Op-Amp input/output impedance

Discussion in 'Homework Help' started by star, Feb 22, 2007.

  1. star

    Thread Starter Member

    Dec 18, 2006
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    I've read a lot in textbooks about how ideal op-amps have infinite input impedance and zero output impedance. My question is: why would we want a high input impedance and a low output impedance? i haven't been able to find an explanation in the txtbooks i've been looking at..
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    The higher the input impedance of an opamp stage the less that stage will load the circuit feeding it. The lower the output impedance of an opamp stage the less its output will be affected by the stage it feeds.

    hgmjr
     
  3. star

    Thread Starter Member

    Dec 18, 2006
    19
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    hmm.. what does that mean? sorry.. i'm still trying to get up to speed with basic electronics
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    Put another way, when feeding the signal from one opamp stage to the next, the signal is least attenuated when the output signal originates from a low impedance and feeds into an opamp stage with a high input impedance.

    hgmjr
     
  5. sohcahtoa

    Member

    Nov 7, 2006
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    I'm not sure if this might help further, or if it's 100% accurate. Someone please correct me if it's misleading.

    Think about a circuit that you're connecting to the op-amp input. It is currently "on" and has a current running through it. Say that circuit has an output terminal. You attach a resistor with very high resistance to that output terminal and tie it to ground. Current is not going to "want" to flow down that 10 Mohm or so resistor, meaning that the current flow throughout the original circuit is still fairly close to what it was before attaching the additional resistor. So your voltages in your original circuit are about the same as before. This has the added advantage that you can still use the voltage at the output terminal, as the voltage at the node that connects the resistor to the original circuit is still of the same voltage as the output terminal.

    Now, say instead you attach a resistor of small resistance value, say 10 ohms and tie it to ground. Since you've provided a path of presumably least resistance, most of the current in the original circuit is going to flow through that resistor, which greatly affects the voltage drops in your original circuit.

    Thus, you want a high input impedance so you do not "load" your original circuit.

    You can think of the low output impedance of the op-amp similarly, but wanting the opposite result. If you have less output resistance, then attaching an additional circuit to do something with that output signal will use almost all of that output signal from the op-amp effectively. If you had high output resistance, you'd have a hard time using that output.

    I hope that helped, and didn't make things worse. ;)
     
  6. Salgat

    Active Member

    Dec 23, 2006
    215
    1
    Input impedance is comparable to putting a resistor in in parallel with a circuit. The higher the value, the closer it becomes to acting like it doesn't exist. If you put a low input impedance device on a circuit, its the same as putting something more resembling a short in your circuit. For output impedance, its the same as putting a resistor is series. The lower the output impedance, the less effect it has on the total resistance of the circuit.
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    The easiest example would be to construct or calculate two voltage divider circuits. One circuit should be of low impedance [say 100 ohms each] and the other should be of higher impedance [say 10k each].

    Measure or calculate the circuit parameters for each divider circuit.

    Add a low impedance load to each circuit [say 100 ohms].

    Measure or calculate the circuit parameters for each divider circuit noting the changes.

    Replace the low impedance load with a high impedance load [say 1 Meg ohm].

    Calculate the circuit parameters for each divider circuit noting the changes.
     
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