Re: Aw: Re: AW: Re: Re: AW: [EVGRAY] Re: buck plug ...

Body

Hi Sven,

"When D1 is loaded there is the positive half wave and at D3 the negative half wave. This means for me if D1 is no longer loaded, then D3 is then loaded because the polarity yes changes. In principle, I load with both half-waves like a FWBR. And, of course, the charged inductor L1 continues to draw current through diode D3 until it is discharged. OK."

The diode plug is not connected like an ordinary full bridge rectifier. Each half of the circuit is connected in common to one of the AC legs. The normal bridge rectifier DC output use the positive and negative legs only and the AC input is connected to the other two legs only. When D3 is conducting (current going through it) the current cannot come from AC as there is no return path for the current. The only return paths to AC are through D1 or D2. No one of these paths are possible.

When charging C1 from AC the current takes the path through D1, L1 the diode without name which isn't really needed then through C1 and back to AC. As the current goes through L1 it magnetizes L1. When the voltage from AC decreases towards zero L1 will start demagnetizing. This makes the current continue at the level it reached when the voltage starts to decrease from its peak value (or actually later because of the phase shift). When AC reaches 0 Volt D3 isn't reverse biased any longer. When AC gets negative U_pos is limited to the forward conduction voltage of around 0.7V to 1V or more of D3. U_pos is then -0.7V to -1V. This voltage is coming from the current through L1 that now decreases as L1 is demagnetizing. L1 now has positive voltage at its right end and negative U_pos (= -0.7V) at its left end. When the demagnetizing current of L1 stops running the voltage across L1 is zero. If the diode without name wasn't there (shorted) the voltage at both ends of L1 is then same as the voltage across C1 (U_pos = .voltage across C1). This makes the negative voltage across D3 disappear. Even when the diode without any name is there the voltage U_pos across D3 can't be negative. This can be used as information for knowing when the capacitor C1 is fully charged during a half cycle from AC.

The end of the signal where U_pos is negative then indicates when C1 has reached its peak voltage. Thus a proper designed trigger circuit will be able to fire the SCR T1 from monitoring the voltage across D3: The same applies for the bottom half part of the circuit but having the signs reversed. The line connecting D3, D4 C1 and C2 is the reference point for both half part trigger circuits.


Regards
Ole
 

---In [email protected], <s.friedrich@...> wrote :

   Hi Ole,
 
 I do not quite get along with it now. We have an alternating voltage of positive half-wave and negative half-wave.
 
 When D1 is loaded there is the positive half wave and at D3 the negative half wave. This means for me if D1 is no longer loaded, then D3 is then loaded because the polarity yes changes. In principle, I load with both half-waves like a FWBR. And, of course, the charged inductor L1 continues to draw current through diode D3 until it is discharged. OK.
 
 I think so soon when I discharge the thyristors so, it may be that they recharge themselves and the SCR does not turn off and everything collapses, you would have to delay again, perhaps through suppressor diodes possibly!?!
 
 I can connect the Scope to both diodes, to be honest, I just see that you can replace the 4 diodes with a FWBR what makes everything a little more compact, I think, I did not even recognize that. laughing
 
 regards
 
 Sven Gesendet: Donnerstag, 15. Februar 2018 um 03:41 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: Re: Aw: Re: AW: [EVGRAY] Re: buck plug ...
    
 Third time.
 
 Hi Sven,
 
 Here is one of your schematics which will do for describing an idea for peak detection. I hope this message doesn't get truncated.
 
 The current through the freewheeling diodes can be monitored by measuring the sign of the voltage across them. By watching when the voltage (U_pos) across D3 is negative this tells that the current through D1 has stopped and the stored energy of inductor L1 is being 'discharged' or actually demagnetized. When the inductor L1 is fully demagnetized the U_pos voltage stops being negative and the capacitor C1 is fully charged. This is the time for discharging C1 by triggering the T1 SCR.
 
 A similar circuit is used for triggering the negative cycle. By doing it this way the triggering point will always happen when the charging current to the capacitors stop flowing.
 
 Regards
 Ole
 
 
 ---In [email protected], <s.friedrich@...> wrote :
    
 Hello Ole, thank you for your comments. I would like to switch to the peak first, unfortunately, the circuit works together then I could imagine if I switch later, I get back to a clean sine wave without distortion. I would like to switch the voltage almost to zero a short peak, which is possible with reactive energy. By impedance reduction, of course, I can switch later or is it now only the impedance that makes it dependent? I think that I discharge the capacitors so now when the current is in zero crossing, otherwise the energy would be reflective to the load because additional current is charged into the capacitors. Greetings Sven
  
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