Gotoluc’s Real Self Runner... Gentlemen: “Let us Compute!â€
The data used for the following calculations is taken from:
Self Running Coil Test #7 video – (Search for gotoluc on youtube.com)
Since we can’t experiment physically, we will do some theoretical
calculations.
The first calculation finds the equivalent resistance that would discharge
the gain from the bulk capacitors at (about) the same rate the of voltage
gain from the circuit is charging the bulk capacitors. Per second. Taken
from the video.
The voltage on the bulk capacitors: = 16.64volts
Total size of the bulk capacitors = 2 x 3900uf = 7800uf
Approximated change rate in voltage on the bulk capacitors when circuit
is tuned taken from video #7 = 1/100 or .01volts per each (1) second.
Calcs #1: 1/1664 in one second = time period giving = 0.000601Hz.
filter “corner frequency"
Calcs #2: RC (filter spreadsheet) calculator web link;
http://www.muzique.com/schem/filter.htm C= 7800uf cap., plus .000601Hz freq. gives an RC time constant
R = 34Kohms
This means that when the circuit is charging at a maximum rate that would
be balanced by 34Kohms of additional resistance connected across the
capacitors. This means that our transistor could only have a gate drive
coming through 34Kohm total. I can tell by estimating this amount is
*problems* relative to the gate drive power of the mosfet transistor
IRF640 we are using. Even worse for the Buz11 mosFet.
Now lets find current through 34K ohms at 16.64Volts
E=IR : ma = 16.64V/34. = 0.5ma total available current.
Math Calculator web link;
http://www.math.com/students/calculators/source/scientific.htm 0.5 ma gain current is (x10) too small to supply the SG3525 mosfet osc.
drive circuit estimate of 5.0ma.
---
The second calculation finds the equivalent resistance of two resistors
connected in series.
Rs = R1 + R2. where R1 is the (variable) resistor that converts the total
bulk voltage down to the gate drive voltage where R2 is the resistor
equivalent of the parasitic capacitance of the mosfet being driven at the
frequency of the input signal.
Capacitance gate to output taken from IRF640 spec sheet
Cisd (capacitance of input , to source-drain) = 1560pf = 1.56nf
Frequency of input signal F = 33.3KHz =~ 32KHz.
Bulk Voltage drive to optoisolator VB = 16.64Vdc pp.
Gate drive Voltage to mosfet VG = 10.94Vac pp.
Calc #1: Capacitive impedance calculator web link;
Btw the equation is: [Zohms = 1/(2pi * F * C)]
http://www.cvs1.uklinux.net/cgi-bin/calculators/cap_imp.cgi C= 1.56nf , at F= 32.0KHz gives an equivalent resistance of about
IRF640 gate @ 32Khz Z equiv: R = 3.2Kohms
Find R1 and R1+R2 where R2/(R2 + R1) :=: VG/VB
Calc #2: R2+R1 = 3.2K/( 10.94V/16.64V) = 4.8Kohms (total drive circuit
impedance)
Calc #3: R1 = 4.8K – 3.2K = 1.6K so the variable resistor should be 5Kohms
total, set at about 1/3 up, for 1.6Kohms
Math Calculator web link;
http://www.math.com/students/calculators/source/scientific.htm ---
This suggests that a better transistor type (a non-power) mosfet
might give better results. If we can get a transistor with Cisd down
below 156pf and the Rsd = .15ohms is not critical, then we will be
balancing transistor drive with circuit overunity gain. I think this is
doable. As long as the optoisolator is isolating the AC led drive
through a very small cap 2pf who necessarily cares where the
drive signal originates...For now. Thanks.
:S:MarkSCoffman