Precise Half Wave Rectifier Equation Challenge

Discussion in 'General Electronics Chat' started by MrAl, Jun 30, 2015.

  1. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello there,

    We've all seen what a half wave rectifier circuit looks like, but this one is ideal. By that i mean that every component in it is completely ideal:
    Source: zero impedance, zero tolerance
    Cap: zero ESR, zero ESL, zero tolerance
    Resistor: Exact value, zero tolerace
    Diode: Perfect short when 'on', perfect open when 'off.

    The interesting question is:
    What is the value of C required to obtain exactly 10 percent ripple, or put another way, the maximum discharge of the capacitor is 100*K in percent? (Use K=0.9 for 90 percent, as a numerical example).

    The input is 1vac peak, the value of the resistor is 1000, the line frequency is 60Hz.
    Since the input is 1vac peak, that means the capacitor is only allowed to discharge to 0.9 volts for K=0.9 and 0.8 for K=0.8, etc., so K is a variable but we can use 0.9 for now if you like. Actually R is a variable too, but we can use 1000 ohms for now if you like. The frequency is also a variable but we'll use 60Hz for now too if you like.
    So either use variables K,R,F, or use those numerical values above, to calculate the exact value of C to say 6 decimal places or better.

    To help show how precise we want this, if the cap discharges to 0.90002 when the next charge cycle starts that's too high, or if it discharges to 0.89998 that's too low, but acceptable might be 0.90001 or 0.89999, and i want to get this out of the way before we start thinking about capacitor tolerances and stuff like that, because in the problem description we dont allow any tolerances so it has to be exact to at least 6 decimal places, but better would be an equation to solve for this capacitor value.

    I had solved this in my own way and am looking for other solutions to compare with. Note that this is a purely theoretical question, one that might be asked in a classroom, not a practical application question where there are tolerances in everything we deal with.

    Should you decide to accept the challenge i want to wish you good luck. For such as simple circuit that we've all dealt with many times in the past, it's not quite a simple as it seems, maybe that is what makes it so interesting :)

    Alternately, come up with a simulation that is precise.
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    146.6uF
     
  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Why?
     
  4. #12

    Expert

    Nov 30, 2010
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    That leaves me out. :(
     
  5. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    Ha ha, ok, so how did you come up with that, besides reading other forums? <he he>
    Care to try a simulation that is very precise for this?
     
  6. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    Just to we can test the results.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    Maybe it's obvious, but is the resistor the load?

    I worked out the math to do this for a full-wave rectifier. I can revisit it for a half-wave if it's worth the time.

    There is NOT an analytical solution, if that's what you're after. Well maybe a good mathematician could solve it but I couldn't. But anyway you can solve it numerically by iteration to any level of precision you want.
     
    Last edited: Jun 30, 2015
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Off course I used LTSpice to come up with the 146.6uF. Do you think I would waste my time doing it analytically?

    84.gif
     
  9. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    I was surprised you would waste your time doing even that. Am surprised I read this one to the end.
     
  10. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Actually, I did it to yank Mr Al's chain...
     
    #12 and GopherT like this.
  11. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello again,

    Mike, trying to turn my light on? :)
    Actually im not sure what you mean by 'yank my chain' but you did what i was hoping to see. That wasnt so hard right?
    I had been after the analytical solution, which i found for the switching times, and the capacitor is solved for using those results. I came out with 146.604uf which matches your simulation as well as i had hoped. I believe i have found the exact solution now, which we were talking about on another site.

    If you are into theory then you can appreciate this, but if you are not then you will never appreciate the beauty of some of these theories. I took engineering science, not applied engineering, so i was more interested in the pure science behind many theories in the universe not as much in the practical applications. I found value in the practical later in life however when i had to build my own stuff, for real, using those theories :)

    There is also an interesting article on the internet which is for a full wave solution, which involves a big equation to be solved for the switching times. The solution for this came out much simpler however.

    Id also like to see equations.
     
    Last edited: Jun 30, 2015
  12. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Somewhere a while back I posted the exact analytical solution to this problem, but the search option has gone to rat_ _ _ _
     
  13. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    Oh sorry to hear that, i would have liked to see your solution.
    I'll post mine at some point, but i'd like to see others first.
     
  14. studiot

    AAC Fanatic!

    Nov 9, 2007
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    I'll write it out again when I have a spare century...
    :)
     
  15. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello again,

    Well the solutions that have been found so far arent too bad really. A little involved, but can be shown in one page. I'd like to see yours too, or at least how you went about it, what things you had to consider.
     
  16. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello again,

    I forgot to mention, everyone seems to be viewing this as very hard to solve exactly. But consider:
    Vs=cos(wt)
    and since Vs is across both R and C, we can write:
    Ir=cos(wt)/R
    Ic=C*d[cos(wt)]dt

    and sum those to get Is:
    Is=Ir+Ic

    and that leads us to the time delay at the start of the cap charge cycle, as the departure point is where Is=0.

    Once we have that, we can solve for the time from the peak to the start of the charge cycle, then use that to find the actual charge time by subtracting the time where Is goes to zero.
    From there we can find the cap value, after correcting for the exponential due to the start voltage is not exactly 1 anymore.
     
  17. wayneh

    Expert

    Sep 9, 2010
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    I believe that would be a publishable paper if it you really have it.
     
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