7/16/14
UNDER CONSTRUCTION.
NOT READY FOR PRESENTATION.
This section is a pure technical section for those who want to know more about how the RC filter parts were chosen.
The DCC bus has DUAL functionality. It has BOTH Digital Data and High Power present at the same time on the same two wires.
It is not just a data bus. So using a pure terminator design and installation solution will not address its needs..
It is not just a power bus. So using a pure snubber design and installation solution will not address its needs.
Hence the design of the RC circuit and it rules of installation must go back to basics. The solution needs to address both the data and the power requirements by striking a balance between the two.
1) What We Know, Data Collection.
A) DCC is an AC digital signalling system that uses FSK (Frequency Shift Keying) to communicate binary data.
B) The highest frequency used by DCC is the frequency use to send a "1" bit.
C) The DCC standard defines the "1" bit frequency by it pulse time duration. The frequency can be calculated from the pulse time: T = 116uS. F = 1/T F = 8.6KHz
D) The DCC waveform is a squarewave when there is no DC locomotives in operation on DCC power track.
E) Squarewaves are rich in "high frequency harmonics" that are higher than the DCC frequency. For more information, see: Squarewave Harmonics. These frequencies are needed to be maintained in the wiring to preserve the fidelity of the DCC squarewave waveform.
F) Sending digitral data that has
2) Things we can rule out or take off the table in terms of design.
2A) The DCC layout wiring is NOT a transmission Line.
Why? Although there is digital data being sent down the track, the data waveform contains high frequencies AND the track rails carry the data are parallel "giving the basis" of a transmission line concept, the following is also true:
A1) The wiring carrying the current from the booster to the track is random in its design. Simple Reality Fact. No to layouts are wired the same. No two people who wire the layouts have the same electrical wiring skills nor knowledge. Some are simply wired that way because it worked.
A2) There is no electrical goal required in the wiring for DCC operation. Nothing is discussed in the DCC standards regarding terminators, wiring impedance, capacitance nor inductance. DCC is advertised to work on any layout regardless of how it is wired.
A3) The current path going down the wiring is not constant. The locomotives moves on the track constantly changing the effective capacitance and inductance involved in the DCC current path. Simply put, electrically this is a moving target.
CONCLUSIONS:
There are no transmission line requirements because there are no constant nor defined transmission line parameters to base one on. The lack of a being able to determine a proper transmission line termination value means:
I) The DCC waveform will have waveform distortion.
II) The DCC waveform will have "ringing" oscillations on riding on top of the DCC waveform.
2B) Snubber Design cannot be used.
One of the goal was to design a RC filter using common parts. Then based on the common parts, make a design the meets the requirements for filtering out the voltage spike and noise that is not part of the base DCC signal found on the tracks.
How were these parts. 100 Ohm and 0.1uF chosen?
The design goals were chosen based on the information about:
1) Highest DCC base operating frequencies. We do not want to filter out these signals.
2) Booster Output Impedance. We want the filter to be strong relative to the strength of the booster.
3) Understanding the nature of the noise and voltage spikes we are attempting to eliminate or get rid off.
The part goals of the design are:
1) Use parts that are easy to get. Any electronics parts store should have these parts in stock and be inexpensive to buy.
2) Keep the heat generated low. Do not want it to cause any potential damage as it get to hot that it might burn something.
The RC circuit is best understood from a filter point of view since we are dealing with electrical energy both random (Shorts), which is our noise that we do not want, and repetitive (DCC waveform) at the same time.
The nature of the noise that we do NOT want is that it tends to consist of energy (Voltage & Current) that is much higher in frequency content than the base DCC frequencies. We can use that information in the design of the RC filter. So what we want is a filter that can pass the low base DCC frequency but filter out everything else that is at a higher frequency. This is called a low pass filter because we want the filter to ignore the lower frequencies and filter out the high frequencies.
DCC’s highest base frequency is t = 116uS. See S-To translate that into frequency
F = 1 / t
F = 1/ 116uS = 8.62KHz.
The output impedance of a Booster is a function of it.
We also want to have a strong low pass filter that has a low enough impedance that it can support enough current to be useful as a DCC filter.
The corner frequency where the filter starts to be useful is
F = 1 / (2 * Pi * R * C) =
F = 1 / (2 * Pi * 100 * 0.1uF) = 15.9KHz or 16KHz to keep is simple.
So the filter start to become active at a frequency that is about 2x the highest DCC frequency. Hence we meet our goal of passing DCC base frequencies and start filtering out anything above that.
What this means is that any noise who’s frequencies contents is above 16KHz will see the RC filter as a load. The current passes through the capacitor and resistor at the same time. However since capacitors do not dissipate any heat of any consequence compared to what the resistor does,
If we lower the resistance value or increase the capacitor value, we start getting closer to the base DCC frequency and start attacking the DCC base frequency which means the amount
X) POWER CALCULATION
For 14V of track power, the RC filter will dissipate about:
1) 0.33W of power worse case (All 1 bits sent continuously)
2) 0.20W of power easiest case (All 0 Bits sent continuously)
The equation that calculates the Power Dissapation in the Resistor of a RC filter is as follows
Pd(resistor) = 1/2 * C * Vp-p^2 * F
Where:
Vp-p = 2 * Vtrack = 2 * 14V = 28V
C = 0.1uF = 1E-7F
F = DCC frequency. Digital Data pattern dependent. 1 Bit = 116uS or 8.62KHz & 0 Bit = 200uS or 5.00KHz
The load is only seen by the booster during the DCC voltage's (AC) transition time between one voltage polarity (phase) and the other polarity (phase). After the transition is complete, the current literally disappears with the booster seeing nothing until it is time for the polarity to change again.