

    so when the switches are in the other position, would power be flowing into terminal #5?
     
  14. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    Don't look at the datasheet labels for the IC pins. Look at the "internal picture" in your schematic above.

    +9V goes to one side of a switch, then back out to a capacitor, and continues.
     
  15. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    yes i can follow that one. +9V comes in #8, then to #2, through a cap, back in #1, up to #3, then up to V- on the op amp

    now when the switch is in the other position

    +9v comes up to #8 and just sits there. the two capacitors are now in parallel and are supplying the V- of the op amp. the 2 caps in series are equal to 4.4uF, but how do I know what kind of voltage they are supplying? would both of them be storing a full 9V charge?
     
  16. bertus

    Administrator

    Apr 5, 2008
    15,647
    2,346
  17. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    im sorry i meant to say "the two capacitors in parallel". but still the C total would be equal to 4.4uF.
    [​IMG]
     
  18. bertus

    Administrator

    Apr 5, 2008
    15,647
    2,346
    Hello,

    How will the voltage distribution be when the capacitors are in series?
    Try to follow the voltage path from V+ to ground.

    Bertus
     
  19. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    [​IMG]

    i drew arrows in where i think the current path is. where does it go from where i left off?
     
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    As I see it, the way the schematic is drawn, V- will equal +4.5V. The way the 7660 datasheet is drawn, it will yield -9V.
    I think either pins 3 and 5 are erroneously swapped, or it is a trick question. I am leaning toward the latter. It can sort out the student who is thinking, and understands the concept, vs one who just looks at the datasheet and parrots the answer.
     
    Last edited: Oct 31, 2010
Loading...