AW: [EVGRAY] Re: "Impressive"!

Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven

Von Samsung Mobile gesendet

-------- Ursprüngliche Nachricht --------
Von: "[email protected] [EVGRAY]" <[email protected]> 
Datum:01.03.2018  08:22  (GMT+01:00) 
An: [email protected] 
Betreff: [EVGRAY] Re: "Impressive"! 

Sven,


I don't recall if you are using the diode plug with the inductor or without.

You want to switch fast here is something:
http://www.ti.com/lit/ds/symlink/uc1710.pdf

Hi Sven,

"I want to turn off the scr faster and use no fets."
If forcing them to disrupt the current through the inductors flyback pulses are generated unless shorting the inductors. Actually shorting the inductors would be a way to divert the current from the the SCRs to make them turn off. Then the current stays in the inductors as a magnetic field. I.e  this make the inductors permanent magnets.

Regards
Ole


 

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

 

 Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven
 

 Von Samsung Mobile gesendet



-------- Ursprüngliche Nachricht --------
Von: "mkjekyll@... [EVGRAY]" 
Datum:01.03.2018 08:22 (GMT+01:00) 
An: [email protected] 
Betreff: [EVGRAY] Re: "Impressive"! 

   Sven,
 

 I don't recall if you are using the diode plug with the inductor or without.
 

 You want to switch fast here is something:
 http://www.ti.com/lit/ds/symlink/uc1710.pdf http://www.ti.com/lit/ds/symlink/uc1710.pdf

Empty body

Hi Sven,

"how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?"

A diode reverse biased in parallel with the inductor will delay demagnetization of the inductor. See left side of the image.

 The image is copied and pasted from here: https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch

As the diode as well as the inductor both have resistance the energy is dissipated as i^2 x R heat. But it does prevent the energy being released as a high voltage flyback pulse or actually it is a very slow pulse of low voltage. The voltage is only the voltage across the diode in forward conduction mode. If the inductor is the coil of a relay this method makes the relay release much slower. For a fast release (dissipation of the stored energy) a resistor can be used to dissipate the energy at a much faster rate. As the current at shut off is known the resistance can be chosen to limit the voltage to a level just below the max rating of the semiconductor. U=IR where U is the voltage I is the current and R is the resistance.

Regards
Ole


 

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

 Hi Ole,
 
 how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?
 
 regards
 
 Sven   Gesendet: Donnerstag, 01. März 2018 um 13:22 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "I want to turn off the scr faster and use no fets."
 If forcing them to disrupt the current through the inductors flyback pulses are generated unless shorting the inductors. Actually shorting the inductors would be a way to divert the current from the the SCRs to make them turn off. Then the current stays in the inductors as a magnetic field. I.e  this make the inductors permanent magnets.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
    
 Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven
  
 Von Samsung Mobile gesendet

 
 
 -------- Ursprüngliche Nachricht --------
 Von: "mkjekyll@... [EVGRAY]"
 Datum:01.03.2018 08:22 (GMT+01:00)
 An: [email protected]
 Betreff: [EVGRAY] Re: "Impressive"!
 
   Sven,
  
 I don't recall if you are using the diode plug with the inductor or without.  
 You want to switch fast here is something:
 http://www.ti.com/lit/ds/symlink/uc1710.pdf http://www.ti.com/lit/ds/symlink/uc1710.pdf

Hi Sven
 If you want to short circuit the inductors, short them with very low impedance switching (mechanical brush commutator will work fine)
 

 to ":short" I mean to simply touch the coil leads together, shorting it out!
  
 put a FWBR AC legs across the coil being shorted. (this will also be across the switch too since switch shorts the coils leads together!)
 

 DC side of FWBR goes into capacitor (not too big of UF this capacitor is a "resonant chamber collector capacitor" - not to be a resistive load the system sees which kills everything)
 

 Fill capacitor with no load on it.
 

 Dump capacitor :by itself: to load after it fills up to level you choose for it to fill up to
  ("two stage capacitor output circuit" similar to diode pluig because capacitor is decoupled when it discharges)
 

 Time the switching that shorts coil to occur at the voltage-peak as seen on scope and keep pulse width of the peak coil short fairly narrow too...you will be happy with what you see happen I think expect X20 or X50 the voltage in cap than what the inductor makes with no coil shorting...there should be no reflection back to input draw if you time the coil shorting at the peak, and keep the pulse width short, and do not have huge capacitors too big to collect the hyper-ringing AC that the coil short creates during switch closure... (not switch opening like backemf/recoil collapsing field)
 

 ciao
 Kone

Hi Sven,

This only works if the current through the thyristors are stopped from outside the thyristors by diverting the current around the thyristors for long enough time to remove the charge carriers in the crystal inside the thyristors. The diode then will prevent the destructive flyback voltage. If waiting until the thyristors automatic shut off there is no more current (magnetism) in the inductors and thus no flyback voltage is generated. Thus the diode protection can be omitted.

Diode protection is only necessary if shutting off the thyristors faster than they do when left alone.

Regards
Ole


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

 Hi Ole,
  whether this really works with thyristors, which only switch off when no or little current flows. I can not imagine that this will switch off faster as long as there is still energy flowing. The modern semiconductors are shut down so that the energy can then flow into the inductor.
 
 regards
 
 Sven   Gesendet: Freitag, 02. März 2018 um 00:26 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: Aw: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?"
 
 A diode reverse biased in parallel with the inductor will delay demagnetization of the inductor. See left side of the image.
 
 The image is copied and pasted from here: https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch
 
 As the diode as well as the inductor both have resistance the energy is dissipated as i^2 x R heat. But it does prevent the energy being released as a high voltage flyback pulse or actually it is a very slow pulse of low voltage. The voltage is only the voltage across the diode in forward conduction mode. If the inductor is the coil of a relay this method makes the relay release much slower. For a fast release (dissipation of the stored energy) a resistor can be used to dissipate the energy at a much faster rate. As the current at shut off is known the resistance can be chosen to limit the voltage to a level just below the max rating of the semiconductor. U=IR where U is the voltage I is the current and R is the resistance.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
   Hi Ole,
 
 how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?
 
 regards
 
 Sven   Gesendet: Donnerstag, 01. März 2018 um 13:22 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "I want to turn off the scr faster and use no fets."
 If forcing them to disrupt the current through the inductors flyback pulses are generated unless shorting the inductors. Actually shorting the inductors would be a way to divert the current from the the SCRs to make them turn off. Then the current stays in the inductors as a magnetic field. I.e  this make the inductors permanent magnets.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
    
 Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven
  
 Von Samsung Mobile gesendet
 
 
 
 -------- Ursprüngliche Nachricht --------
 Von: "mkjekyll@... [EVGRAY]"
 Datum:01.03.2018 08:22 (GMT+01:00)
 An: [email protected]
 Betreff: [EVGRAY] Re: "Impressive"!
 
   Sven,
  
 I don't recall if you are using the diode plug with the inductor or without.  
 You want to switch fast here is something:
 http://www.ti.com/lit/ds/symlink/uc1710.pdf http://www.ti.com/lit/ds/symlink/uc1710.pdf

Empty body

Hi Sven,

To turn off the thyristor the current just has to be prevented from going through the thyristor. This can be done by discharging a capacitor across the anode-cathode legs to make the current go through the capacitor for a brief moment of time. This is class C and D here: https://www.electronicshub.org/scr-turn-off-methods/ https://www.electronicshub.org/scr-turn-off-methods/

A transistor can be used as a switch to discharge the capacitor instead of a second thyristor. If using a transistor directly in parallel to the thyristor the transistor has to have much less conduction resistance than the SCR which makes the transistor probably better as a switch instead of the SCR.

Regards
Ole


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

 Hello Ole, that's right, but how do I turn it off faster? I have already looked at everything with SCR Commutation. The only thing I think that can work is a capacitor in series at the cathode and a bleeder resistor in parallel, so I'll change the whole signal and bring an additional active share.
 
 regards
 
 Sven   Gesendet: Freitag, 02. März 2018 um 18:22 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: Aw: Re: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 This only works if the current through the thyristors are stopped from outside the thyristors by diverting the current around the thyristors for long enough time to remove the charge carriers in the crystal inside the thyristors. The diode then will prevent the destructive flyback voltage. If waiting until the thyristors automatic shut off there is no more current (magnetism) in the inductors and thus no flyback voltage is generated. Thus the diode protection can be omitted.
 
 Diode protection is only necessary if shutting off the thyristors faster than they do when left alone.
 
 Regards
 Ole
 
 
 ---In [email protected], <s.friedrich@...> wrote :
  
 Hi Ole,
  whether this really works with thyristors, which only switch off when no or little current flows. I can not imagine that this will switch off faster as long as there is still energy flowing. The modern semiconductors are shut down so that the energy can then flow into the inductor.
 
 regards
 
 Sven   Gesendet: Freitag, 02. März 2018 um 00:26 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: Aw: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?"
 
 A diode reverse biased in parallel with the inductor will delay demagnetization of the inductor. See left side of the image.
 
 The image is copied and pasted from here: https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch
 
 As the diode as well as the inductor both have resistance the energy is dissipated as i^2 x R heat. But it does prevent the energy being released as a high voltage flyback pulse or actually it is a very slow pulse of low voltage. The voltage is only the voltage across the diode in forward conduction mode. If the inductor is the coil of a relay this method makes the relay release much slower. For a fast release (dissipation of the stored energy) a resistor can be used to dissipate the energy at a much faster rate. As the current at shut off is known the resistance can be chosen to limit the voltage to a level just below the max rating of the semiconductor. U=IR where U is the voltage I is the current and R is the resistance.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
   Hi Ole,
 
 how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?
 
 regards
 
 Sven   Gesendet: Donnerstag, 01. März 2018 um 13:22 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "I want to turn off the scr faster and use no fets."
 If forcing them to disrupt the current through the inductors flyback pulses are generated unless shorting the inductors. Actually shorting the inductors would be a way to divert the current from the the SCRs to make them turn off. Then the current stays in the inductors as a magnetic field. I.e  this make the inductors permanent magnets.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
    
 Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven
  
 Von Samsung Mobile gesendet
 
 
 
 -------- Ursprüngliche Nachricht --------
 Von: "mkjekyll@... [EVGRAY]"
 Datum:01.03.2018 08:22 (GMT+01:00)
 An: [email protected]
 Betreff: [EVGRAY] Re: "Impressive"!
 
   Sven,
  
 I don't recall if you are using the diode plug with the inductor or without.  
 You want to switch fast here is something:
 http://www.ti.com/lit/ds/symlink/uc1710.pdf http://www.ti.com/lit/ds/symlink/uc1710.pdf

Empty body

Empty body

Empty body

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 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,

It only works until enough energy is stored in the coil to saturate it. By putting a lamp in series with the diode the energy will be dissipated faster and thus demagnetize the coil.

Is the coil in series with the output transformer? I seem to have forgotten that. Else there may be nothing gained here.

Regards
Ole


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

   Hi Ole,
 
 That would really work by attaching a freewheeling diode to the winding to feed the magnetizing current back into the coil.
 
 It is worth a try. I can switch a light bulb in series with the diode so you can visually see if energy flows through the diode.
 
 regards
 
 Sven Gesendet: Freitag, 02. März 2018 um 00:26 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: Aw: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?"
 
 A diode reverse biased in parallel with the inductor will delay demagnetization of the inductor. See left side of the image.
 
 The image is copied and pasted from here: https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch
 
 As the diode as well as the inductor both have resistance the energy is dissipated as i^2 x R heat. But it does prevent the energy being released as a high voltage flyback pulse or actually it is a very slow pulse of low voltage. The voltage is only the voltage across the diode in forward conduction mode. If the inductor is the coil of a relay this method makes the relay release much slower. For a fast release (dissipation of the stored energy) a resistor can be used to dissipate the energy at a much faster rate. As the current at shut off is known the resistance can be chosen to limit the voltage to a level just below the max rating of the semiconductor. U=IR where U is the voltage I is the current and R is the resistance.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
   Hi Ole,
 
 how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?
 
 regards
 
 Sven   Gesendet: Donnerstag, 01. März 2018 um 13:22 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "I want to turn off the scr faster and use no fets."
 If forcing them to disrupt the current through the inductors flyback pulses are generated unless shorting the inductors. Actually shorting the inductors would be a way to divert the current from the the SCRs to make them turn off. Then the current stays in the inductors as a magnetic field. I.e  this make the inductors permanent magnets.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
    
 Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven
  
 Von Samsung Mobile gesendet
 
 
 
 -------- Ursprüngliche Nachricht --------
 Von: "mkjekyll@... [EVGRAY]"
 Datum:01.03.2018 08:22 (GMT+01:00)
 An: [email protected]
 Betreff: [EVGRAY] Re: "Impressive"!
 
   Sven,
  
 I don't recall if you are using the diode plug with the inductor or without.  
 You want to switch fast here is something:
 http://www.ti.com/lit/ds/symlink/uc1710.pdf http://www.ti.com/lit/ds/symlink/uc1710.pdf

Empty body

Empty body

Hi Sven,

OK. Some of the power is then taken out on the primary side of the output transformer by the lamp. The impedance transformation through the transformer has the wrong transformation ratio from input to output then. Or the load isn't of high enough power to dissipate the energy fast enough at the given secondary voltage. Add more load in parallel and that's not by just paralleling transformers. It has to be resistive load to be able to dissipate the power. A motor with a load that brakes it also takes active power. At impedance match is where the product of the output voltage and current is max.

Regards
Ole


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

 Hi Ole, the coil ????
 
 you mean the Variacs ???
 
 At the push-pull transformer I have 2 primary windings which I connect with each thyristor only at this winding I put the free-wheeling diode with a light bulb in series. It works wonderfully.
 At low load on the secondary side of the transformer, it shines brighter than at high load, then only a slight glow is visible. A very clear sign that the whole magnetic energy is transferred to the secondary side.
 
 regards
 
 Sven   Gesendet: Sonntag, 04. März 2018 um 16:45 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: Aw: Re: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 It only works until enough energy is stored in the coil to saturate it. By putting a lamp in series with the diode the energy will be dissipated faster and thus demagnetize the coil.
 
 Is the coil in series with the output transformer? I seem to have forgotten that. Else there may be nothing gained here.
 
 Regards
 Ole
 
 
 ---In [email protected], <s.friedrich@...> wrote :
  
   Hi Ole,
 
 That would really work by attaching a freewheeling diode to the winding to feed the magnetizing current back into the coil.
 
 It is worth a try. I can switch a light bulb in series with the diode so you can visually see if energy flows through the diode.
 
 regards
 
 Sven Gesendet: Freitag, 02. März 2018 um 00:26 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: Aw: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?"
 
 A diode reverse biased in parallel with the inductor will delay demagnetization of the inductor. See left side of the image.
 
 The image is copied and pasted from here: https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch https://electronics.stackexchange.com/questions/31014/where-should-i-put-the-kickback-diode-in-a-transistor-switch
 
 As the diode as well as the inductor both have resistance the energy is dissipated as i^2 x R heat. But it does prevent the energy being released as a high voltage flyback pulse or actually it is a very slow pulse of low voltage. The voltage is only the voltage across the diode in forward conduction mode. If the inductor is the coil of a relay this method makes the relay release much slower. For a fast release (dissipation of the stored energy) a resistor can be used to dissipate the energy at a much faster rate. As the current at shut off is known the resistance can be chosen to limit the voltage to a level just below the max rating of the semiconductor. U=IR where U is the voltage I is the current and R is the resistance.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
   Hi Ole,
 
 how should I short-circuit the inductors?
 
 With a magnetic shunt or semiconductors?
 
 regards
 
 Sven   Gesendet: Donnerstag, 01. März 2018 um 13:22 Uhr
 Von: "onielsen@... [EVGRAY]" <[email protected]>
 An: [email protected]
 Betreff: Re: AW: [EVGRAY] Re: "Impressive"!
   Hi Sven,
 
 "I want to turn off the scr faster and use no fets."
 If forcing them to disrupt the current through the inductors flyback pulses are generated unless shorting the inductors. Actually shorting the inductors would be a way to divert the current from the the SCRs to make them turn off. Then the current stays in the inductors as a magnetic field. I.e  this make the inductors permanent magnets.
 
 Regards
 Ole
 
  
 
 
 ---In [email protected], <s.friedrich@...> wrote :
    
 Hi Mick, with adjustable inductance a variac to work for any diode plug. I want to turn off the scr faster and use no fets. Greetings Sven
  
 Von Samsung Mobile gesendet
 
 
 
 -------- Ursprüngliche Nachricht --------
 Von: "mkjekyll@... [EVGRAY]"
 Datum:01.03.2018 08:22 (GMT+01:00)
 An: [email protected]
 Betreff: [EVGRAY] Re: "Impressive"!
 
   Sven,
  
 I don't recall if you are using the diode plug with the inductor or without.  
 You want to switch fast here is something:
 http://www.ti.com/lit/ds/symlink/uc1710.pdf http://www.ti.com/lit/ds/symlink/uc1710.pdf

Hi Ole,

that is not so much, the incandescent lamp with 116 watts light up slightly with a 400 watts load on the secondary side, at 1200 watts load with a 3x 400 watt incandescent bulbs the 116W incandescent light only slightly, so you can once again shorten the switching times minimal perhaps If necessary, install a small capacitor and the light bulb parallel to it. It is signaled anyway how the impedance behaves, a great indicator. The thyristors switch off really fast, the actual 5 milli seconds I have to unload is not nearly needed. I think that with me the synchrony of the components and the length of the lines is a problem.

I recently connected the rotating iron meters for electricity differently or took the other line to the capacitor bank and measured slightly less current there than in the other current path.

The problem of how should I change that apparently does not flow everywhere at the reactive power the same electrical currents.

Of course I have to split the lines through the capacitor bank with the switches so that I can switch several capacitors in parallel. Can the switch contacts of the switches cause this difference. Or is it due to the cable length at the transverter. I have these differences without diode plug or other observed and measured.

regards

Sven

Hi Sven,

"Of course I have to split the lines through the capacitor bank with the switches so that I can switch several capacitors in parallel. Can the switch contacts of the switches cause this difference. Or is it due to the cable length at the transverter. I have these differences without diode plug or other observed and measured."
I've been looking for your schematic. If it's still like this pasted one the positive and negative sides don't trig at the same voltage. To be able to do that the triggering circuits of each side have to see the same potential. The tolerances of the capacitors can also be the problem. If switching place of the capacitors also shifts the asymmetry it is because the capacitors aren't of the same value.

The left hand side has the filter resistor go to the anode of the thyristor. The right hand side has its filter resistor go to the opposite pole of the main capacitor and thus only see the voltage across that capacitor. The left side ought to be coupled that same way for symmetry. I.e. move the 1k resistor to the other side of the primary winding to connect the triggering input directly across the left side 15uF capacitor.

Regards
Ole

 

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

 Hi Ole,

that is not so much, the incandescent lamp with 116 watts light up slightly with a 400 watts load on the secondary side, at 1200 watts load with a 3x 400 watt incandescent bulbs the 116W incandescent light only slightly, so you can once again shorten the switching times minimal perhaps If necessary, install a small capacitor and the light bulb parallel to it. It is signaled anyway how the impedance behaves, a great indicator. The thyristors switch off really fast, the actual 5 milli seconds I have to unload is not nearly needed. I think that with me the synchrony of the components and the length of the lines is a problem.

I recently connected the rotating iron meters for electricity differently or took the other line to the capacitor bank and measured slightly less current there than in the other current path.

The problem of how should I change that apparently does not flow everywhere at the reactive power the same electrical currents.

Of course I have to split the lines through the capacitor bank with the switches so that I can switch several capacitors in parallel. Can the switch contacts of the switches cause this difference. Or is it due to the cable length at the transverter. I have these differences without diode plug or other observed and measured.

regards

Sven

Hi Sven,

"Of course I have to split the lines through the capacitor bank with the switches so that I can switch several capacitors in parallel. Can the switch contacts of the switches cause this difference. Or is it due to the cable length at the transverter. I have these differences without diode plug or other observed and measured."
I've been looking for your schematic. If it's still like this pasted one the positive and negative sides don't trig at the same voltage. To be able to do that the triggering circuits of each side have to see the same potential. The tolerances of the capacitors can also be the problem. If switching place of the capacitors also shifts the asymmetry it is because the capacitors aren't of the same value.

The left hand side has the filter resistor go to the anode of the thyristor. The right hand side has its filter resistor go to the opposite pole of the main capacitor and thus only see the voltage across that capacitor. The left side ought to be coupled that same way for symmetry. I.e. move the 1k resistor to the other side of the primary winding to connect the triggering input directly across the left side 15uF capacitor.

Regards
Ole 

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

 Hi Ole,

that is not so much, the incandescent lamp with 116 watts light up slightly with a 400 watts load on the secondary side, at 1200 watts load with a 3x 400 watt incandescent bulbs the 116W incandescent light only slightly, so you can once again shorten the switching times minimal perhaps If necessary, install a small capacitor and the light bulb parallel to it. It is signaled anyway how the impedance behaves, a great indicator. The thyristors switch off really fast, the actual 5 milli seconds I have to unload is not nearly needed. I think that with me the synchrony of the components and the length of the lines is a problem.

I recently connected the rotating iron meters for electricity differently or took the other line to the capacitor bank and measured slightly less current there than in the other current path.

The problem of how should I change that apparently does not flow everywhere at the reactive power the same electrical currents.

Of course I have to split the lines through the capacitor bank with the switches so that I can switch several capacitors in parallel. Can the switch contacts of the switches cause this difference. Or is it due to the cable length at the transverter. I have these differences without diode plug or other observed and measured.

regards

Sven