# Thread: Can I send you a power supply?

1. Originally Posted by turkey3_scratch
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. I bought a couple. I'm not going to do a full on review, but I'm curious to see what they do.

3. 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. Originally Posted by Samueru Sama
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. 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. Originally Posted by Microflop
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. Flux Capacitor User
Join Date
Jul 2009
Location
Germany
Posts
3,716
Thanks
6
Thanks
72
Thanked in
66 Posts
Originally Posted by turkey3_scratch
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. 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. Perhaps at first they couldn't get certification due to radio transmission on the Earth line.

12. Flux Capacitor User
Join Date
Jul 2009
Location
Germany
Posts
3,716
Thanks
6
Thanks
72
Thanked in
66 Posts
Or they do it because they've always done it....

#### Posting Permissions

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