Electrical – kW to kWh Conversion

I think I spotted it:

DecToBase7(0.25) = (0*7^0)+(2*7^-1)+(5*7^-2) = (0) + (0.2857142857) + (0.1020408163) = 0.387755102 or by specs 0.3877.

Yes, that is how I convert a 0.25_(base 7) to a base-ten decimal value.

To convert 8.8750_(decimal) to a base-7 value, you need to go the other way:

8.8750_(decimal)                            1
                                            1
                                            .
0.8750_(decimal) * 7 = 6.1250_(decimal) --> 6
0.1250_(decimal) * 7 = 0.8750_(decimal) --> 0
0.8750_(decimal) * 7 = 6.1250_(decimal) --> 6
0.1250_(decimal) * 7 = 0.8750_(decimal) --> 0
0.8750_(decimal) * 7 = 6.1250_(decimal) --> 6
0.1250_(decimal) * 7 = 0.8750_(decimal) --> 0
0.8750_(decimal) * 7 = 6.1250_(decimal) --> 6
...                               ...

So 8.8750_(decimal) is equal to 11.60606060606...(base 7).

With a few special exceptions, converting a fraction in any one base to a fraction in any other base usually leads to such repeating patterns. (Perhaps the most notorious such unexpected repeating pattern happens when converting 0.1_(decimal) to binary. A similar repeating pattern occurs when converting 0.1_(base 3) to decimal).

An inductor with inductance \$L\$, operating at frequency \$f\$, has an impedance \$Z = 2 \pi j f L\$. For \$\mu\$H inductors at 50 Hz, the impedance would be in the m\$\Omega\$ range. But your power supply can't source enough current to maintain 12 V over such an inductor. Try measuring the voltage on the power supply side when the circuit is running - I bet it is close to 0 V.

To get the voltage step-down/up you are expecting, you therefore need the inductance of the primary to be much larger than \$ V_\mathrm{PSU}/(2 \pi f I_\mathrm{PSU})\$, in your case approximately 100 mH.