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Hi Sven,
"If I could connect the scope somehow so I can see where I have to flow back into the reactive power with the energy from the transformer ?"
I'm not sure that I understand this question. It is possible to see if the power is reactive or active by using the scope in X-Y mode and having the voltage across a component in the horizontal direction and the current in the vertical direction. A pure resistive load is then shown as a tilted line. The slope of the line is the resistance. If showing an ellipse there's a phase shift between the voltage and the current which is then reactive power. If the ellipse is tilted there is both real and reactive power in the component. I.e. the component both stores and dissipates energy at the same time like a resistive coil or a capacitor having big resistance.
An even better way is to use a scope with math functions to multiply the channel representing the voltage by the channel representing the current and having it show the curve representing the power. This way it can be seen if the component is consuming or delivering power. E.g. this can be used to see if a motor is actually working as motor or generator.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hello Ole, in the video I have 3x 400 Watt spotlight and the DC motor connected. I could slow down the engine with my hand a big difference did not make that. The total active power dropped by 30 watts at this load. Finally, I have a 2 kw fan heater connected now you can hear the scr switch hard. Soon, only the ohmic resistance of the transformer will work. If I could connect the scope somehow so I can see where I have to flow back into the reactive power with the energy from the transformer ?. Greetings Sven
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Datum:13.02.2018 00:04 (GMT+01:00)
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Hi Sven,
The impedance is reduced by mechanically loading the motor or if it is a transformer the secondary is loaded which cancels some of the field in the primary coil.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hi Ole,
Here is my video what I did back then. At high load, the lamp also went out and at idle it shone brightly. So you could extract some of the energy from the primary coil to reduce the impedance. Or am I wrong.
regards
Sven
https://www.youtube.com/watch?v=yTORuC57G0g
Gesendet: Montag, 12. Februar 2018 um 13:10 Uhr
Von: "'Sven Friedrich' s.friedrich@... [EVGRAY]" <[email protected]>
An: [email protected]
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Hi Ole,
I had rather thought of a snubber, about a freewheeling diode, I could unload the primary coil immediately partly.
I tried that with a DC motor and a self-built IGBT push pull inverter. At idle, the coils were disposed of energy via the freewheeling diodes and the capacitor in series and a light bulb. The lighter the secondary, or the lighter the DC motor was, the brighter the bulb was. Under load, the impedance dropped and the energy thrown back, I make a drawing or cut.
regards
Sven Gesendet: Montag, 12. Februar 2018 um 11:59 Uhr
Von: "onielsen@... [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Re: AW: Re: AW: Re: AW: Re: Aw: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
Hi Sven,
The boost and buck converters change the transformation ratio by changing the duty cycle or ratio between input and output part of the cycle. When using thyristors this gives the problem of shutting them off while current are going through them. Shut off has to be done externally by bypassing the current through them. Transistors are usually used instead of thyristors as well as much higher switching frequencies for being able to use smaller transformers for the same power.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hi Ole, I have already connected a 400w spotlight and a 116w bulb. Is not it possible to directly discharge the primary winding like a boost or buck converter to lower the impedance? Greetings Sven
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