AW: Re: AW: Re: Aw: Re: Re: Re: Re: AW: [EVGRAY] Re: "Impressive"!

Hello Ole, it is spoken only by a parallel capacitor to scr path. Additional scr are not provided or with this control not so feasible. Or a capacitor is meant which is simply placed on the anode and cathode. Theoretically, when the scr is switched on, the capacitor should charge so that the cathode side of the capacitor heats up positively, the load would probably have to be on the cathode side. Actually, I only need to generate a fast potential change but how and dynamically depending on the impedance of the load and without electronics. I have already thought of class A commutation. Greetings Sven

Von Samsung Mobile gesendet

-------- Ursprüngliche Nachricht --------
Von: "[email protected] [EVGRAY]" <[email protected]> 
Datum:03.03.2018  12:01  (GMT+01:00) 
An: [email protected] 
Betreff: Re: AW: Re: Aw: Re:  Re:  Re:  Re: AW: [EVGRAY] Re: "Impressive"! 

Hi Sven,

As your circuit has two SCRs two more SCRs are needed for switching off the original two SCRs. The switch-off SCRs have to have the cathodes in common and the shorting capacitor between the anode of a main SCR and its switch-off SCR.

If using this scheme of not totally discharging the capacitors it must be ensured that the voltage from the ferro resonant transformer doesn't keep charging the capacitors to some destructive voltage. That is the output from the ferro resonant transformer must not work as a current source. Perhaps this isn't a problem. Else some controller or protection circuit is needed to prevent the voltage from increasing above the component ratings.

Regards
Ole

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



Hi Ole, a capacitor inverts the polarity in the AC circuit in my case not suitable for commutation. you could not cancel the inversion by switching 2 capacitors in the row and there parallel to the Scr anodes. I switch negative and positive with a parallel scr inverter is always switched positive. Greetings Sven

Von Samsung Mobile gesendet


-------- Ursprüngliche Nachricht --------
Von: "onielsen@... [EVGRAY]" 
Datum:02.03.2018 22:03 (GMT+01:00) 
An: [email protected] 
Betreff: Re: Aw: Re: Re: Re: Re: AW: [EVGRAY] Re: "Impressive"! 

 
Hi Sven,

"To get SCR easy to go to 0 current: parallel a second capacitor to SCR circuit that discharges parallel to SCR path as internal current goes to 0 SCR switches off "

This is class C switching here: https://www.electronicshub.org/scr-turn-off-methods/#Class_C_Commutation
When SCR1 is turned on the capacitor C is charged through the resistor R. Then when SCR2 fires the positive side of the capacitor is brought to the cathode potential or close to that potential. The negative side of C is then at a negative voltage with regards to the cathode voltage. This makes the current through RL go through the capacitor until the capacitor is discharged and then charged in the reverse direction which shuts of SCR1. SCR2 is then conducting keeping the positive side of the capacitor at the cathode voltage while the negative side is charging to the positive Vdc voltage through RL. Next time SCR1 fires we are back at the beginning once again which shuts off SCR2 by bringing the positive side of C that was at the cathode voltage down to a negative voltage with regard to the cathode voltage.
"Hector has written that you can discharge such a thyristor quickly."
That is the capacitor or the inductor that is discharged quickly. This is done by consuming the stored energy quickly which is done by choosing the proper impedance. For a capacitor to discharge fast a low impedance is needed. I.e. this gives big current. For an inductor to 'discharge' or demagnetize quickly the impedance must be high giving big voltage.

"I can not imagine anything circuit-technically under this description where this capacitor should be inserted."
The capacitor goes between the two anodes to be able to short the opposite SCR with the discharge pulse of the capacitor bringing the anode of the SCR to shut off to a negative voltage. This reverse biases the SCR thus forcing it to break the current.

""In PLUG switching we need to tailor CHOKE (reactor) to get switching to SCR at Max capacitor Plug
The VARISTOR alike to TUNE was in the Max Voltage turn on phase
peak ,,,
The Idea is to trigger at the top of the sine peak, were not reflective to AC Radiant source .. (reread Transverter Basics) "


Maybe you can explain these sentences to me."
This says the SCRs are to be triggered at max voltage which is when the capacitors are charged to max voltage and charge. This is also when the current reaches zero and if the connection is kept (no diodes) the current reverses to discharge the capacitors. Instead this is when the energy stored in the capacitors can go to the load and thus when the SCRs are to be fired. They must be finished discharging the capacitors before the cycle repeats. Either they must be shut off which gives a flyback voltage because of the inductance or the energy has to be consumed. To consume the energy requires the impedance of the load to be low enough to discharge the capacitors before the nest cycle starts.

If the load is too small (too high impedance) for the power generated the SCRs won't turn off by themselves. Higher load is then to be added (less impedance) for a faster discharge.

Regards
Ole


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


 
Sorry this ist right sentence:
To get SCR easy to go to 0 current: parallel a second capacitor to SCR circuit that discharges parallel to SCR path as
internal current goes to 0 SCR switches off .
 
regards
 
Sven
 
Gesendet: Freitag, 02. März 2018 um 20:23 Uhr
Von: "'Sven Friedrich' s.friedrich@... [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Aw: Re: Re: Re: Re: AW: [EVGRAY] Re: "Impressive"!
 
 

Hi Ole,

a transistor or mosfet makes the control more complicated and vulnerable, or you have to invest more in hedging.

Hector has written that you can discharge such a thyristor quickly.

SCR easy to go to 0 current: parallel to second SCR circuit that discharges parallel to SCR path as
internal current goes to 0 SCR switches off "

I can not imagine anything circuit-technically under this description where this capacitor should be inserted.

This description I also noticed that he uses a choke reactor here. So a coil to additionally influence the triggering of the SCRs!?!

"In PLUG switching we need to tailor CHOKE (reactor) to get switching to SCR at Max capacitor Plug
The VARISTOR alike to TUNE was in the Max Voltage turn on phase
peak ,,,
The Idea is to trigger at the top of the sine peak, were not reflective to AC Radiant source .. (reread Transverter Basics) "


Maybe you can explain these sentences to me.

regards

Sven
  ...

Hi Sven,

When the turn-off thyristor SCR2 is turned on like a switch being closed the capacitor C is placed in parallel to the main thyristor SCR1. This makes the current go through the capacitor that reverse biases the main SCR by making its anode negative when referenced to its cathode. This turns of this main SCR1. Something has to connect the capacitor in parallel to SCR1 at the moment it is to be turned off as the capacitor can't do that by its own being a passive component. Else the load will have to quickly discharge the main capacitors which will also stop the current through the thyristors.

The class A commutation puts conditions to the load that must be part of an under damped RLC tank. Then why not just have the condition to have the load discharge the main capacitors fast enough to turn off the SCRs quickly? That way the circuit of yours needs no modification except that the load impedance has to be low.


Source: https://www.electronicshub.org/scr-turn-off-methods/#Class_C_Commutation https://www.electronicshub.org/scr-turn-off-methods/#Class_C_Commutation

Regards
Ole


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

 

 Hello Ole, it is spoken only by a parallel capacitor to scr path. Additional scr are not provided or with this control not so feasible. Or a capacitor is meant which is simply placed on the anode and cathode. Theoretically, when the scr is switched on, the capacitor should charge so that the cathode side of the capacitor heats up positively, the load would probably have to be on the cathode side. Actually, I only need to generate a fast potential change but how and dynamically depending on the impedance of the load and without electronics. I have already thought of class A commutation. Greetings Sven
...

Hi Ole,

you just have to create a low impedance at the push-pull transformer. The problem is that this transformer is just a isolating transformer so the voltage is not down-converted to 24 volts.

The 230/24 volt transformers I have to connect to the secondary side of the pushpull generates a higher impedance which ensures longer switching times at the SCR.

I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance.

The only question is when the 24V winding is connected with a FWBR to the battery, the battery will act against it with its own voltage and create additional impedance in the push-pull transformer.

The easiest way would be to connect the secondary winding directly to the battery and to charge it with the energy peaks.

I'm just not comfortable because the batteries are in excellent condition and I do not want to destroy them.

regards

Sven

HI SVEN ,I had the same problem running a motor from the RV and connecting to battery the battery would start the motor ,SO what I did was put a large diode in circuit to direct voltage!hope that helps .
 

    On Saturday, March 3, 2018 8:45 AM, "[email protected] [EVGRAY]" <[email protected]> wrote:
 

     Hi Ole,

you just have to create a low impedance at the push-pull transformer. The problem is that this transformer is just a isolating transformer so the voltage is not down-converted to 24 volts.

The 230/24 volt transformers I have to connect to the secondary side of the pushpull generates a higher impedance which ensures longer switching times at the SCR.

I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance.

The only question is when the 24V winding is connected with a FWBR to the battery, the battery will act against it with its own voltage and create additional impedance in the push-pull transformer.

The easiest way would be to connect the secondary winding directly to the battery and to charge it with the energy peaks.

I'm just not comfortable because the batteries are in excellent condition and I do not want to destroy them.

regards

Sven  #yiv3502285781 #yiv3502285781 -- #yiv3502285781ygrp-mkp {border:1px solid #d8d8d8;font-family:Arial;margin:10px 0;padding:0 10px;}#yiv3502285781 #yiv3502285781ygrp-mkp hr {border:1px solid #d8d8d8;}#yiv3502285781 #yiv3502285781ygrp-mkp #yiv3502285781hd {color:#628c2a;font-size:85%;font-weight:700;line-height:122%;margin:10px 0;}#yiv3502285781 #yiv3502285781ygrp-mkp #yiv3502285781ads {margin-bottom:10px;}#yiv3502285781 #yiv3502285781ygrp-mkp .yiv3502285781ad {padding:0 0;}#yiv3502285781 #yiv3502285781ygrp-mkp .yiv3502285781ad p {margin:0;}#yiv3502285781 #yiv3502285781ygrp-mkp .yiv3502285781ad a {color:#0000ff;text-decoration:none;}#yiv3502285781 #yiv3502285781ygrp-sponsor #yiv3502285781ygrp-lc {font-family:Arial;}#yiv3502285781 #yiv3502285781ygrp-sponsor #yiv3502285781ygrp-lc #yiv3502285781hd {margin:10px 0px;font-weight:700;font-size:78%;line-height:122%;}#yiv3502285781 #yiv3502285781ygrp-sponsor #yiv3502285781ygrp-lc .yiv3502285781ad {margin-bottom:10px;padding:0 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Hi Sven,

For transformation between output to load impedance the ratio of the transformer has to fit the impedance ratio to be able to do its job. Else the output voltage and impedance must be chosen to fit the load voltage and impedance or vice versa.

Or the load has to be disconnected from the capacitors even when the capacitors haven't fully discharged which gives the problem of switching off the SCRs. A load that can absorb the energy before the next charging cycle requires the proper resistance for the capacitance value chosen. This resistance can be calculated when the other parameters of the circuit are known. The other circuit parameters can also be calculated for a given load and worked backwards if the load is known as everything affects each other.

"I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance."

Or use a power resistor to dissipate the excess power as heat. This can transfer the energy of the capacitors into heat at almost any rate when correct chosen. The series inductance will be what limits the current and thus must be small enough to be able to discharge the capacitors fast enough to turn off the SCRs before the next capacitor charging. This power resistor can be added in parallel to make the load impedance less or in series to make it greater.

Regards
Ole
 

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

 Hi Ole,

you just have to create a low impedance at the push-pull transformer. The problem is that this transformer is just a isolating transformer so the voltage is not down-converted to 24 volts.

The 230/24 volt transformers I have to connect to the secondary side of the pushpull generates a higher impedance which ensures longer switching times at the SCR.

I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance.

The only question is when the 24V winding is connected with a FWBR to the battery, the battery will act against it with its own voltage and create additional impedance in the push-pull transformer.

The easiest way would be to connect the secondary winding directly to the battery and to charge it with the energy peaks.

I'm just not comfortable because the batteries are in excellent condition and I do not want to destroy them.

regards

Sven

Empty body

Empty body

Hi Sven,

"If I switch 2x 230V / 24V transformers primarily in parallel I reduce the impedance since the energy is distributed to both transformers."
No this doesn't change the impedance. If the primaries are put in parallel and the secondaries are put in series (and in phase) this transforms the impedance to another ratio that delivers the same power to a load having four times the impedance of a load with only one transformer. The output voltage is doubled to 48V and thus the current must be halved for the same output power which quadruples the load impedance.

A switched mode converter can also transform the voltage but a transformer is a simple way of doing that. The transformer just has a fixed ration between the input and output sides. Transformers come in many different ratios and can be custom made for the required ratio.

"I also wonder if I should twist the lines back and forth all together."
This is a must when the frequency is increased and the wires else becomes a coil with one turn and thus stores the current as a magnetic field going through the formed loop. When the wires are twisted they become more like a transmission line that doesn't store the energy but transfers it to some other place.

Regards
Ole

 

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

 Hi Ole,
 
 with a load resistor I waste the energy yes, I would like to push this into the battery.
 
 If I switch 2x 230V / 24V transformers primarily in parallel I reduce the impedance since the energy is distributed to both transformers.
 
 You mean to cache the energy after the push-pull transformer and then discharge it. Help that gets more and more complicated.
 
 There are also people who have loaded the capacitors directly into the battery, apparently the lead-acid batteries keep that out. They are only sharp pulses. It would be nice if you could install a kind of electrical sink, so that the energy of the transformer falls almost without resistance, perhaps with large capacitors and then discharged in the battery with a separate circuit or the capacitors swallow the energy peaks and give them with lower voltage from.
 
 I'm a bit overwhelmed, and working on a new constellation with shorter and equal connections between the diodes. I also wonder if I should twist the lines back and forth all together.
 
 regards
 
 Sven   Gesendet: Samstag, 03. März 2018 um 19:57 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: Re: AW: Re: Aw: Re: Re: Re: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 For transformation between output to load impedance the ratio of the transformer has to fit the impedance ratio to be able to do its job. Else the output voltage and impedance must be chosen to fit the load voltage and impedance or vice versa.
 
 Or the load has to be disconnected from the capacitors even when the capacitors haven't fully discharged which gives the problem of switching off the SCRs. A load that can absorb the energy before the next charging cycle requires the proper resistance for the capacitance value chosen. This resistance can be calculated when the other parameters of the circuit are known. The other circuit parameters can also be calculated for a given load and worked backwards if the load is known as everything affects each other.
 
 "I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance."
 
 Or use a power resistor to dissipate the excess power as heat. This can transfer the energy of the capacitors into heat at almost any rate when correct chosen. The series inductance will be what limits the current and thus must be small enough to be able to discharge the capacitors fast enough to turn off the SCRs before the next capacitor charging. This power resistor can be added in parallel to make the load impedance less or in series to make it greater.
 
 Regards
 Ole
 
 
 ---In [email protected], <s.friedrich@...> wrote :
   Hi Ole,
 
 you just have to create a low impedance at the push-pull transformer. The problem is that this transformer is just a isolating transformer so the voltage is not down-converted to 24 volts.
 
 The 230/24 volt transformers I have to connect to the secondary side of the pushpull generates a higher impedance which ensures longer switching times at the SCR.
 
 I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance.
 
 The only question is when the 24V winding is connected with a FWBR to the battery, the battery will act against it with its own voltage and create additional impedance in the push-pull transformer.
 
 The easiest way would be to connect the secondary winding directly to the battery and to charge it with the energy peaks.
 
 I'm just not comfortable because the batteries are in excellent condition and I do not want to destroy them.
 
 regards
 
 Sven

WernerI find with my RV I must not allow the load on the capacitor tuning circuit to be in circuit until the motor is up to full speed, regardless of tuning.I shut off the load by shorting across the two leads from the transformer or whatever load I have, so it is no longer in series with the capacitors. The full speed motor is robust enough to allow plenty of latitude for tuning, including the load on the capacitor circuit, in my case, an electric outboard boat motor, or another 3hp RV motor.
Starting the motor with the load in circuit is impossible.Cheers Warren
Sent from Yahoo Mail on Android 
 
  On Sat, Mar 3, 2018 at 1:13 PM, werner held [email protected] [EVGRAY]<[email protected]> wrote:       

HI SVEN ,I had the same problem running a motor from the RV and connecting to battery the battery would start the motor ,SO what I did was put a large diode in circuit to direct voltage!hope that helps .
 

    On Saturday, March 3, 2018 8:45 AM, "[email protected] [EVGRAY]" <[email protected]> wrote:
 

     Hi Ole,

you just have to create a low impedance at the push-pull transformer. The problem is that this transformer is just a isolating transformer so the voltage is not down-converted to 24 volts.

The 230/24 volt transformers I have to connect to the secondary side of the pushpull generates a higher impedance which ensures longer switching times at the SCR.

I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance.

The only question is when the 24V winding is connected with a FWBR to the battery, the battery will act against it with its own voltage and create additional impedance in the push-pull transformer.

The easiest way would be to connect the secondary winding directly to the battery and to charge it with the energy peaks.

I'm just not comfortable because the batteries are in excellent condition and I do not want to destroy them.

regards

Sven  

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Hello Ole,

When I turn coils inductors in parallel, the inductance reduces in the same proportion as when I switch capacitors in series. I've tried it myself 2 equal to first at the secondary terminal of the push-pull transformer with full load secondary works. One alone unfortunately not. In the event of a short circuit, the impedance of the individual transformer on the primary side is still too large.

Of course you can connect the secondary windings in series or operate individually on a FWBR and feed together a DC circuit.

Maybe there is also a translation error.

I've already started with the new platform for the diode plug a few parts are still missing to get started.

I have already considered using low sensitive auxiliary thyristors via a voltage divider, so one could use less energy for the ignition and less influence the phase shift. I want to make everything even more modular build shorter lines, preferably so if you look on it can continue to build after the circuit. The cooling problem is no longer a problem with the elements. Luckily I got that for very little money as a scrap commodity.

regards

Sven

in series capacitance voltage increases x 2, capacity is divided by  2  current is halved /2.

inductance

series....      inductance increases X 2  current decreased /2 

parallel....    inductance decreases /2  current increases X 2  


in transformers  , LOAD = low impedance impedance  hi current as LOAD increases so impedance value decreases in OHMs (lower resistance ) 

Yes I know , it fries the brain as you cannot think in mere "linear terms"

So we have to self document ourselves to prevent our own personal & unique brain fucks ....  (yes I am not immune to it ) sometimes I had shlonged my brain dead !   confusing one thing for the other .

 https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/series-and-parallel-inductors/ https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/series-and-parallel-inductors/

here are some helping references , 

(H)    



 

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

 Hello Ole,

When I turn coils inductors in parallel, the inductance reduces in the same proportion as when I switch capacitors in series. I've tried it myself 2 equal to first at the secondary terminal of the push-pull transformer with full load secondary works. One alone unfortunately not. In the event of a short circuit, the impedance of the individual transformer on the primary side is still too large.

Of course you can connect the secondary windings in series or operate individually on a FWBR and feed together a DC circuit.

Maybe there is also a translation error.

I've already started with the new platform for the diode plug a few parts are still missing to get started.

I have already considered using low sensitive auxiliary thyristors via a voltage divider, so one could use less energy for the ignition and less influence the phase shift. I want to make everything even more modular build shorter lines, preferably so if you look on it can continue to build after the circuit. The cooling problem is no longer a problem with the elements. Luckily I got that for very little money as a scrap commodity.

regards

Sven

Help I have no more transformers to lower my primary impedance than parallel connection. laughing.

If you already short-circuit the secondary winding, you can not lower the impedance on the primary side by this method.
These transformers that I have are also highly overloadable, probably the reason for the high primary impedance at full secondary load.

You could also pass the spikes into large DC capacitors via a FWBR and then to the battery. The capacitors swallow the tips and store them in capacity and return them to the battery in smoothed form.

regards

Sven

Hi Sven,

"If you already short-circuit the secondary winding, you can not lower the impedance on the primary side by this method."
I don't think shorting is a good idea to transferring the highest power. For max power transfer to discharge the capacitors as fast as possible the impedance of the load must be the same as that of source. Having higher or lower impedance slows tings down. If working with 5kW or whatever the load must dissipate 5kW. Impedance match gives the best power transfer. You only need the proper load to dissipate the energy before the next cycle starts.

Regards
Ole
 

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

 Help I have no more transformers to lower my primary impedance than parallel connection. laughing.

If you already short-circuit the secondary winding, you can not lower the impedance on the primary side by this method.
These transformers that I have are also highly overloadable, probably the reason for the high primary impedance at full secondary load.

You could also pass the spikes into large DC capacitors via a FWBR and then to the battery. The capacitors swallow the tips and store them in capacity and return them to the battery in smoothed form.

regards

Sven

Empty body

You forgot about the laminate mass  ( it also determines coil impedance ) 

maybe like sex size does matter  , (use a larger transformer) 

(H)