Is the electric field stronger in resistors with higher resistance?

[Mr.Bomzh]

another similar schematic , I wanted to ask , isn't the amount of charge flow in one cycle limited to the 1uF/250v capacitor in series with the transformer primary in tis schematic?
would seem logic that the amount of work/charge flow this circuit can accomplish in one cycle to be limited by that capacitor is that correct?
A dc path , but where is it , is it through the capacitor in series with the primary ?
I know this cirucuit is different from what I was talking , already said that in my previous post , I was rather interested in some details of the work of this circuit that I asked about :)

 

[sophiecentaur]

There is a DC path around your second circuit too. Do you really want to disbelieve everything I tell you? You are so convinced you are right that it is not good for your ability to progress your learning. Pleas stop asking these silly questions in an effort 'not to be wrong'.

 

[Mr.Bomzh]

Sophie , I am really not tryng to prove myself right , now your just being judgmental of me because of my previous questions and the fact that I couldn't let go for a while.I said that now I wanted to gain understanding on some aspects of this circuit. you are a good help but your aslo very judging towards me on this one.just please don't think about what I have said sometime ago , ok a dc path around this circit too, I see that as it is almoust 1:1 simiolar to the one i posted before but where is that dc path ? It can;'t be through the mosfets so where else it is , I assume through the capacitor leading to the large filter caps on the mains input ? But why is that a dc path ? I'm not disbelieving you I just want to know why a dc path can go through a capacitor ?
as I thought only AC can pass through a capacitor
also was I rigt about the fact that in these circuit the amount of current in each cycle is limited by the capacitance of the capacitor in series we are talking baout now ?

 

[sophiecentaur]

I am fed up with your responses. I told you why the first circuit wouldn't work. You didn't believe me. I told you why the second circuit doesn't work your way. You didn't believe me. Now you refuse to see why the third circuit doesn't work your way.
Take your finger to the mains input, trace it through the diodes, then through a mosfet, then then back, through the other mosfer, to the other mains terminal. THAT is a DC path for the mains to supply current through, each cycle. The output is AC coupled to the transformer - but that is fine (most audio amplifiers do that), you can still draw power from the supply into the amplifier -

I must apologise for saying there is a DC path through the transformer in those circuits - that is not needed. What counts, is the DC path from the supply - which was why your first idea would not work as you thought.

Why don't you look at some simpler circuits and get to understand how they function before going to this complexity? Just because you may have wired some stuff up successfully doesn't mean you understand it. It can lull you into a false sense of 'knowing' when things 'just work'.

 

[Mr.Bomzh]

i didn't said that this should work my way , I was just asking where is the DC path which took you two posts to answer because you were so " fed up" with me that you first had to explain me that and only then we could get down to business again. :)If you would be my grandfather or something I would just clap you on the sholder and tell you it's ok , no need to be angry I understand. Let's just say I'm too much of an optimist when it comes to my ideas. :D

Ok I now understand what DC path you were talking about , if one would switch both of the mosfets on at the same time , assuming they could whitstand this , a short circuit dc path would form, from the mains rectifier , through the mosfets back to the rectifier diodes.In such a scenario almoust no current would flow through the transformer as it has a capacitor in series with it.I guess I'm correct on this one?

But because both of the mosfets are not switched on at the same time they act as a push pull design right? when one turns on the current flows through the transformer primary to the cap until the cap is charged and when the other turns on the cap is being recharged and the current flows the other direction and now back to the maisn rectifier, is that correct?

Also at each cycle the design operates with half the rectified mains voltage , that would be because the transformer primary is tied to the middle point between the filter caps which form a voltage divider and in the middle you have rectified mains/2 is that correct?

And finally was I right in asking that in each cycle the work done through the transformer is equal to the capacitance of that capacitor in series wtih the primary ? I'm asking this because you didn't answer this one.

Thank you Sophie, Simon is also welcomed :)

 

[sophiecentaur]

Haha. I am probably as old as your GF.
Yes you have the DC path right.
Energy in. Capacitor is half C Vsquared so energy in and out each cycle is not just "capacity" - more complicated. If no load on output the charge in and out of C is zero, I think, because all volts are dropped across tra nsformer primary. That charge is only due to power consumed by load.

 

[Mr.Bomzh]

Not sure about the grandpa , mine is 86 years old.we still live togethr and I help him out , but he walks fine on his own too.
But why would thee be no charge flow if there is no load on the transformer?
the transistors switch on and off one at a time so that means that the cap is introduced to +ve rail and then the +ve rail is switched off and the groud rail is switched on , doesn't the positive charge flow away from the cap each time the lower transistor comes on? To my understanding it should as the lower transistor has a direct path to the negative rail of the power supply, and before that the positive rail was connected to the cap while the other plate of the cap is attached to the middle point , so the cap is connected to 230 x1.414 =325v DC and since the filter caps are a voltage divider then 325v /2=162v DC.
So if the cap gets charges to a potential which equals 162v everytime the upper transistor comes on ,hmm I think I kinda start to get why you say there are no charge flow if the transformer is at idle , ok let me ask this , if the traffo is at idle is there any PD across that cap on each cycle? and if there is does it change from cycle to cycle or no?

 

[sophiecentaur]

Don't forget that the primary of the transformer has very high (ideally, infinite) self inductance so, unless some current is allowed to pass through the secondary, no current will pass through it.* Hence, the 2.2uF capacitor will have a steady DC potential (zero) across it of zero; it doesn't actually charge / discharge until some current is taken by the secondary. No power out gets no power in (ideally).
Yes - your GF is a bit older than me. (Older, maybe but not uglier!)

*This is standard transformer theory and applies, in practice to all set ups with properly designed transformers. Your Electricity company's transformers only take appreciable power in when people start to demand power. Anything else would make the whole power supply industry impracticable. Google transformer theory.

 

[Mr.Bomzh]

Ah right , I was soo focused on the capacitors and transistors that I almoust forgot about the back emf that arises in all kinds of coils, transformers, electric motors , inductors.
So unless no load is applied the back EMF is ideally as strong as the applied voltage/current, then the transformer acts like a mirror to any incoming charge?
So a copper wire has some low resistance but if it is wound on a transformer , then ideally for the first instant it acts like a infinite resistor, only in this case the current is resisted by field rather than real resistance of a non ideal conductor?

Ok so if the transformer is under load , is the maximum amount of energy under load in each cycle delivered through the transformer limited by the capacitance of the capacitor is that correct?
As you said under load current flows both in primary and secondary as they are coupled , but the current in the pirmary can flow only until the cap is full and then it stops so at that moment if the switch would still be on and the cycle long enough the current would just drop until become zero?

 

[sophiecentaur]

The series LC circuit would be resonant at the design frequency, I think. The load would be specified so that there would be no limiting.
What do you mean by the Capacitor being 'full'? How can you fill a capacitor? Q =CV, until it breaks down.

 

[Mr.Bomzh]

Ok I see , you can increase the PD across a cap until the charge build up per given plate area is so huge that the caps safety margin is crossed and breakdown through the dielectric may and with time happens.

 

[sophiecentaur]

Ok I see , you can increase the PD across a cap until the charge build up per given plate area is so huge that the caps safety margin is crossed and breakdown through the dielectric may and with time happens.

The cause and effect are the other way round, I would say. As the capacitor is charged, the PD increases. It's the work done on the charges that equals the PD.

Not all that dramatic. haha There are capacitors with only 12V operating voltage. The higher the capacity, the closer the plates - so the lower the operating voltage (for a given form of construction). Think of charging a capacitor through a resistor.

It's a long time since I was involved in detailed circuitry but the secret is always in spotting the important components and ignoring the 'supporting parts'. Once you eliminate the decoupling Cs and the biasing Rs, the circuit becomes more understandable. Of course, you have to be careful because some components perform more than one function.

Aren't we 'done' with this now?

 

[Mr.Bomzh]

you wish we would be don't you ? :D

Well we are kinda done in terms of that I do understand caps a lot better now and the theory also, so even with all the fed up and repeating stuff that was going on here we managed to make a increase in knowledge here:)
But it's not the end of the line , I will probably have some threads sooner or later about my crazy ideas.Although knowledge base helps in whenever you hava an idea so that you can check it for faults and inconsistencies.
Anyway thanks.