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input impedance output impedance

Discussion in 'Electronic Basics' started by [email protected], Jul 12, 2007.

  1. Guest

    I'm a newcomer to analog electronics.
    Currently, what I'm studying is Op-amp circuits.
    The term I'm confused about is Input & Output impedance concept.
    For example, input impedance is differenf for inverting amplifier and
    non-inverting amplifier.
    How do you define the input impedance or output impedance?
    and what does the term "Impedance matching mean?"
     
  2. Robin

    Robin Guest

    If an input is "high impedance" it means that is easy to "drive" i.e.
    easy to push around, in the sense that powered steering is easy to
    handle, you don't need much muscle power to move it.

    So generally speaking, inputs will tend to be high impedance because
    then they will have more tendency to do what they are told as opposed
    to a lower impedance input e.g. unpowered steering.

    A "high impedance" output is a "weak" output, a weak drive. If you
    used a high impedance output to drive a low impedance input (old man
    trying to turn non-powered steering) then the movement transferred
    will be reduced or to use the jargon, the ouput will be loaded by the
    input, and subsequently the signal will be loaded and reduced in size.

    So it is usual to have e.g. the output of a HiFi amplifer (0.0001
    Ohms) drive a loudspeaker of 4 Ohms. By this design, the loudspeaker
    will do exactly as it is told being driven by an impedance very much
    lower than its own.

    Now a big confusion is "matched impedances". When a strong guy turns
    powered steering, he does not expect to get tired and he won't because
    his low impedance is not loaded by the high impedance reaction of the
    wheel. There is no intention to depend on the man's strength. The
    power is provided by the petrol. But what if there is no petrol? As in
    e.g. in a bicycle now it is important to get *power* from the man to
    the machine. When the road speed is low, a gear selection is required
    to match the maximum output of man to machine i.e. the impedances are
    "matched" but as he road speed increases so the peddle speed reaches a
    maximum that the man can manage and no power is transferred (the
    impedances have become mis-matched) so a new gear is selected to match
    the impedances again, to reduce the pedal speed to the "power zone" of
    the man and so on, up through the gears.

    Under certain circumstances it is necessary to match impedances to get
    the maximum power from input to output. Different requirements might
    need maximum voltage transfer (low driving high) while others require
    maximum current transfer ( high driving low ).

    Cheers
    Robin
     
  3. default

    default Guest

    The "ideal" opamp would be one with infinite input impedance and zero
    output impedance. There is no intrinsic difference between inverting
    and non inverting inputs on the integrated circuit - if you look at
    the data sheets and the internal circuit you will see that both inputs
    are identical - same type of differential amplifier.

    Op amps have very high input impedance - or to state it another way -
    very little power is required to impress a signal on the input.

    Op amps have relatively low output impedance. They have the ability
    to transfer more power to the output load than the input consumes.
    Some "power" op amps can even drive loads like loudspeakers or motors
    directly.

    In a circuit there will/may be some impedance difference due to the
    gain setting resistors and capacitors - so you'd have to be more
    specific with your question if that's the case.

    "Impedance matching," in the context of op amps, is harder to grasp.
    Typically, with a series of op amps used to amplify the same signal,
    you have a low impedance output driving the high impedance input of
    the next stage . . . a mismatch but desirable in that case since the
    input won't overload and change the signal of the output stage driving
    it.

    Matching an output stage to a load is more critical, but even there,
    with low frequency signals the output stage will be lower than the
    load as a rule.

    You couldn't drive a low impedance load with a high impedance output -
    most of the available power is dissipated in the op amp as heat, the
    load will see little power.

    A little ohms law will show you how power transfers between a source
    and load. Arbitrarily assign an impedance to a source (battery,
    opamp, signal generator, whatever) and see how much energy (power)
    goes from the source to the load, with the load one quarter of the
    source, the same and four times the source impedance. How much power
    is dissipated in the source versus the load for each scenario?

    The rules (or perhaps the goals) change in some cases when dealing
    with radio frequencies - in those cases you usually want a close
    match. Source, transmission line, and load (antenna) at the same
    impedance or matched with transformers. With RF, the source that
    isn't matched to the load - doesn't drive the load, but signal is
    reflected back to the source creating high voltages or currents along
    the way, and causing problems in the system or source. With many RF
    applications you pay a premium in power and parts cost to generate it
    in the first place, so wasting it makes no sense.
     
  4. Eeyore

    Eeyore Guest

    Which completely ignores how they're used in practical circuits.

    Graham
     
  5. Eeyore

    Eeyore Guest

    Let's start at the beginning.

    Do you know what impedance is ?

    Graham
     
  6. BobG

    BobG Guest

    Just pretend its input resistance and output resistance and do all the
    analysis using DC. Rules of thumb: A low output resistance puts out
    the same constant voltage no matter how much load you hang on it.
    (example: audio amp) A high output resistance can be loaded down by a
    load resistance less than about 10 times the output resistance.
    (Example: 150 ohm microphone) They call a high load resistance a
    'bridging load'. You get maximum power transfer from source to load
    (watts) when the input and output resistance match.
     
  7. Guest

    Thanks for everyone who answered to my question.
    Yeah..I know what impedance is.And I know the voltage source
    and current source concept a little bit.
    But input impedance and output impedance are new concepts to me.

    It seemed that input and output impedance concepts are related to
    those "source" concept.
    For example..
    Let's suppose that there is a circuit that has input and output( ex)
    amplifier)

    How do you define input impedance?
    Impedance between input terminal and ground?
    or Impedance between signal source and input terminal?

    How about output impedance?
    Impedance between output terminal and ground ?

    I was confused about that.
    I also can't know why value of impedance changes
    amplifier's function.
    Is it really related to some "source" concept "?
     
  8. Eeyore

    Eeyore Guest

    Input voltage / input current

    That's normally a wire !

    It's mostly related to the resistor values used around the op-amp.

    Graham
     
  9. First lets get impedance, in general, out of the way. It is
    a concept closely related to resistance. It has the same
    units, volts per ampere, as resistance, but includes the
    frequency dependence. But, in simplest terms, high
    impedance is a lot like high resistance, and low impedance
    is a lot like low resistance.

    When applied to an amplifier circuit made with an opamp,
    input impedance refers to how many volts it takes to drive
    an ampere of current through the input of the amplifier (or
    microvolts to drive a micro ampere, etc.). Output impedance
    is how many volts the output signal is loaded down by, if
    you pull an ampere out of it. A high input impedance reacts
    to a signal without requiring much current from the signal
    source, while a low output impedance drives a wide range of
    load resistances without the output voltage drooping much.
    The impedance of the amplifier, configured with programing
    resistors is a combination of the opamp impedances and the
    resistor impedances. The reason the inverting configuration
    has a lower input impedance is that the output (which swings
    in the opposite direction from the inverting input) is
    connected back to that input through a feedback resistor.
    This causes the inverting input to be held at an almost
    fixed voltage (a voltage that matches whatever you have
    connected to the non inverting input. So, whatever resistor
    you connect between the signal source being amplified and
    the inverting input is what determines how many volts per
    ampere (ohms) the signal source must drive, since the entire
    input signal voltage will be dropped across that resistor.

    If the opamp is configured as a non inverting amplifier
    (lets say, a follower with a gain of 1 and the output
    connected directly to the inverting input) the inverting
    input is forced to match whatever voltage you apply to the
    non inverting input, by the output, so there is no other
    path for input current than just what the opamp inputs leak.
    The opamp is designed to keep that input leakage current
    as low as possible, so the amplifier input impedance (volts
    required to force an ampere of current through the input) is
    very high.
    I have defined input resistance and output resistance, but
    I'll leave it there, for the moment as an approximation of
    input and output impedance. The full definition involves
    including how the volts per ampere changes with frequency.
    It is a signal concept that generally does not apply to
    opamps, where lots of gain is available and gain precision
    is important. It is an energy transfer concept. If a
    signal source is loaded with a load that has a matching
    impedance, the largest possible energy transfer is achieved.
    This can be a useful thing to strive for when the signal
    energy is th limiting factor. For instance, if you are
    trying to force the maximum energy out an antenna you might
    try to impedance match the antenna impedance to the
    amplifier output impedance. In that case, you are not
    trying to get some specific voltage across the antenna
    terminals, but are just trying to radiate as much power as
    possible from a given amplifier.

    There are other reasons to be concerned with source and load
    impedances that do not involve matching. Some components,
    like transformers, have a frequency response that depends on
    the impedances of the sources and loads connected to them.
    So you have to design the rest of the circuit to produce the
    impedances that will give the transformer the best frequency
    response.
     
  10. John Fields

    John Fields Guest

    ---
    Impedance is the opposition to the flow of current.

    In a DC circuit or in an AC circuit which is resistive, resistance
    and impedance are the same. In a reactive AC circuit (a circuit
    containing capacitive and/or inductive elements) the impedance is
    more complex than that.

    For the moment, let's say we have a circuit with a resistive load
    and a resistive generator that looks like this: (View in Courier)


    10VRMS
    10mA--> /
    +----->>------+
    | |
    [GENERATOR] [LOAD]
    | |
    +----->>------+
    |
    GND

    In this case, since there is 10 volts across the load and 10mA
    passing through it, and it's resistive, we can use Ohm's law to find
    the impedance of the load, like this:

    E 10V
    R = Z = --- = ------- = 1000 ohms
    I 0.01A

    Now suppose that instead of a resistor the load was an amplifier
    with a resistive input, and that with a 10VRMS signal into it the
    input took 10mA of current from the generator.

    Using Ohm's law again:


    E 10V
    R = Z = --- = ------- = 1000 ohms
    I 0.01A

    We see that the results are identical to the case with a resistor,
    and since that was the load the amplifier's _input_ presented to the
    generator, we say the amplifier has an input impedance/resistance of
    1000 ohms.


    Now, since the generator isn't perfect, it has resistance through
    which the current into the amplifier's input must pass, and since
    it's on the generator's _output_, it's called the output impedance.

    Modifying the circuit above to show the generator's resistance, we
    have:
    10VRMS
    10mA---> /
    +------>>------+
    | |
    [Rg] |
    | [1000R]
    [GENERATOR] |
    | |
    +------>>------+
    |
    GND


    In order to find out what the generator's output impedance is, we
    can disconnect the load, measure the generator's output voltage,
    reconnect the load, measure the voltage again and (knowing the
    current into the load) calculate the output impedance of the
    generator.

    In this case, let's say that with the load disconnected the
    generator's output voltage went to 20V, and when the load was
    reconnected the voltage fell to 10V.

    Since, when the load was disconnected, the generator was supplying
    no current, (except to the voltmeter) its output voltage rose to
    essentially its no-load value. Then, since its voltage dropped to
    10V when the load was connected, 10 volts was being dropped across
    the generator's internal resistance and, using Ohm's law, the
    resistance had to be:


    E 10V
    R = --- = ------- = 1000 ohms
    I 0.01A

    ---
    ---
    Yes.


    For a non-inverting amplifier:


    Vin>--------|+\
    | >--+-->Vout
    +--|-/ |
    | |
    +--[R1]--+
    |
    [R2]
    |
    GND

    The input impedance is going to be, essentially, the very high input
    resistance of the opamp, while for an inverting amplifier:


    +--[R2]--+
    | |
    Vin>--[R1]--+--|+\ |
    | >--+-->Vout
    vref--> +--|-/
    |
    0V


    It's going to be the resistance of R1 because Vout will swing to
    whatever voltage it has to to make the voltage on the + input equal
    the voltage on the - input. In this case that's 0V, so the input
    resistance, as far as Vin is concerned, will be R1.

     
  11. You measure the impedance by applying a voltage and
    measuring the current that voltage drives through the input.
    the ratio of voltage to current is the input impedance.

    The definition of input impedance is the ratio of input
    voltage to input current.
    Yes, unless the input is applied between two nodes, neither
    of which is ground. Balanced differential signal lines do
    not use ground as one of the signal nodes.
    No. If the source and input are directly connected, there
    is zero impedance in that connection, regardless of the
    source impedance and the input impedance.
    Since the output is a source, you measure the voltage change
    caused by a load current and divide that voltage change (the
    drop across the output impedance) by the load current.
    Ideally, this would be the same ratio you would find if you
    divided the open circuit output voltage (no load voltage) by
    the short circuited current, since a short circuit (zero
    impedance load) drops all the output voltage across the
    output impedance.
    Yes, if the output is ground referenced, and not a floating
    transformer winding or something else not related to ground.
    You will have to think that question through a bit before
    you can ask it in a way that makes sense.
    The output is a source concept. The input is a load concept.
     
  12. default

    default Guest

    I'm not following you. Your point is what?

    We almost never use op amps as they come - typical op amp has an open
    loop gain of 200K how many times have you encountered an op amp that
    actually uses a tenth of that? (it is done . . . but not often) -
    great way to sense current for instance.
     
  13. neon

    neon

    1,325
    0
    Oct 21, 2006
    on a dif amps there are usualy two bases of transistors whereby their inpedance charatheristics are he same. the differential voltage is nill or virtualy the same. the output the amplifier must be relatively low otherwise you could not drive any load pluss any F/B will bereduced by output inpedance drop to the input source.
     
  14. PeteS

    PeteS Guest

    The whole point of the opamp is to deliberately 'throw away' the excess
    gain to get other desirable characteristics. What characteristics we are
    after depends on application, of course.
    A typical cheap opamp with inputs shorted might easily have it's output
    pinned at one of the supply rails (Vo = Vos x G) which would hardly be a
    desirable thing ;)

    Cheers

    PeteS
     
  15. PeteS

    PeteS Guest

    The input impedance (for low frequency signals) for the ordinary
    inverting configuration is simply the value of the input resistor to the
    inverting input. That's because the effective potential at this point
    (provided the device is in it's linear range) is the same as at the
    non-inverting input (It's all because of the feedback loop - you really
    need to understand the basics of feedback theory before you'll
    understand an op amp or virtually any other amp for that matter).
    Normally, that's ground (at signal frequencies). This is a basic part of
    op amp theory and operation.

    Basically, the inverting input (which is the feedback point) is changed
    to follow the potential at the non-inverting input. If it's fixed, the
    inverting input will remain (roughly) fixed while the device is in it's
    linear range.

    A non-inverting configuration varies the potential at the non-inverting
    terminal, forcing the output to drive the inverting input until the
    difference between the inputs is 0 - the input impedance (at low
    frequencies) is Vin/Ib (where Ib is input bias current). That's a fairly
    high impedance.

    There's a whole lot more to op amps than this, of course.

    Cheers

    PeteS
     
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