Kapanadze Cousin - DALLY FREE ENERGY

[TinselKoala]

If your scope has a trace "add" or A+B function you can also use the two probes as a "differential", with both reference leads hooked at circuit ground, and the tips across your CSR. The voltage difference between the probe tips gives your current by Ohm's Law, so you "add" the more negative reading to the inverse of the other one, effectively performing the subtraction and displaying the differential voltage signal on a single trace.

[Сергей В.]

to itsu

You have problem with your LCR meter. Never mind, use RF-genarator and Oscillograph to precisely measure inductance L and interwinding capacitance CL. Take a look on picture.

Turn on the switch P (close the switch) and sweep frequency of signal generator to catch resonance in measuring coil. Track envelope on oscillograph. It's a frequency Fr1. Than off the switch (open it) and repat the same. Catch the frequency Fr2. Now calculate L and CL. Use good quality standard capacitor with tolerance 0.1% or better.  Check it and measure once before measurements.

Take care !!

Regards Сергей В.

[d3x0r]

Sorry I am still learning, but...


if I have a capacitor with 12V and 1uf; that is the supply for a npn transistor(call it U1) that can handle 8A at 100V with a 5V drive; this is then fed to the base (gate) of my E13009 npn(call it u2) which can handle 6A drive on the base for 12A current conducted...


without a resistor; the 1uf capacitor could be 12A in 1us...


so if I put a 2ohm resistor between the emitter of U1 to gate of U2... how do I know it won't get 0.5V and still 12A instead of 12V and 0.5A?


basically I need to cut the current in half, but I never know if resistors aren't just voltage sinks, and current stays the same or if they drop current and leave voltage the same...

[verpies]

Top trace of attached photo 11 is voltage C/E. Lower trace is C/E current taken with a 0.1R CSR. Probe x10.

Do you have your CSR positioned like shown here or here?
The trace seems to indicate the first option.

The collector current seems to rise at the rate of 7A/μs and the collector supply voltage seems to be 65V  which calculates to 9μH of primary (W1) inductance, if the secondary (W2) is not loaded (or is it 70A/μs and 650V because of the "probe x10" ?)
The horizontal time scale is too dense to see any meaningful change in the slope of this curve, thus the 9μH is probably a mix of the saturated and unsaturated inductances.

Photo 12 shows transformer primary current taken with 0.1R CSR. Probe x10.

Pretty messy but that trace nicely shows the long decay current in the primary winding. This long decay should be caused be the Flyback Diode (D1) working quite well.
I don't like the polarity of that short spike in the beginning, because it indicates that the current is reversing in the CSR.  That should not be possible if the secondary (W2) is disconnected.

[verpies]

if I have a capacitor (C1) with 12V and 1uf; that is the supply for a npn transistor(call it U1)

How is C1 supplying the transistor?  e.g.: in common emitter or common collector mode?  Is C1 in the emitter path or in the base path?
Please don't call a transistor U1 (that's reserved for integrated circuits). Let's call your transistors Q1 and Q2, from now on.

that can handle 8A at 100V with a 5V drive; this is then fed to the base (gate) of my E13009 npn(call it u2) which can handle 6A drive on the base for 12A current conducted...

Is that a BJT transistor?  If "yes" then it does not have a "gate". Only FETs have "gates".  BJTs have bases. It's not only  a difference in the name, they also act differently. The base of a BJT acts like a diode (or a resistor) while the gate of a MOSFET acts like a small capacitor.

Also BJTs are not voltage driven - they are current driven. Thus 5V of forward voltage applied into the base will damage Q1 immediately if the current is not limited by a resistor.  BTW: The 5V is most likely the maximum reverse voltage that will not break down your transistor's base-emitter junction in reverse.
Also transistor's max ratings are not indicative of the transistor's transconductance.  To figure out how much collector current (I_C) you will get for a given base current (I_B), use the equation: I_C=I_B*h_FE.
The h_FE parameter is also known as beta (β).  See this article.

The resistor in series with the base works like a voltage to current converter.
For example if you apply 5V to a base through a 2Ω resistor (R_B) then I_B=(V-0.7V)/R_B of current will flow. The 0.7V is the constant voltage drop across the the base-emitter junction.  This calculates to I_B=4.3V/2Ω = 2.2 Amps.
If a transistor, that has a β or h_FE equal to 100, is fed with this 2.15A of current into the base then 215 Amps will flow in its collector (in common emitter configuration).

Does that answer your questions?