I used both sine and square wave, showing both the sine wave resonance of 3.19MHz.
You were lucky. A square wave has many frequency peaks (lobes) that can confuse the measurement. The 5uH was calculated with an online calculator using the now known 500pF and 3.19Mhz.
That's even better if the 500pf is trustworthy. Right, calculated in an earlier post from the measured peak voltage.
Peak voltage at what frequency? Good info, i have some copper tape which i could use to do some experiments with.
Remember that you have 1kV pulse and that is very different from the 1V pulse on that video. You don't need a 1Ω transmission line at that voltage So basically, its not that simple to get this nanopulse transferred to the cell without losses or signal corruption and probably would take lots of experiments.
If your pulse is 10ns wide then it means that is similar to a 50MHz square wave. (20ns period). For 50MHz square wave, a 5μH inductance has an inductive reactance of 1273Ω. X _{L}=5.09*f*L for square voltage source ( here is why) X _{L}=6.28*f*L for sine voltage source The coil is a low pass filter. It has low reactance at low frequencies and high reactance at high frequencies. The capacitance of the cell behaves oppositely. It is a high pass filter. It has high reactance at low frequencies and low reactance at high frequencies. In this case, a 500pf capacitor has 8Ω capacitive reactance to a 50MHz square wave. Of course, your waveform is not square but a low DUC rectangular waveform, consisting of a dense comb of frequencies from 1kHz to 50MHz (see attachment) Also, the transformer action decreases the primary winding's inductance of any loaded transformer (Lenz Law). P.S. The frequency at which the inductive reactance is equal to the capacitive reactance is your LC resonance frequency (in your case 3.19MHz).
« Last Edit: 20130509, 13:40:33 by verpies »
