Hi all,
I have a new idea and would like people to have a look at as it is quite simple. I have seen some experienced electronic guys on here so hope to get some feed back before building.
See attached circuit diagram. From my calcs the capacitor & inductor can store just over 1 joules per cycle. The current circuit would only use 50 % of the cycles but another circuit could be added to use the full wave cycle.
The circuit should work by triac T1 firing on the first half cycle which should start the resonance circuit. In theory the parallel LC circuit shouldn't draw any current from outside its own circuit. During the second cycle T1 stops conducting decoupling from the mains and T2 should fire on the negative cycle produced by the LC tank and power dissipated through the dampening load. The cycle then starts again.
Thanks for looking, and I welcome any comments.
Cheers
Jim
Is this for OU or for some other purpose?
Cheers, Pomo 8)
If you want some help you need to reduce the size of your drawing to the size recommended for this forum. Otherwise no one can see all of it without scrolling back and forth. You can edit your post and remove the attachment and then insert a smaller drawing the does not go way beyond the normal width of the forum. There are several free picture editing programs on the net you can get if you need one to help you reduce the size. I use one called "Free Photo Converter". Very easy to use and works great.
Quote from: Jim36 on February 16, 2016, 05:56:33 PM
See attached circuit diagram. From my calcs the capacitor & inductor can store just over 1 joules per cycle.
1 Joule is a lot in electronics.
But why do you think that the parallel LC circuit will give up more energy then it took to energize it?
Quote from: Jim36 on February 16, 2016, 05:56:33 PM
During the second cycle T1 stops conducting decoupling from the mains and T2 should fire on the negative cycle produced by the LC tank and power dissipated through the dampening load. The cycle then starts again.
Please compare the forward voltage drop of a Triac vs. a MOSFET, while keeping in mind that the power lost inside the Triac is equal to the product of the current flowing through it and its voltage drop.
here
Pomo, yes it's a circuit for OU (I hope). :)
Verpies,
Quote from: verpies on February 16, 2016, 07:33:29 PM
1 Joule is a lot in electronics.
But why do you think that the parallel LC circuit will give up more energy then it took to energize it?
Please compare the forward voltage drop of a Triac vs. a MOSFET, while keeping in mind that the power lost inside the Triac is equal to the product of the current flowing through it and its voltage drop.
Yes I know 1 joule is a large amount for electronics, this is a power device utilising mains rated power electronics to switch on and off close to 0V on each half cycle. Actually I calculate it at 1.44Joule though there will be parasitic losses not accounted for like you said across the Triac (I will calculate this). And also losses within the inductor / capacitor parallel circuit though I will size the wire to be able to handle large current and the inductor will be from a transformer primary, capacitor will be a motor run type.
This is what I want to run by you. Why does a parallel LC circuit (tuned to resonance) draw any current on start up when an AC voltage is applied? I haven't practically tested it and probably missing something as it does seem too simple. My thoughts are the capacitor and inductor will require equal current at the same time but of opposite direction as they are 180 degrees out of phase, therefore the current cycles between them selves and does not draw from the mains?
SmOky2, thanks for editing the picture :)
Cheers
Quote from: Jim36 on February 17, 2016, 05:35:06 AM
SmOky2, thanks for editing the picture :)
Cheers
no problem. If you go back and edit your original post, removing the (large) photo, it will resize the thread page and make it easier for people to view.
the problem I see here, is going to be a difference in Impedance between the tank circuit and the other parts of the circuit.
namely T2, and the common lead back to source.
this creates a reactance, to the source.
which will alter the measurement of power input if not properly accounted for.
Quote from: Jim36 on February 17, 2016, 05:35:06 AM
Why does a parallel LC circuit (tuned to resonance) draw any current on start up when an AC voltage is applied?
Because the phase offset between the external AC voltage and the internal AC voltage across the parallel LC circuit is not 0º.
This phase offset occurs not only during startup but also any time that phase of the external AC voltage is changed.
Most importantly, the current draw happens anytime the voltage across the capacitor is smaller than the instantaneous voltage of the the external AC voltage.
Smoky2, I can't edit the original post only my later replies for some reason? I'm new to the forum so having a few teething issues! If you could point me in the right direction :)
Verpies, Thanks for your explanation. You say that a phase offset between the source voltage and voltage across the circuit
cause an initial current draw to charge the parallel circuit which would be exactly the same amount of current as a capacitor only circuit
or inductor only circuit? What phase angle is this initial current with the source voltage? Have you seen this under test
circumstance yourself? I can't find any information on this phenomenon so would appreciate any links to help explain.
Thanks
Quote from: Jim36 on February 17, 2016, 09:46:00 AM
You say that a phase offset between the source voltage and voltage across the circuit cause an initial current draw to charge the parallel circuit
Yes
Quote from: Jim36 on February 17, 2016, 09:46:00 AM
which would be exactly the same amount of current as a capacitor only circuit or inductor only circuit?
No.
The initial current drawn by a single capacitor (or single inductor) would be less than both.
Quote from: Jim36 on February 17, 2016, 09:46:00 AM
What phase angle is this initial current with the source voltage?
180º
Quote from: Jim36 on February 17, 2016, 09:46:00 AM
Have you seen this under test circumstance yourself?
Yes
Quote from: Jim36 on February 17, 2016, 09:46:00 AM
I can't find any information on this phenomenon so would appreciate any links to help explain.
For example
this link (http://falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+10.20027730826997+50+5+43%0Ar+128+48+192+48+0+1000%0As+64+48+128+48+0+1+false%0Ac+160+192+160+96+0+0.00001015+3.274573213447716%0Al+224+96+224+192+0+1+-0.029960052660409516%0Av+64+48+64+240+0+1+50+10+0+0+0.5%0Aw+160+96+192+96+0%0Aw+224+96+192+96+0%0Aw+160+192+192+192+0%0Aw+224+192+192+192+0%0Aw+192+96+192+48+0%0Aw+192+192+192+240+0%0Aw+64+240+192+240+0%0Ag+64+240+64+272+0%0Ao+4+64+0+299+10+0.00625+0+-1%0Ao+2+64+0+299+10+0.05+0+-1%0A).
You can click on the SPST switch to close it. You can double click on any component to edit it.
The green scope trace represents voltage and the yellow trace represents current.
Hovering over a scope trace highlights in blue the component that the trace refers to.
Where are the Output Terminals?
.
Quote from: FatBird on February 17, 2016, 11:28:28 AM
Where are the Output Terminals? .
To which circuit are you referring to?
Hi Verpies, the only problem with the simulation is the series resistor as I was looking at 0.3 ohms due to wiring only, the dampening resistor would be in parallel and switched in on the
second half of a cycle.
Have you actually seen this happen in a physical circuit? I suppose in electronics this is something you might come across often enough. If so I think this thread has been answered and
you have saved me some time!!
Thanks for all the information you have given me.
Cheers
Jim
Quote from: FatBird on February 17, 2016, 11:28:28 AM
Where are the Output Terminals?
.
The output terminals would be across the load as that is where I was looking to dissipate power.
Quote from: Jim36 on February 17, 2016, 02:16:05 PM
Hi Verpies, the only problem with the simulation is the series resistor as I was looking at 0.3 ohms due to wiring only, the dampening resistor would be in parallel and switched in on the second half of a cycle.
You can double-click on that resistor and change its value.
Also you can right click on the black background and add another component (e.g. a 2nd resistor).
Pressing the space bar will allow you to move existing components.
Remember, that in reality the parallel LC tank will also have some resistance in series with the inductor ...unless your inductor is superconducting. The same can be said about the ESR of the capacitor...
Thanks for the information Verpies, that puts that circuit to bed for me. I have another idea that I will try out on the simulation software before posting it for comments.
Can anyone tell me why falstad or LTspice don't simulate the current cycle going into the negative half for an inductor with an ac sine wave source voltage?