RLC help

Discussion in 'Homework Help' started by Fluke289, Aug 9, 2013.

  1. Fluke289

    Thread Starter New Member

    Aug 8, 2013
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    I have the following circuit that I’m supposed to calculate the voltage drop of R1. I’m not too worried about the answer because I got a 100% on the assignment. The answer is 2.72V @ 36.97 degrees so no one has to bother working it out.
    Screen Shot 2013-08-08 at 9.17.48 PM.png

    Now for my question, I decided to build the circuit to learn more. When I built the circuit in the lab I got much different results than my calculated results.

    For instance, if I put channel 1 on my oscilloscope on the positive side of R1, I get a reading of 4.4 Vp-p @ 2 kHz (I was expecting to see 5 Vp-p @ kHz). When I put channel 2 on the negative side of R2 I get a reading of 2.89 Vp-p. Assuming I’m using it right, my fancy Agilent MSO X3000 tells me that the voltage drop across R1 is 2.09 Vp-p @ ~50 degrees.

    My wave gen seems to be working properly. When I connect channel 1 with no circuit I get 5Vp-p @ 2 kHz, the minute I connect it to the circuit I see a 600 to 700 mV drop. Would that be typical? I was expecting to still see pretty close to 5V because channel 1 is still connected to wave gen alligator clip, I just attached it to the positive side of the resistor.

    I got to thinking, let’s assume the 600 to 700 mV drop is normal, so I redid my recalculations with 4.4 Vp-p instead of 5 Vp-p and I got a calculated result of 2.39 Vp-p @ 36.97 degrees. That’s much closer to what my oscilloscope is telling me, but the phase is still 13 degrees different.

    A long story short, would this be typical, or am I doing something wrong? I would expect the numbers to be a little closer. I’m using 1% resistors and my Fluke 289 tells me that they’re only a few ohms off. I have no idea how accurate my capacitors and inductors are. I have to assume they’re what the label tells me.
     
  2. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    What is the output impedance of your function generator? Are you accounting for it?
     
  3. Fluke289

    Thread Starter New Member

    Aug 8, 2013
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    I’m not sure what you mean by output impedance on my function generator.

    I’m using the wavegen on the Agilent x3000 oscilloscope and set it to “High-Z”. I’m using a Pomona RG58 BNC to Alligator clip on the function generator.

    I’m not sure if this is correct, but after heading over to Google, I kept coming up with a number of 50 Ohms for output for my alligator clips.

    On a hunch, I took everything out of the circuit except for my 100 Ohm resistor to simplify the testing. I measured the 100 ohm resistor on my oscilloscope with the function generator set to 5 Vp-p @ 2 kHz and I got a reading of 3.42 Vp-p @ 2 kHz. Next, I measured my 100 ohm resistor with my Fluke 289 and I got 99.62 Ohms. After that, I took 99.62 Ohms + 50 Ohms and got 149.62 Ohms. Then, I took 5V / 149.62 ohms and I got 33 mA. I took 33 mA * 50 ohms and I got 1.67V. Finally, I took 1.67V and added that to my 3.42 Vp-p reading on my oscilloscope and I got 5.09 Vp-p, which seems pretty close to the 5 Vp-p that I set it for. – Would that be how I account for output impedance?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Yes, it sounds like your function generator has a 50Ω output impedance, so you need to take that into account. You might also check what kind of current output it is rated for.

    I'm not following how you are making your measurements. You said you are attaching channel 1 to the positive side of R1 and channel 2 to the negative side of R2. Since your diagram isn't marked, I'm assuming that the positive side of R1 is the left side on the node that connects it to the function generator and that the negative side of R2 is the bottom side that connects it to the 10mH inductor. So what are you doing form there to get your voltage and phase angle measurment?

    While your resistors might be within 1%, your capacitors and inductors are probably not. Inductors commonly have tolerances of 10% and non-electrolytic capacitors are commonly about that, too.
     
  5. Fluke289

    Thread Starter New Member

    Aug 8, 2013
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    When taking the output impedance into account, what is the best way to do that?

    Yes, I’m doing exactly that. I’m using the math function on the oscilloscope to give me the voltage and phase angle measurement. Of course, those would be off, since I have 50 Ohms of impedance from the function generator.

    I’ll have to take that into consideration. With circuits with phase angle measurements, what is the best way to calculate if it’s within 10% of the calculated result?
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    But if those are the two points where you are connecting your probes, than the Math function is calculating the voltage across several components, not just across R1. Why aren't you putting the two probes on the two ends of the component you are measuring the voltage across?
     
  7. Fluke289

    Thread Starter New Member

    Aug 8, 2013
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    A picture is worth a thousand words. This is how I have the circuit built and I'm testing it. Channel 1 is the probe on the bottom and Channel 2 is the probe on the top.

    circuit1.jpg
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Okay, that looks like Channel 2 is on the negative side of R1, not R2 as stated in your original post and reaffirmed in Post #5.

    Those capacitors look like electrolytics, which means that they do NOT like to be reverse biased. You need to be using non-polarized caps.
     
  9. Fluke289

    Thread Starter New Member

    Aug 8, 2013
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    Sorry, my mistake, I didn’t notice I had written R2 instead of R1 for the negative side.

    I understand that electrolytic don’t like to be reverse biased, but if I put the negative side of the capacitor on the negative from the function generator. Then I hooked up the positive side of the capacitor on the function generator, I assumed that would be ok. I’ll order some non-polarized capacitors and give it a try.
     
    Last edited: Aug 11, 2013
  10. WBahn

    Moderator

    Mar 31, 2012
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    That's okay -- and like you said, a picture is worth a thousand words!

    Not unless your function generator is limited to only putting out positive voltages relative to it's negative terminal. I would be surprised if this is the case. It's easy enough to tell. Just hook your scope probe to the positive (and the ground to the ground) and be sure to have the scope DC coupled and then see if the output is unipolar or bipolar.
     
  11. Fluke289

    Thread Starter New Member

    Aug 8, 2013
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    I’ll order some non-polarized capacitors. I assume that the use of electrolytic capacitors effect the measurements that I’ve been taking then?
     
  12. LDC3

    Active Member

    Apr 27, 2013
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    In some cases, yes. Electrolytic capacitors usually have a larger tolerance (most are 20% or more) and they have a high leak current (the capacitor cannot hold the charge). For some situations, the equivalent capacitance is much less than what is stated on the label.
     
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  13. WBahn

    Moderator

    Mar 31, 2012
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    The capacitance of electrolytics is pretty much undefined when reverse biased because they breakdown almost immediately and start conducting. If they are in a circuit that can support any significant current flow, they tend to fail explosively (particularly tantalums!). The circuit you have (or, more to the point, the function generator you are using) is probably not capable of delivering enough current for this to happen; you are likely seeing what would appear to be very high leakage currents and low capacitance in the reverse part of the waveform and you may quickly be doing permanent damage to the caps.
     
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