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Understanding Electronics Basics #2

Discussion in 'General Electronics Chat' started by cowboybob, Mar 26, 2012.

  1. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Hey, doesn't 8√3 look just like the long division symbol? :rolleyes:
     
  2. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Looks like LateX is working again.

    Don't know the convention of leaving off the top bar of the square root symbol (when you use the root symbol from the character chart to the right).

    But, then, I'm no math wiz. :confused:

    Doing that does make it look like a division function.
     
    Last edited: Mar 31, 2012
  3. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Found this:

    RLC 180° phase issue in Spice sims

    very interesting.

    Tried their solution and no change to the usual (both traces in sync) outcome.

    Although, I must say, the op implies that the 180° shift was unexpected !?!

    Not even sure if the phase shift observed is the one we've been discussing, since he didn't include a schematic.

    To truly test your RLC scenario, looks like I'll have to breadboard a real circuit, apparently.

    That is, unless someone else knows of a solution for the sim method.
     
  4. dave

    Dave New Member

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  5. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    I think they are saying that the inductor has polarity. Not sure though.

    You saw the shift, but not the equal magnitude across L & C, right?
     
  6. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    This looks right:

    Phase, then V:

    View attachment 62843 View attachment 62845

    I'm thinking:

    Ind V dips at resonance (≈ 158Hz) then returns to 1V just as Cap V drops to (essentially) 0 V at around 1.6kHz.

    What say you? That agree with the math?
     
    Last edited: Mar 31, 2012
  7. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    I was looking at ωC = 1/ωL; ω=2∏f
     
  8. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Finally decided to go back to the books.

    I started here.

    With a L=1H and C=1μF, the resonant freq of the series RLC circuit we've been looking at is 159Hz, right?, (the R has no effect on the frequency, just amplitudes and the bandwidth of the current peak (or voltage drop across the LC portion).

    So, at sweep with V response:

    View attachment 62850

    This confirms the 0V across the cap, at resonance as expected and this:

    View attachment 62851

    showing 180° phase differential at 166Hz (close enough to 159Hz to convince me).

    Now, this is not the circuit you described. I'm just trying to get my head straight on RCL resonant circuits.

    The circuit you described is this:

    View attachment 62853 (V) and this (Phase): View attachment 62854 and, of course, no resonance.

    So I guess I'm curious what you were expecting, or why you thought you'd see something different than what the sim is giving us.

    Inquiring minds want to know...;)
     
    Last edited: Mar 31, 2012
  9. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    I got it backwards (head again - didn't look it up) 1/ωC = ωL

    In reality, i was trying to set the Xc to XL which should mean that Xc should totally cancel XL. There is a sign missing there: e.g. -j and +j.

    If you remember the complex plane where R is the +Y axis. -R is the negative Y-axis and j is the positive X-axis and -j is negative X-axis.

    With me so far.

    The vector sum of Xc, XL determine the phase. R is immaterial. I was hoping the capacitive vector along the x-axis would TOTALLY cancel the inductive vector on the x-axis, MUCH like power factor correction. Isn't that the idea, when you have lots of inductance from motor loads, you add capacitance to cancel them out.

    I think a light bulb went off here. That is a parallel LC circuit, not a series LC circuit. Thanks for that. Bunches.

    Now we know we have to cancel the effect of L with a parallel capacitor or vice versa. That's what i failed to see earlier. I don't see how that impedance triangle lets you see if it's parallel or series and I never thought about or was taught that, but it makes sense.

    Now does R have to be in parallel too? Probably the answer is yes!

    Boy, I like when I learn things.
     
  10. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Me too.

    I'm working on the parallel RLC now. Using the same tutorial and the RLC values listed. I'm getting a fine resonance trace View attachment 62860 but having trouble with phasor and other analyses.

    Graham, don't be put off by what is, admittedly, some pretty esoteric ramblings here between KISS and myself.

    This bears on the main topic of the "switcher" circuit that started this whole thread.

    Remember the little inductor on the board you removed from the power plug you had?

    It's part of a series RLC that allowed the circuit to oscillate (giving us the "AC" of the DC to AC to DC).

    It's one thing to create a reliable oscillator using a 555 timer IC or some OpAmps but another thing altogether to do it with an RLC and some transistors.

    The minutia you've seen discussed here is actually important to understanding just exactly how and why these circuits work.

    But I will say that in the end, all I'll remember is how to determine the resonant frequency since that's what really counts when designing an oscillator.

    Or even better yet, I'll remember exactly what question to ask Google to find the answer.

    Very important to know what question(s) to ask.
     
    Last edited: Apr 1, 2012
  11. Muttley600

    Muttley600 New Member

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    Hiya CBB
    I'm buried in end of year stuff at work at the mo, working 7am-9pm on O/T as desperate to get work done by end of today, don't think I've got lost.
    I haven't been taking in conversation between your Good selves as last week has been hectic, finally (read hopefully, the way things work at our place) I should be getting some help soon as we have been a man down for ages, but I should get some time this coming week. At least when I get to this point of working/sleeping I know it can only get better, can't think why people are leaving work with no jobs to go to.lol

    I still need to work on triangles so I'm a bit behind but will catch up when I get a breather :)
     
  12. Ratchit

    Ratchit Well-Known Member

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    I haven't checked this thread lately, but it looks like you guys are really getting wrapped around the axle.

    cowboybob,

    What on Earth are you guys talking about?? The voltage across C is not zero at series resonance. You can get tremendous voltages across the reactive components at series resonance, provided you don't load them down too much. The current in a resonance circuit is limited only by the resistance. Therefore, since Vxc = I*Xc, if the current is high, so will be the voltage. Another thing, you are not specifying whether the phase you are measuring is V vs. I or Vout vs. Vin. In any case, for for a RLC circuit, you cannot obtain a phase difference greater than 90° for the phase measurements described above.

    KISS,

    Reactances Xc,Xl, and capacitance, inductance and resistance are not a vectors. Capacity, inductance and resistance are scalers. Reactance is a complex number with no real component. Voltage is not a vector, it has magnitude but no direction. Voltages across reactances and resistance can be described with phasors. R does matter when computing phase. The larger the R, the smaller the phase.

    Ratch
     
  13. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Last edited: Apr 1, 2012
  14. Ratchit

    Ratchit Well-Known Member

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    KISS,

    Wiki is known for being inaccurate sometimes. I think they are calling a phasor a vector because it has some vector like quantities like addition. But it is not a spacial vector. A phasor is used to represent time-varying quantities, a vector does not.

    From Wiki:
    A phasor does not provide frequency information. Wiki seems to confuse the exponential representation of a voltage/current with phasor representation.

    Ratch
     
  15. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Yep. That's wrong. No frequency component. So, why don't you correct it? You can, you know.
     
  16. Ratchit

    Ratchit Well-Known Member

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    KISS,

    I get my info out of good textbooks, and quote Wiki only when I agree with what they aver. I am not in the proof reading business.

    Ratch
     
  17. Muttley600

    Muttley600 New Member

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    Nearly finished work for the day, I'm starving :)
    Not back in til 9am, s'wat you get for flunking school.lol
     
  18. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Rachit,

    Thanks for dropping by. We can always use the input.

    My error. I meant to write the voltage across the "LC" portion (VF2 LC - V trace below) of the series RLC circuit. Perhaps not exactly zero, but for certain applications, close enough:

    View attachment 62872

    Across C, yes, as evidenced by the "VF1 C - V" trace above.

    Again, thanks for for the catch and your help.
     
    Last edited: Apr 1, 2012
  19. Muttley600

    Muttley600 New Member

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    Still working on post #46, I can see the numbers work but for the life of me, can't understand the logic, I will try an learn what factoring means, why there is always the √3 symbol for underneath
     
    Last edited: Apr 2, 2012
  20. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Factoring by example: 2=2*1 (prime), 3 = 3* 1 (prime); 4=2*2; 5=5*1(prime); 6=2*3; 7=7*1 (prime); 8=2*2*2; 9=3*3;10 =2*5;11 = 11*1(prime); 12=2*2*3; 13= 13*1 (prime); 14=2*7; 15=3*5; 16=2*2*2*2; 17=17*1 (prime)

    The (prime) means it's prime number. Divisible by itself and 1

    Some other things (√ ): √4=2; √8=2√2; √9=3; √10=√10; √11=√11; √12=2√3; √13=√13; √14=√14=√2√7; √16=4
     
  21. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Muttley600,

    Onward and upward in our quest for understanding of the simple things.

    Here's a really great little SMPS circuit I found at This site.

    Outstanding 2 transistor rig that worked in TINA first effort (well, not the first time - had Q1 upside side down, but...).

    Relatively low current capability, but the least complicated one I've seen thus far and it is such a simple circuit that it ought to be perfect for demonstrating thw how and why they work as they do.

    SETUP View attachment 62897 Steady State Traces View attachment 62898

    This is something that's cheap, easily constructed and understandable.

    Play to your heart's content.
     

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