Kapanadze 5


Posted on Feb 6, 2014

Emitter voltage completely drops to 0V from 42V during the pulse (transistor conducting or transistor saturated). That in contrary with the pure resistive load (the 4. 7 Ohm resistor in the last video), where the Collector - Emitter voltage NOT completely drops to 0V from 12 V during the pulse (transistor not fully conducting or not fully saturated) The difference is the pure


Kapanadze 5
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resistive load (4. 7 Ohm) only with low 12V and the combination of toroid (+ flyback) with 1 Ohm resistor with 42V collector voltage. LOL, probably, what i mean is that the Collector - Emitter voltage completely drops to 0V from 42V during the pulse (transistor conducting or transistor saturated). So if i again modify my nano-pulser and use the 100KOhm pot (instead of the 10K now) perhaps i can increase the pulse width even more to fully saturate the transistor at even higher collector voltage. Almost; the probe tip /ground is reversed, so the probe tip is at the junction csr / filter cap, and the ground on the junction flyback diode / csr, see at 13:59 in the video. The slope of the green line represents your unsaturated inductance and the slope of the yellow line represents your saturated inductance. The blue dot placed at the knee of the current trace represents what is known as the "saturation point". The slope of the green line should evaluate to 173 H (your 6-turn inductance) and the slope of the yellow line should evaluate to 30nH (as if the 6t winding was wound over the air, which has r = 1 ). You can disprove this terrifying hypothesis by temporarily shorting out the winding/inductor. When you do this, the current pulse should become rectangular, if the BJT is switching quickly enough. I only query this as a coloured ferrite typically indicates an iron powdered core and these have fairly low permeability. Also...




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