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Op Amps

Discussion in 'Electronic Basics' started by Periproct, Mar 17, 2006.

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  1. Periproct

    Periproct Guest

    This has got to be simple for someone who knows.

    I'm a beginner in all this but I've found Livewire software that allows me
    to simulate my circuit including blowing up components.

    I'm trying to detect a voltage dropping below battery voltage in a car.
    Rightly or wrongly I thought an Op Amp (741) with the voltage I want to
    monitor going to the + input and the full battery voltage going through two
    resistors as a potential divider giving me the voltage I want the op amp to
    switch at, going to the - input. The output is going to the input of an opto
    isolator via a current limiting resistor. I'm powering the op amp directly
    from the battery voltage.

    Simulating this in Livewire gives me exactly what I want. When the monitored
    voltage on the + input drops below the voltage supplied to the - input via
    the potential divider I get an positive output from the op amp. Playing
    around with this I've found that if I drop the battery voltage that powers
    the op amp by as little as half a volt below the voltage on the + input,
    Livewire shows both the op amp and opto isolator as destroyed. In reality
    this situation should never arise as there is theoretically no way the
    monitored voltage can get above the battery voltage but half a volt seems a
    very fine margin.

    Would the op amp be destroyed in reality by having half a volt higher
    voltage on one of its inputs than the voltage than the voltage that is
    powering it or is this just a glitch in Livewire?
     
  2. redbelly

    redbelly Guest

    Is that a typo? You should get a low voltage out when +in is lower
    than -in. Might be 1 or 2 volts, assuming the "- supply" to the amp is
    simply ground in your circuit.
    I don't know from experience, but according to the datasheet (National
    Semiconductor version):

    "For supply voltages less than ±15V, the absolute maximum input
    voltage is equal to the supply voltage."

    Regards,

    Mark
     
  3. Periproct

    Periproct Guest

    Is that a typo? You should get a low voltage out when +in is lower
    than -in. Might be 1 or 2 volts, assuming the "- supply" to the amp is
    simply ground in your circuit.

    Yes, that was me getting it backwards.

    I know the op amp should be powered from positive and negative rails but
    seems to work fine with just my battery and ground supply.
    I don't know from experience, but according to the datasheet (National
    Semiconductor version):

    "For supply voltages less than ±15V, the absolute maximum input
    voltage is equal to the supply voltage."

    Now I feel stupid. I though I'd read the data sheet thoroughly but I must
    have missed that bit. I guess I'll have to find a way of dropping the
    voltage I'm monitoring a shade 'just in case'.

    Thanks for clarifying that
     
  4. Chris

    Chris Guest

    Good morning, Periproct. Where to begin?

    First, most of what you need to know about an IC is usually in the
    manufacturer data sheet. If you hunt around at supplier websites, you
    will probably find a link to the datasheet there. After you have some
    familiarity, you'll find yourself going directly to the manufacturer's
    website for the data sheet. One you'd look for is:

    http://www.national.com/ds/LM/LM741.pdf

    The LM741 data sheet states that, for supply voltages less than +/-15V,
    the absolute maximum input voltage is the supply voltage (Note 4). Abs
    max means the manufacturer doesn't guarantee the chip will survive
    anything beyond this.

    Most ICs are set up so that all inputs are made to operate between the
    supply rails. Due to the way they're made, there's a hidden connection
    between inputs and the substrate of the chip, which acts like an SCR if
    supply rails are exceeded at an input. This latchup will cause high
    continuous power supply current to flow inside the chip, even if the
    input is removed. This is called parasitic latchup, it destroys the
    IC, and is avoided at all costs.

    If you consider that all wires have inductance and are inherently noise
    pickups, and there's trace capacitances just about everywhere, it's
    trivially easy to get ringing or noise on signal lines that bring
    inputs past the supply rails in an electrically noisy automotive
    environment. Given enough of a pulse, a microsecond is all it takes.

    As a practical matter, parasitic latchup is a current-driven event. If
    you can limit the voltage and then limit the current if you get an
    excursion above or below the supply rails, there isn't enough current
    to trigger the SCR. Here's one way to avoid this (view in fixed font
    or M$ Notepad):

    |
    | VCC
    | +
    | 1N4001|
    | - VCC
    | ^ 220 +
    | Vin | ___ |\|
    | o----o-|___|--|-\
    | | | >-
    | - .-|+/
    | 1N4001^ | |/|
    | | | ===
    | === VCC | GND
    | GND + |
    | | |
    | .-. |
    | | | |
    | | | |
    | '-' |
    | | |
    | o--'
    | |
    | .-.
    | | |
    | | |
    | '-'
    | |
    | ===
    | GND
    |
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


    Now, the LM741 isn't really made for automotive applications, anyway.
    It's made for split supplies (e.g. +/-15V). You generally have to keep
    your inputs more than 2V away from the power supplies (that means
    between +2V and +10V for a single 12V power supply). If you don't
    operate the inputs within that range, the op amp may not operate
    correctly, even if it won't smoke itself. Also, the output is also not
    made to go to the supply rail, so the output will typically go between
    1V and 10V for a 12V supply with a 10K load resistor to GND.

    Another series of op amps was made to operate on single supplies. The
    LM324 (quad) and LM358 (dual) op amps became popular because they are
    made to operate on single supplies, the IC will still operate properly
    with the inputs going all the way down to GND, and the output can go
    down to GND with a pulldown resistor at the output. This will give you
    a lot more latitude for automotive and single-supply applications.
    Both these ICs are available everywhere -- the LM324 is even sold at
    Radio Shack.

    The LM741 is a bit of a relic, and has only a couple of manufacturers
    left. The LM324 and LM358 are made by many manufacturers, and cost
    significantly less in part because of volume. They're truly gumball
    parts.

    Also, you should know that there's another type of IC similar to op
    amps, that's made specifically for comparing two voltages and giving a
    "1" or "0" output. They're called comparators, and are made for high
    gain, and fast transistions between output logic states. The single
    supply equivalents to the above op amps are the LM393 and LM339
    comparators. For most comparators, you need a pullup resistor from the
    output to the power supply. Using op amps as compators can be a
    problem with most optoisolators, because the slow transitions can mean
    intermediate voltage levels may be read as a "1" or a "0". However,
    they do have some advantages, particularly if you can live with lower
    speed. They're made to operate in a linear mode, are less susceptible
    to oscillations, particularly if you design in a little hysteresis, and
    may be able to sink more current than a comparator. For instance, if
    you're driving a relay, they can be very useful because relays always
    operate in the millisecond time range, and power supply glitches from
    the relay turning on and particularly off can result in oscillation,
    causing relay chatter.

    Still, let's see if there's a safe way to use a 741 as a voltage
    comparator driving an opto for your automotive application, following
    all the rules for the 741:

    |
    | VCC VCC
    | + + 4N28
    | VCC | '----. o
    | + R2.-. LM741 | |
    | D| 10K| | | |/
    | - | | VCC V ~ .-|
    | ^ R1 '-' + - ~ | |>
    | Vin | ___ | |\| | | |
    | o----o-|___|-o--------|-\ ___ | | o
    | D| 10K | | >---o-|___|---' |
    | - R3.-. .---|+/ | R7 |
    | ^ 10K| | | |/| | 680 o
    | | | | | GND |
    | === VCC '-' | ___ |
    | GND + | o---|___|--'
    | | === | 1M
    | R4.-. GND | R6
    | 10K| | |
    | | | |
    | '-' |
    | | |
    | o--------'
    | |
    | R5.-.
    | 10K| |
    | | |
    | '-'
    | |
    | ===
    | GND
    |
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    Let's assume for the sake of discussion that you want input voltages
    from the power supply rail down to GND, with your comparator switching
    point to occur at 1/2 the supply voltage, or 6V (based on a +12V
    supply). Let's start out by ignoring R6. So, with no input, R2-R3 and
    R4-R5 bias both inputs at 6V, or 1/2 the power supply voltage. But
    there is an input, limited by the two diodes to -0.6V and 12.6V (under
    normal conditions, 0 to 12V). This is summed at the inverting input,
    so the voltage there will be 4V to 8V. As the input voltage goes past
    6V, the output state of the op amp will change.

    Now, let's look at R6. Assume for a minute that the 741 output is high
    (12V, for the sake of discussion). Then the reference voltage at the
    non-inverting input will be 6.03V. Now, the voltage at the - input
    slowly goes up past 6.03V, and the output goes down to 0V. The
    reference voltage will then snap down to 5.97V because of the voltage
    divider. This ensures that the transition between logic "1" and logic
    "0" takes place quickly, without oscillations. This can be very
    helpful if you get a glitch on the 741 power supply or GND because of
    the current drain from the opto.

    The output of the 741 (which is actually current limited at 20mA, you
    don't absolutely have to have R7 but it's good form) will then drive
    the opto, giving you your clean comparator signal.

    The extra resistors and diodes seem like a little much, but they are
    there for good reasons. And it's possible to do this a little more
    elegantly, but this basic circuit is easy for newbies to understand,
    and will work.

    I hope this has been of help. If you have further questions, please
    feel free to post again.

    Good luck
    Chris
     
  5. John  Larkin

    John Larkin Guest

    Some opamps have esd (static electricity zap protection) diodes from
    their inputs to their V+ terminals, and most opamps have inherent
    substrate diodes from inputs to V-.

    So if you apply a voltage to an input pin that's more positive than
    V+, current can flow through the diode. If the thing driving in+ has a
    huge current capability, that could damage the amp.

    But where will you get the reference voltage from? If it's some dinky
    voltage reference chip, or a zener, or a forward-biased led or
    something, it probably can't dump enough current to hurt the amp.

    You could always add a series resistor to the pin in question to limit
    the current. If the ref is derived from the battery itself, as you
    note, it just doesn't happen.

    Some opamps, like an LM324, can tolerate inputs way above the + rail.
    An LM324 (crappiest opamp on the planet, maybe) gets very weird if any
    of its outputs are pulled even gently below V-.

    I wouldn't power any IC directly from a car battery. Nasty transients
    happen in cars, including excursions to high + or - voltages, that
    tend to wipe out unprotected ICs.

    John
     
  6. "Periproct" Said
    ----------------------------------------

    Ordinarily you should apply a scaled down version of the monitored voltage
    to the input of the op amp. For instance you could use 1.2 volts from an IC
    voltage reference or a 5 volt zener diode with an appropriate voltage
    divider to get 1.2 volts. Now apply a scaled down version of the 12 volts
    you're monitoring to a voltage divider so that a 1.2 volts input to the op
    amp is actually 12 volts on the input of the voltage divider or 1/10 of the
    monitored voltage. If your monitored voltage is greater than 12 volts the op
    amp will swing positive and if the monitored voltage is less than 12 volts
    the output will swing negative, or to ground if you're not using a negative
    power supply. In this way the op amp input voltage can never exceed its
    power supply value.



    If you're worried about having to use dual power supplies simply use a
    comparator, which is more fitting for this application, instead of an op
    amp.



    Use a large value decoupling capacitor (e.g. 1000 uf) across the op amp
    power supply fed through, let's say, a 47 ohm resistor or better still a
    small choke and this will help filter out any nasty transients from the cars
    electrical system.



    Dorian
     
  7. Rich Grise

    Rich Grise Guest

    If I was serious about something like this, I'd use an isolation diode to
    the + 12V, then say, that 47R, then about an 18V transzorb, then the cap.
    That way, when the battery droops, it at least won't pull the charge out
    of the cap. And everybody's heard of the dreaded Load Dump. ;-)

    Cheers!
    Rich
     
  8. Rich Grise

    Rich Grise Guest

    And I'd use a comparator instead of an opamp, of course. :)

    Cheers!
    Rich
     
  9. Periproct

    Periproct Guest

    Thanks to everybody who took the time to reply. I really appreciate you all
    taking the trouble.

    I'm a great fan of this Livewire software I've found. Really easy to use and
    the fact it will 'blow' a component helps an amateur like me. Unfortunately
    it doesn't include a comparator in its component library.

    I'll be taking on board all your advice and eventually I'll get my little
    circuit up and running.
     
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