Re: Aw: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.

Body

Hi Sven,

The gate trigger current of the main SCR http://ixapps.ixys.com/datasheet/mcd162-16io1.pdf can be up to 200mA. If the voltage across the capacitor is 400V at the point of triggering then a gate resistor at 1.1kOhm makes the gate current around 364mA. With 400V across and 364mA through the gate resistor some 146W of power is dissipated in the resistor during the brief moment of triggering.

"That's right, but you should only give back a certain part of it, part should be free energy that we store in the capacitors, otherwise the circuit makes no sense."
The part given back is the reactive part of the power. The part being consumed is the active (or real) power. As long as only making the free energy into active power the active power then is free energy.

"I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point where this happens explicitly in the circuit could you perhaps do something about it?"
The parts of the circuit having Ohmic resistances are the ones with the phase factor of one. Any reactive impedance which is caused by inductance or capacitance has the phase factor different from one. At resonance the inductive and capacitive reactances cancel each other as one is positive and the other is negative.

"I meant the stray field of the push pull trafos or the energy which is switched on the push pull back into the primary windings. Without secondary load you can not set the diode plug and this load must also be ohmic nature, a transformer with no load on the secondary of the push-pull transformer stops the switching in the diode plug."
Yes there could be a problem here. At no load the power is pure reactive being reflected back to source.

"You could work with freewheeling diodes on each winding and burn the energy through resistors, but that further increases the losses and only work with freewheeling diodes is too risky..."
If the energy is free it could be dissipated as heat or the device could be automatic shut off when there is no demand for power. When the demand for power returns the device then must also start automatic.

"I will try to work with smaller discharge capacitors so that the voltages on the diode plug side and the transverter are pretty much the same."
It's not the capacitors that determine the voltage when to be discharged. It's the triggering circuit parameters of the voltage divider that fires the SCR. Smaller capacitance only makes the capacitors charge/discharge faster and use less cycles for given voltage build up.

Regards
Ole
 

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

 Hi Ole,
 
 The voltage divider is at signal level. I do not think power resistors are necessary for triggering the power thyristor. The break down voltage is still quite high for the small signal thyristor which makes the voltage decrease.
 
 That's right, it worked very well only until I had set the diode plug started to heat the 2 watt resistors, so abe I replaced it with 2x 9Watt ceramic resistors. Before I had 2.2 kohm now 1.1 kohm. I can not imagine that the resistors are responsible for the positive phase shift, but in fact they consume active power.
 
 
 "While the circuit is not reflective of the source, a lot of extra power is applied to charge and discharge the capacitors"
 Charging the capacitor is real power when not given back the charge to the source. Only by giving back the charge makes the power reactive.
 
 That's right, but you should only give back a certain part of it, part should be free energy that we store in the capacitors, otherwise the circuit makes no sense.
 
 
 "Somewhere the power factor shifts again direction 1 but how do I find out?"
 
 The power factor is given by which way the phase shift is. A power factor of one means no phase shift between the voltage and current. If the voltage increases / decreases before the current does the power is inductive reactive. If the phase shift is opposite the power is capacitive reactive.
 
 I meant only with the shift of the power factor, the additional active power is taken from the reactive circuit, if one finds the point where this happens explicitly in the circuit could you perhaps do something about it?
 
 
 
 
 
 
 "The stray field energy is so a problem I can not get rid of them so easily because of this converter works with alternating polarities for the supply."
 
 To keep stray fields low avoid current running through big loops. This can be done by twisting the wires in pairs. E.g. twist the forward current carrying wire. The twisting makes the small loops left alternate in different directions thus canceling the fields from each other. Alternative use coaxial cables which therefore cancel the fields.
 
 I meant the stray field of the push pull trafos or the energy which is switched on the push pull back into the primary windings. Without secondary load you can not set the diode plug and this load must also be ohmic nature, a transformer with no load on the secondary of the push-pull transformer stops the switching in the diode plug.
 
 You could work with freewheeling diodes on each winding and burn the energy through resistors, but that further increases the losses and only work with freewheeling diodes is too risky, because I switched through directly with a defective diodes without load and the power thyristor was defective.
  
 I will try to work with smaller discharge capacitors so that the voltages on the diode plug side and the transverter are pretty much the same. So always creates a sink, that is 400V at the transverter and 250V at the diode plug more create the coupling capacitors not, that is principle that I try too much energy to extract.
 
 regards
 
 Sven
  ...