25 minutes of hemmoroidal tests w/scope boring but interesting 

At 5:12 that is the current seen on scope that is affected by the 4uf AC cap in parallel to AC input going to primary of transformer. The current is measured with a .1 ohm resistor in series to one of the AC input lines to primary and the scope probe "tip and clip" of channel A goes right across this .1 resistor. If the tip and clip are "reversed" the phase shift between volts and amps traces on scope will show 180 degrees out of phase when you know it should show volts right on top of amps so check tip and clip placement over shunt resistor first before taking phase shift measurements. Good test is simply measuring phase shift of a common incandescent light bulb and it will show no phase shift at all as is resistive load. If it does show a 180 degree phase shift, the tip and clip are reversed, so switch them around to correct. Current can be calculated knowing the peak to peak in voltage shown on scope and also knowing the ohms of the "shunt" resistor in series (.1 ohm in this case) Channel A is the current trace seen on scope in this video while Channel B is the voltage trace.

NICE! KEEP DOING THE RESEARCH! GOOD LUCK! GOD BLESS YOU!

Great video Doug! This video shows how to tune the BiTT for one's that need help, there's alot of information here for a successful build. Ps, This video is not boring to me. Cheers👍👍👍👍👍

To calculate amperage (and watts) in AC with scope if anyone wants to know: See what the square division setting is on scope. In this case it is .02V per square division. (set by poteniometer on face of scope) This means just one of those squares seen on scope represent .02 volts. There are five "spaces" (AKA minor divisions) within each square division, so to see what each space represents in voltage, divide the square division voltage by five....so .02 / 5 = .004 volts. I counted 18 spaces ("minor divisions") vertical, counting from lower peak to upper peak in the voltage seen in the current-trace on scope (Channel A) (see around 4 minutes into video) . The current trace is the one that has the .1 resistor in series, and is "Channel A" in this case.....while Channel B is the voltage-only trace to be seen on scope. The phase shift is the space betweeen the current trace and the voltage trace. 18 spaces vertical X .004V = .072V (peak to peak voltage)"shown on the scope in the current-trace. So .072 /2 = .036V (You will divide the peak to peak voltage in half when you put it into the amps- formula) .036V x .707 = .025V (this is to convert voltage to "RMS" Root Mean Square) Now multily this.025V by the shunt-resistance in ohms that the scope probe "tip and clip" go across . So, using iohms law: .025V x .1 ohm resistance = .0025A in current Now this is unbleievable, (2.5 milliamps) so figure something must be wrong, And in this case my scope probes have a slide switch that sets what voltage is, to be X 10 (or not) and the probe had its setting at X10 voltage, so X 10 this .0025 Amps to be .025 Amps which is 25 milliamps, and this is right in between 44ma I got with meter in previous video, the 44m without a paralleled AC cap across primary when using grid to power the hemmoriodal transformer, and the 14ma shown on meter when the AC cap WAS in parallel to the primary feed to the transformer (in previous video powering the transformer with the grid) ....so looks pretty good for what the current is in reality.as it almost double-checks with a meter. Formula in nutshell for amperage from looking at current trace on scope: : Volts measured peak to peak on scope with the channel that has the current-probe , then divide this number by two X .707 to convert to a RMS value X the ohms resistance of the shunt resistor (.1 in this case) that is in series with that AC line, which is the one that has the tip and clip of scope probe across that .1 shunt resistor that is in series with the current feed to the primary of the transformer.. For phase shift: Count how many horizontal "spaces" minor divisions) there are in a full sinewave. Start from the left, right at the zero line, the wave goes up then the wave goes down then comes back to the zero line. So count from the start of the full sinewave to left, and the end of the sinewave to the riight, amd count where the sinewave crosses the zero line. IF it happens to be 8 spaces horizontal, that means each of the five spaces (minor divisions) with a square division represent 45 degrees. Because: One sinewave = 360 degrees and 360 / 8 = 45 degrees. Lets say you count just two spaces seperating the current-trace from the voltage trace. This is the phase shift, or also called the phase angle...if each space is 45 degrees, then the phase shift is 90 degrees. Lets say there are 10 spaces to count in a full sinewave, instead of 8 spaces.: 360 degrees / 10 = 36 degrees "per space" And lets say it is also still 2 spaces seen between the current and voltage traces. that is now a 72 degree phase shift. When you know the phase shift, look up what the COSINE of the 72 degrees is, and this cosine number is the POWER FACTOR (PF) Watts in AC it is Volts x Amps x Power Factor = Watts (not just volts x amps) You can just use google to find what the cosine is of whatever phase angle you find. 72 degree phase shift happens to be .309 (Note 90 degrees happens to be ZERO so go figure on that and why 90 degrees is the "target" you want to find with phase shift) So lets say there is a 72 degree phase shift you see to the AC input to a transformer, and you also see 25ma of current X 120VAC input .025 Amps x 120Vac = 3 "VA" (not watts but Volts-Amps) 3 VA x .3 PF = .9 WATTS input "unbelievable" thats for sure especially when there is about 5 watts of REAL power going into the 50 ohm resistive load from the secondaries at same time. Even more unbelievable if the phase shift shows 90 degrees, as now the watts is V X A X PF (PF being zero) so watts is therefore ZERO) In these tests I am measuring INPUT in AC.....the OUTPUT is REAL power with no phase shift at all to "worry about" since it is a "lump" resistive load of a 50 ohm resistor straight across the AC output of the secondaries of transformer. So output is V X A = Watts, while the input is V x A X PF = Watts (since the input is an inductive load of the primary coil and that creates a phase shift, being an inductive load)

Your Oscilloscope seems to be drifting a bit. That can be dialed in a bit, but opening up your scope can be dangerous. I have tuned several. The Mu symbol signifies Micro Capacitance is measured in Farads so it is pronounced Micro-Farads. Mu is a Greek letter, not a small u.