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

Body

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