Page 4 of 9 FirstFirst ... 23456 ... LastLast
Results 31 to 40 of 84

Thread: Can I send you a power supply?

  1. #31
    Join Date
    Mar 2018
    Location
    Maracaibo, Venezuela
    Posts
    59
    Thanks Thanks Given 
    25
    Thanks Thanks Received 
    0
    Thanked in
    0 Posts

    Default

    Quote Originally Posted by turkey3_scratch View Post
    Wouldn't the impedance of the capacitor and inductor have not real but purely imaginary value? If we do the operation "0.1/(1000+0.1)" we are treating the capacitor and inductor with real impedance which is not true. While the error on this calculation in particular is small, doing this the same way could result in more error in less extreme cases.

    Instead, in this case, it should be along the lines of "-0.1/(1000-0.1)" which gives 0.0001V/V of gain. Obviously not a big difference here but would have an effect elsewhere.



    Also seen here: https://www.allaboutcircuits.com/tex...ies-r-l-and-c/

    Or if I happen to be misled in some way here, please inform me.

    Edit: This also looks like something that would have resonance.

    Edit 2: Well my diagram I realize isn't the same anyway. Without the load resistor it changes things a little. Mine assumes it is infinite resistance. Well, yours does too.
    Yes, IIRC if it was a RC or RL circuit you needed to plot the reactance of L/C and the resistance of R and Pythagoras the matching point to get impedance of L/C for the voltage divider. However, since this one is LC I'm not sure how to proceed. xd

    Something like this:

    Vo = -0.1 / (1000-0.1)

    (The numbers in the parenthesis need to be converted to this √(1000^2 - 0.1^2))

    So, Vo = -0.1 / √(1000^2 - 0.1^2)
    Vo = -0.1/ 999.999995
    Vo = -0.0001V

    Something along those lines. Notice that it is the same result as yours, just that now it is negative.

    -------------------------------------------------------------------------

    Edit: The PSU increased its price to 60$ (the 60% discount is still available).
    Last edited by Samueru Sama; 04-08-2018 at 01:49 PM.

  2. #32
    Join Date
    Oct 2006
    Location
    jonnyGURU forums, of course!
    Posts
    16,305
    Thanks Thanks Given 
    545
    Thanks Thanks Received 
    333
    Thanked in
    243 Posts

    Default

    I bought a couple. I'm not going to do a full on review, but I'm curious to see what they do.

  3. #33
    Join Date
    Oct 2017
    Location
    Grand Junction, CO
    Posts
    2,127
    Thanks Thanks Given 
    117
    Thanks Thanks Received 
    70
    Thanked in
    63 Posts

    Default

    Many, many thanks; I suspect a lot of us are wondering.

    I'd like to know if it can carry full load, and if it fails gracefully on overload.
    Last edited by ashiekh; 04-08-2018 at 05:07 PM.

  4. #34
    Join Date
    Nov 2015
    Location
    U.S.A.
    Posts
    1,264
    Thanks Thanks Given 
    327
    Thanks Thanks Received 
    0
    Thanked in
    0 Posts

    Default

    Quote Originally Posted by Samueru Sama View Post
    Yes, IIRC if it was a RC or RL circuit you needed to plot the reactance of L/C and the resistance of R and Pythagoras the matching point to get impedance of L/C for the voltage divider. However, since this one is LC I'm not sure how to proceed. xd

    Something like this:

    Vo = -0.1 / (1000-0.1)

    (The numbers in the parenthesis need to be converted to this √(1000^2 - 0.1^2))

    So, Vo = -0.1 / √(1000^2 - 0.1^2)
    Vo = -0.1/ 999.999995
    Vo = -0.0001V

    Something along those lines. Notice that it is the same result as yours, just that now it is negative.

    -------------------------------------------------------------------------

    Edit: The PSU increased its price to 60$ (the 60% discount is still available).
    Actually I did get a negative sign in mine, I just omitted it from the answer since I was talking about magnitude.

    However, I believe both of us are wrong with the negative sign, because we still need to do Pythagorean theorem for the numerator, not just the denominator. I think it should be like this:

    √((-0.1)^2) / √(1000^2 - 0.1^2)

    Which would result in 0.1 / √(1000^2 - 0.1^2) =0.0001V/V. This makes more sense.

    Regardless, I was still a little confused pertaining to the LC circuit myself. I took it upon myself to solve for the transfer function and magnitude of the transfer function for the RLC circuit; I included the load resistor in all these calculations, because if I exclude that resistor I will get different results. On the left side of the paper I also threw in an RC filter just to show how the procedure would be replicated with the RLC filter.



    I do successfully get a lowpass filter for both, so that is good.

    In the final result, we can set the resistor to an infinite resistance; this should give us the magnitude of the transfer function now for the LC filter with no resistor:



    So in the case where it's just an LC circuit with no resistor, it's still low pass but at the resonant frequency there is a massive overshoot (goes to infinite voltage). With the resistor being taken into account, there is still overshoot but not as much. As a matter of fact, if the resistor is tuned to precisely the proper value, there will be no overshoot at all and the behavior will be similar to that of a 1st order lowpass filter, but with steeper decline in voltage later on.

    I've created a playable graph here where the values of R, L, and C can be adjusted and the gain in V/V (or dB if it shows 20log of the result) can be seen on the y axis. The x axis is the input frequency (in radians / second).

  5. The Following 2 Users Say Thank You to turkey3_scratch For This Useful Post:

    Jon Gerow (04-08-2018), Samueru Sama (04-08-2018)

  6. #35
    Join Date
    Feb 2015
    Location
    Germany
    Posts
    110
    Thanks Thanks Given 
    59
    Thanks Thanks Received 
    4
    Thanked in
    4 Posts

    Default

    The ferrite core from the photo is placed at the earth wire, so there is no load current through this wire.

    Are you mixing it up with the neutral wire?

  7. #36
    Join Date
    Nov 2015
    Location
    U.S.A.
    Posts
    1,264
    Thanks Thanks Given 
    327
    Thanks Thanks Received 
    0
    Thanked in
    0 Posts

    Default

    Quote Originally Posted by Microflop View Post
    The ferrite core from the photo is placed at the earth wire, so there is no load current through this wire.

    Are you mixing it up with the neutral wire?
    Maybe I'm wrong about this but don't both neutral and ground go to earth but earth is just designated for chassis grounding as a safety protocol? Maybe I'm wrong, let me know!

  8. The Following User Says Thank You to turkey3_scratch For This Useful Post:

    Samueru Sama (04-12-2018)

  9. #37
    Join Date
    Jul 2009
    Location
    Germany
    Posts
    3,716
    Thanks Thanks Given 
    6
    Thanks Thanks Received 
    72
    Thanked in
    66 Posts

    Default

    Quote Originally Posted by turkey3_scratch View Post
    Maybe I'm wrong about this but don't both neutral and ground go to earth but earth is just designated for chassis grounding as a safety protocol? Maybe I'm wrong, let me know!
    Depends on what grid is used, what type of Installation.

    Here in Germany, when I was last doing Building Installations, TT seemed to be the Standard.
    In my House from the 1920 or so, it seems more like its TN-C-S. TN-C is not allowed for decades right now. But the old installation don't have to be upgraded. I know only of one case where what we called "Bestandschutz" did not apply and that was with the 4 pin/3 Phase Connectors where you file off a nose you can plug it the other way around and have one of the Phases on the Case...

    https://en.wikipedia.org/wiki/Earthing_system

  10. #38
    Join Date
    Oct 2006
    Location
    jonnyGURU forums, of course!
    Posts
    16,305
    Thanks Thanks Given 
    545
    Thanks Thanks Received 
    333
    Thanked in
    243 Posts

    Default

    So it may be that the inductor on the Earth ground is for EMI suppression, but I've yet to see proof of that. I guess it depends on the application. Would love to see some lab data on it.

  11. #39
    Join Date
    Oct 2017
    Location
    Grand Junction, CO
    Posts
    2,127
    Thanks Thanks Given 
    117
    Thanks Thanks Received 
    70
    Thanked in
    63 Posts

    Default

    Perhaps at first they couldn't get certification due to radio transmission on the Earth line.

  12. #40
    Join Date
    Jul 2009
    Location
    Germany
    Posts
    3,716
    Thanks Thanks Given 
    6
    Thanks Thanks Received 
    72
    Thanked in
    66 Posts

    Default

    Or they do it because they've always done it....

Similar Threads

  1. Cannot send messages
    By Folktale in forum Site Suggestions & Forum Issues
    Replies: 4
    Last Post: 08-13-2018, 02:42 PM
  2. Replies: 4
    Last Post: 01-28-2014, 06:35 AM
  3. How psu send power to a graphic card?
    By diezjackdaniels in forum PC Power Supply Discussion
    Replies: 14
    Last Post: 09-14-2013, 05:00 AM
  4. Which to keep......which to send back....PCP&C Silencer 610 vs. Corsair HX620
    By C'DaleRider in forum PC Power Supply Discussion
    Replies: 10
    Last Post: 02-18-2007, 05:18 PM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •