Why is impedance considered when connecting devices?

Discussion in 'General Electronics Chat' started by CiaranM, Jul 17, 2012.

  1. CiaranM

    Thread Starter New Member

    Jul 15, 2012
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    Hello, there's a question that I've been wondering for ages. Why is impedance considered when connecting circuits/devices? (I know that impedance is the square root of [R^2 + Xl^2 - Xc^2], so don't worry about that).
    I've read that input impedance should be lower than the output impedance of the circuit connected, why is this? The way I see it, a high output impedance will lower the current, which reduces signal strength; which leads to inefficiency.
    Also, how are manufacturers able to define impedance? Take an 8R speaker, for example. The impedance will obviously change when an alternating signal is applied to it!
     
    Last edited: Jul 17, 2012
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  2. steveb

    Senior Member

    Jul 3, 2008
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    Where did you read that? Maybe this is true for current source and current loads, but usually devices are voltage driven, and there you want high input impedance relative to output impedance of the preceding stage. Other times you may want impedance matching, and still other times you may want to match the load with the complex conjugate of the source impedance.

    One needs to be careful to define the circumstance, but the rule of thumb for voltage driven audio circuits is input impedance should be greater than 10 times the output impedance of the stage that drives it.


    Impedance is defines relative to sinusoidal voltages and currents. If you drive a device with a sinusoidal voltage source, it will respond with a sinusoidal current with the same frequency, different amplitude and a phase shift (this assumes linear circuits). The impedance is simply the ratio of voltage to current, using complex numbers to represent the amplitude and phase of the sinusoidal signal ( that is, using phasors). Note that generally impedance is a function of frequency.
     
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  3. Audioguru

    New Member

    Dec 20, 2007
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    The output impedance of a modern audio amplifier is extremely low so that it can drive one 8 ohm speaker and adding a second 8 ohm speaker in parallel does not reduce the signal level, and also damps resonances of speakers. The amplifier has an output impedance of 0.04 ohms or less.
     
  4. CiaranM

    Thread Starter New Member

    Jul 15, 2012
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    oh right, thanks. I see that input and output impedances can vary.. why is it important to get the input and output configurations right? What happens if they are not correct?
    If you make a filter with several poles, should each successive pole's resistance be multiplied by 10?
    - - -
    Thanks Audioguru, I didn't that!
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    A filter with many poles is usually made with a few opamps so the poles do not affect each other. The output impedance of an opamp is extremely low.
     
  6. kevin.cheung19

    New Member

    Nov 22, 2011
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    You can understand it via voltage divider. Let's say output of device1 is connected into the input of the device2. Let's say the output impedance of device 1 is 1Ohm and it's output is 100Volt, and the input impedance of device2 is 100Ohm. The perceived signal input, because of the voltage divider, is now only seeing 99.01Volt instead of the 100V that device 1 is outputing. One way to reduce this loss of voltage is to either reduce the output impedance of device 1 or increase the input impedance of device2.

    So in general, in order to minimize signal loss, we want Low output impedance as an output and High input impedance for the input.

    Kevin
     
  7. CiaranM

    Thread Starter New Member

    Jul 15, 2012
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    hey,that's a great explanation! thanks. I just found it hard to understand why higher impedances (seeming to mean lower current) could actually allow for better signal transfer, but now I understand it.
    One thing, why is device two's output considered to be between a potential divider's resistors, and not after R2? Or I am wrong in thinking that device two's input has to go in series across R2; does R2 simply need to be in parallel to create the input impedance?
     
  8. zaman999

    New Member

    Jan 29, 2011
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    The way i know it, input impedance should be sufficiently high so as to avoid the noise in the circuit(to pass through the input). The output impedance is preferred low so that enough current can be supplied to drive the load.

    Zaman.
     
  9. kevin.cheung19

    New Member

    Nov 22, 2011
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    both input impedance of a device and output impedance of the same device are totally separate and not the same, so don't treat them the same. =) I guess this is what you mean by R2?
     
  10. CiaranM

    Thread Starter New Member

    Jul 15, 2012
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    I'm asking if, with your helpful comparison of a potential divider; the top rail represents device 1's output, R1 represents the output impedance of device 1, the second rail represents device 2's input, and R2 represents device 2's input impedance. Is that right? If so, then does this mean that output impedance is in series with the output, whereas input impedance is parallel to the input (going between it and -V, )V, or whatever)?
    sorry if I'm confusing you, I've made a fancy picture to show you what I think is going on..
     
    Last edited: Jul 18, 2012
  11. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    The maximum power transfer theorem tells us that the maximum power will be transferred between a power source and a load when the impedance of the load is the complex conjugate of the source impedance. This theorem works perfectly whether you're using battery jumper cables or a microwave transmitter. It's one of the most fundamental (and elegant) laws in all of electronics.

    Learn it well!

    Eric
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    You showed a voltage divider. Its output level IS NOT the same as its input level because IT DIVIDES!
     
  13. CiaranM

    Thread Starter New Member

    Jul 15, 2012
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    Audioguru, sorry that I didn't mention it, but I meant for R2 to represent a high value of resistance. Thanks for clearing things up though. I didn't know that output impedance is in parallel. I wish my tutors taught me this stuff, makes me wonder if engineering courses are being, er, dumbed-down...

    Eric, if real men don't use voltmeters, what do they use? Their tongues? 'Time to test the ol' Tesla coil' ..
     
  14. vk6zgo

    Active Member

    Jul 21, 2012
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    I still don't think you've quite got it!

    First draw the output impedance* of the drive device as a low value resistor between the two output terminals of the device.

    Now draw the high impedance of the driven device in parallel with the resistor you have just drawn.

    That is how the circuit looks,with the input of the driven device "bridged" across the output of the driving device.

    Audio line amplifiers usually use this form of coupling in most modern equipment,but that wasn't always the case.

    In the past it was common to use the maximum power transfer system Eric describes,with 600 Ohm inputs & outputs,but modern audio amplifiers have great enough power gain to not need this method.

    The "high impedance input bridged across a low impedance drive" trick is not commonly used at high frequencies,as the connecting cables when not correctly terminated,become part of the response determining circuitry,with resulting aberrations in frequency response.

    *
    This should have read "the termination"-- see comments in the next few postings ,sorry!
     
    Last edited: Jul 23, 2012
  15. MrChips

    Moderator

    Oct 2, 2009
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    What VK6ZGO says is correct. At higher frequencies it is critical to match the output and input impedance with that of the connecting cable.

    It is common practice to have a low impedance output amplifier feeding a 50Ω series resistor driving a 50Ω coaxial cable and finally terminating into a 50Ω load at the input of the receiving end. For 75Ω or 93Ω cables you would match those cable impedances.

    Similarly, your TV input should match the 75Ω or 300Ω cable that feeds the signal.

    If this is approach is not adhered to you will end up with reflections and ringing in the transitions of the signal. You can see this on the TV screen as ghosting on the edges of objects in the picture.

    Audio pads are pi and T networks of resistors that provide signal attenuation while maintaining the same input and output impedances.
     
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  16. steveb

    Senior Member

    Jul 3, 2008
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    He may be right about this part, but unless I am misunderstanding him, he is wrong about the location of the source impedance.

    The source impedance of a voltage source is in series with the source, and not in parallel. Then, the load is in parallel with the series combination of the voltage source and the source impedance. This model works at low frequency where you what low impedance driving high impedance, and it works for impedance matching at high frequency. It also works for complex conjugate matching when you want to maximize power transfer.

    It's not clear to me if he is saying that the voltage divider model is incorrect or not. I believe that is a good circuit model to use at any frequency where circuit theory can still be applied, but the choice of input and output impedance conditions for good design is very specific to the application for sure. I mentioned this in my first response in this thread.
     
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  17. vk6zgo

    Active Member

    Jul 21, 2012
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    Yes,I oversimplified the model*,assuming the output was already terminated in its output Z,working on my memory of the audio line amps I have worked on.

    Going back to the ideal circuit, with a generator Vg in series with a resistance Ro.( I will use Resistance rather than impedance,for simplicity)

    (1) If we leave this circuit unterminated,we can measure Vg across the output terminals.

    (2) If we now connect a resistance Rt across the output terminals equal to Ro,we will read an output voltage. Vg/2.

    (3) Leaving Rt in place,if we place a new Resistance Rin >>>Rt across the combination,we will still see a voltage extremely close to Vg/2.

    This is what normally happens with "bridging" inputs.

    It is not normally easy to analyse a circuit back to an ideal generator in series with a value for Ro,(or Zo),in complex circuitry,so usually matching in such cases is pretty much done by "rule of thumb" methods.


    * Actually ,I stuffed it up ! See the edit in my earlier posting,above---I knew what I meant,but didn't phrase it correctly.
    After all,it was the middle of the night for me!
    The rest of this sentence is valid,though
     
    Last edited: Jul 23, 2012
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