Example C leakage Calculations

It shows, using math,  the amount of current the capacitor will pass when presented with the full DCC voltage across it.

This is important in that it shows how excessive capacitance, created by long track and the long wiring to it, can create problems with Occupancy Detection circuits buy giving false occupancy indications.

Below is an equation that calculates the current flow through the capacitor.  The math is only being used to prove that this current is real and gives you a idea of the amount of current flow.

Capacitor Current Law:

The equation governing "Ohms Law for a capacitor" current flow is:

To learn more about this equation,  capacitor Impedance and Current flow, go here: Capacitor Impedance

Example Capacitance Current Calculation:

Track Capacitance = 1300 pF = 1.3nF = 0.0000013 Farads or (1.3 x E-9) Farads
Track Voltage  = 14V
Booster Transition Time = 0.00001 seconds or 10uS

dv = |V(initial) - V(final)|   where:  V(initial) = +14V   &   V(final)  =  -14V
dv = |(14V) - (-14V)| = |14V + 14V| = |28V| 
dv = 28Volts.

dt = |T(initial) - T(final)|   where:   T(initial) = 0 sec  &   T(final)  = 0.00001 seconds
dt = |(0) - (0.0001)| = |0 - 0.0001 | = |-0.0001 | 
dt = 0.0001 seconds

Plugging in "Ohms law for a capacitor":

I  = C * dv/dt = (1.3 x E-9)F * 28V/0.0001S = 0.00364 Amps

I(cap) = 3.64mA

So what is 3.65mA in terms of an equivalent resistor value when placed across the track?

Ohms law: V = I*R --> Solve for R --> R = V/I
Track voltage: 14V

R = 14V/0.00364 Amps = 3,846 Ohms.

R(cap) = 3.8K Ohms

Conclusion: Although 1300pF of track capacitance may not be a load of any consequence to a DCC booster,  it is however more than big enough load for a Block Detector to think there is occupied track.