Impedance and Frequency Demo not what I Expected

Discussion in 'General Electronics Chat' started by Ωhm, Mar 13, 2013.

  1. Ωhm

    Thread Starter New Member

    Feb 1, 2012
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    I enjoy the electronics videos by YouTube user AllAmericanFiveRadio but I came across one the other day that he did on frequency and impedance that seemed completely contrary to my understanding of what impedance does to a signal. In this video https://www.youtube.com/watch?v=TRSNKwtUy8g he increases the frequency of a signal through several components individually and measures the impacted waveform with an oscilloscope.

    When he uses a resistor, the signal amplitude does not change with frequency as expected.

    With the coil (inductor) the signal amplitude is shown to increase as the frequency increases. I thought inductive reactance increased with frequency.

    With the capacitor, the amplitude of the signal decreases as the frequency increases. I thought capacitive reactance decreased as the AC signal frequency increased.

    Am I wrong? Is there some sort of inverse relationship occurring because of how he is measuring this?

    I thought I had a solid grasp on this and now I am questioning my understanding because his demo illustrates the exact opposite of the behavior I expected. Can someone confirm or deny the validity of his demo. If he is correct, can you explain why? Thanks,
    Ωhm
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Yes, you are wrong.
    These two components behave the way you WERE thinking only when they are in the feedback loop of an amplifier.
     
  3. crutschow

    Expert

    Mar 14, 2008
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    Your are correct in you understanding of inductive and capacitive reactance change with frequency. The problem is you are confused about how that affects the signal voltage in the series circuit he used.

    He has two resistors in series with the variable device connected between them and the voltage measured at the junction between the signal generator resistor and the variable impedance. If you think about it, you will realize that an increase in the variable impedance increases the signal voltage at that point and and vice versa, just as observed in the video.

    If he measured the voltage at the other end of the variable impedance, then the behavior would correspond to what you thought.

    Make sense?
     
  4. Ωhm

    Thread Starter New Member

    Feb 1, 2012
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    I re-watched the video and paid closer attention to his schematic this time. So if channel 2 of his scope was measuring after the inductor or capacitor then I would expect my understanding to hold but he was measuring after the 3500Ω resistor and before the coil or capacitor. This is where my understanding falls apart.

    I still don't understand how the inductor and capacitor influence the signal amplitude before the signal even goes through the inductor or capacitor. How does the presence of the component cause the exact opposite effect on the signal when measured in front of the component than it does if you were to measure after the component?

    Is this circuit allowing him to make a capacitor act like an inductor and an inductor act like a capacitor depending based on where the measurement is taken? This does not compute in my mind right now.

    Everything I have understood about electronics to date is that signals are affected by components as they move through them (or attempt to) and that the effect on the signal will be observed after the signal passes through the component, not before. I obviously need more help here than I thought...
     
  5. Ωhm

    Thread Starter New Member

    Feb 1, 2012
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    OK, I missed that he is measuring across the inductor or capacitor because he left out the ground lead of the scope, guess I need more sleep :( Since that's the case though, he is measuring across the component and an inductor should behave as an inductor and attenuate higher frequencies and the capacitor should pass higher frequencies. He is measuring across the component where the signal is presumably being impacted by the component in the characteristic way those passives do. Completely confused.
     
  6. crutschow

    Expert

    Mar 14, 2008
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    He is measuring from that point to ground, not across the component.

    A signal is affected on both sides of a component, not just after it passes through it. I believe you need a better understanding of how voltages are distributed in a series circuit. Here's a start.
     
  7. Ωhm

    Thread Starter New Member

    Feb 1, 2012
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    Been busy for a few days but I did read the suggested page, however, it did not address why the components acted contrary to the way their behavior is always described.

    I think the most difficult part of learning electronics to be that there is often little explanation as to why a circuit behaves as it does. I'm checking out some other electronics sites as I still don't have a handle on this. If anyone else can give a stab at why, it would be much appreciated. Thanks,
    Ωhm
     
  8. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Ok, I apologize for my quick answer in my first reply, I didn't watch the video. Here's my educational answer.

    One thing not clear in the video is that an oscilloscope probe really has two leads for each channel - the second lead is connected to Ground (far right of the 100 ohm resistor).

    Think of capacitors and inductors as if they are resistors that change resistance with frequency while regular resistors don't change with frequency.

    Now, make a series of three resistors in order as if it is DC (using a battery)...
    12v+ - 3500 ohms - 3500 ohms - 100 ohms - Ground.

    Now, if you measure the voltage drop across each resistor, then add them up, the total should be 12. They can also be calculated. First, find the current flow -> 12v/(3500+3500+100) = 0.0017 amps. The voltage measured across each resistor will be (0.0017 x resistance). You do the work and see what you get. Note that your voltage will be smallest for the 100 ohm resistor. Finally, measure like the video did for channel 1 from ground to signal source (12v in this case). Also measure from channel 2 positions (ground to node between the two 3500 ohm resistors.

    Now just a quick experiment (using any 4 resistors you may have). Set up a series network using three reistors. Measure the voltage from ground to the node two resistors away. Now replace the middle resistor with one of slightly lower value (other two unchanged). Did the voltage go up or down? A resistor of lower value should have caused voltage to decrease.


    Now, since most of us don't have a lab power supply with AC output, lets just think through the rest with a capacitor. Then you can work out the inductor case o your oun.

    replace the center resistor with a capacitor (.022uF) which can also be written as 0.000000022 farads. This will be like a resistor but we must know the frequency to calculate the resistance. "C"Resistance = 1/(6.28 x freq x 0.000000022). Calculate one time for freq= 60HZ and one time where freq= 10000Hz. You can see from the formula that that making a bigger value for freq on the denominator of the fraction will make resistance SMALLER (just like 1/8 is smaller than 1/2).

    Now, think about the DC circuit above. What happened to the signal strength when we reduced the value of the smaller resistor? It went down. In this case, the "c"resistance decreases with increasing frequency so, the way the circuit is connected and measured, voltage from ground to the far side of the capacitor goes down as frequency is increased.

    It is unfortunate the way it was demonstrated. As another poster suggested, using only aa capacitor and resistor (with resistor closest to ground) and measuring between those two components would have made the voltage increase with frequency in that circuit. However, it is just a different circuit. The formula above is still true and resistance (actually called "capacitive reactance") decreases as frequency increases.

    Final note, arranging resistor and capacitor one way makes a filter that passes low frequencies better. Flipping them will pass high frequencies better. There was nothing right or wrong about the way the video was presented. You will see both types of RC networks.

    Now, look at the same frequency dependent behavior for inductors. The formula looks similar, just without the 1/(). Inductive reactance = 6.28 x freq x Henries (where Henries is the value of inductance).

    Cheers.
     
  9. Ωhm

    Thread Starter New Member

    Feb 1, 2012
    24
    3
    GopherT,

    Your explanation did the trick. I really appreciate you taking the time to spell this out and straighten out my understanding.

    Thank You, Thank You, Thank You!!! And Crutschow, I appreciate your effort as well, what you said makes more sense now.
    Ωhm
     
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