Summary: Voltage Drop is the loss of voltage caused by the flow of current through a resistance. This loss, measured across the resistor, is directly proportional to the current.
For best performance from your Digital Command Control system a voltage drop of no more than 5% at 5 Amps is preferred. The maximum acceptable voltage drop is 10% at 5A.
Most charts for bus wiring use these rules to determine the maximum length of the power bus for a given wire gauge.
Wire resistance tables available online quote the resistance measured using Direct Current. Tables for the maximum current capacity of a wire often show values for Alternating Current at 60Hz. Since DCC is neither AC or DC, these charts can be misleading. The actual impedance a wire presents to a DCC signal is greater, due to the nature of the signal.
Voltage Drop occurs when current flows through a conductor. It obeys Ohm's Law, where the amount of voltage lost is equal to the resistance multiplied by the current.
A wire has resistance, which in short runs is negligible. Over long distances it becomes an significant. Resistance is typically expressed in 'Ohms per thousand feet' or another similar measure. The resistance is a function of both the diameter and length of the wire. The resistance of a light gauge wire such as 22 AWG is 23 times greater than the 14 AWG typically used in house wiring.
For example, in house wiring, 14 ga is used for 15A circuits, but for 20A circuits 12 ga is specified. This minimizes voltage drop and other losses. These losses appear as heat. Too much current, too much heat, potential for a fire. (Note: these are for copper. Aluminum wire is even heavier for the same current capacity, sometimes called ampacity.)
Electrical codes specify the maximum voltage drop allowable, usually no more than 3%. If the voltage drop exceeds the specifications, changes are required. Load calculations are made prior to wiring to ensure everything meets Code. Even though the circuit is rated for 15A, codes usually only allow a safety rating of 80% of the rated current, limiting the circuit to a maximum of 12A on a 15A circuit. If the load exceeds the maximum allowable current rating, an extra circuit must be installed.
For Digital Command Control voltage drop is a concern. DCC wiring must be more robust than the older Direct Current Block wiring, as one booster may supply the entire layout. Thus, the bus wires must be heavy gauge, to minimize voltage drop in the bus. Unlike old analog configurations, the booster supplies a lot more current to multiple points on the layout, instead of a small amount of current supplied to one block by a small power supply located nearby.
Voltage drop reduces the amount of voltage available to the decoder. To maintain the power output of the motor, less voltage requires more current draw to do the same work, which can result in a fried decoder. A drop of one volt between the booster and locomotive can cause problems. As stated above, a voltage drop at 5A of 5% is acceptable to a maximum of 10%.
A good example is a car battery. It has an internal resistance. A voltmeter it may indicate 13.2V, its typical no-load voltage. When a mechanic tests the battery he connects a device which has a high resistance (the voltmeter) in parallel with a low Ohm / high wattage load resistor. There is a switch which must be depressed to connect the load resistor across the battery. The purpose of that load is to simulate the starter motor's current draw. By pressing the switch, a low resistance is paralleled with the high impedance of the voltmeter, which indicates good or not. Excessive internal resistance reduces current passing through the load resistor, creating a voltage (Ohm's Law: Voltage equals Current times Resistance) below an acceptable level, meaning the car may not start. This is the typical reason why cells die: Their internal resistance increases to a point where their capacity under load is insufficient to power the device.
LEDs have very strict limits on the voltage and current they can withstand. The total current is important, as too much will destroy the device. To limit the current, a resistor is wired in series with the LED. The series resistor limits the current flow to an acceptable amount, and a voltage proportional to the current appears across the resistor. This protects the LED from both excessive voltage and current.
See the page on LEDs for more details.
Since DCC uses a digital signal in the 8 to 10kHz range, wire resistance becomes more complicated. The addition of inductance and capacitance into the mix results in an Impedance, represented by Z, which is also expressed in Ohms (Ω).
DCC uses heavy gauge bus wires to distribute power to the track. At intervals, a wire is connected between the bus and the track, typically every three to six feet in HO. Since the de facto standard today is Nickel Silver rail, which has a much higher DC resistance, the resulting circuit parallels the low impedance bus wires with the higher impedance rails. Ohm's law states that the sum of resistors in parallel is less than that of the smallest resistor. The bus wire effectively negates any impedance in the rail. Current follows the lower impedance bus wire under the track instead of relying on the track to carry all the current.
In Digital Command Control, two things can happen with poor wiring or no bus wiring. The decoder overheats from excessive current draw, damaging or destroying it. Or the circuit breaker may not work as intended, failing to trip on an over current event means damaged locomotives or even wheels being welded to the track.
During the wiring phase, the track should be tested regularly. One test is the Quarter Test, named because a coin is placed across the rails. The booster should cut out immediately. If not, there is a problem with the wiring. Correct it before any damage happens.
Voltage Drop caused by Rail Impedance
One of the reasons for using a track bus with multiple feed points is to reduce the impedance presented by the rail. Compared to an equivalent copper wire, Nickel Silver has 19 times the resistance. Nickel Silver is a copper alloy (60%), with the remainder being equal amounts of nickel and zinc. Silver, which is an excellent conductor, is not present.
In DCC applications, the rail will have two properties: Resistance and Inductance. Some of the inductance is cancelled by the rails being parallel to each other. As a rule, as the diameter of a wire is reduced the inductance increases. The profile of the rail and its physical size (cross sectional area) will impact both resistance and inductance.
Basic Rule of Electricity: Current will always seek out the path of the least resistance. That path can be you!
By paralleling a lower resistance copper wire with a higher resistance wire (such as rail) Ohm's Law states that the resistance is
Rtotal = 1 ÷ (1/R1 + 1/R2)
Effectively making the resistance of the pair almost equal to the value of the lower resistance copper wire.
By not relying on the rail to carry all the current, voltage drop is minimized by allowing current to move to the site where it is needed using a lower resistance copper bus wire.
- For more information on resistance of nickel silver rails, see the Rail Size page.
Equivalent Resistance Equal Lengths of Rail and Wire:
- 5m length of Nickel Silver C100 rail
- 5m 14AWG copper wire.
- Rail = 0.4Ω
- Copper Wire = 0.05Ω
This may be written as 0.4Ω || 0.05Ω
This formula for calculating the equivalent resistance of two resistor in parallel can be expressed as
RTotal = (R1 –1 + R2 –1) –1
- Rail = 1 ÷ 0.4 = 2.5
- Copper = 1 ÷ 0.05 = 20
Sum of 20 + 2.5 equals 22.5, the result of 1/22.5 = 0.044 Ω
- X –1 is the same as 1/X. Calculators can use either symbol for this operation.
Another formula for calculating parallel resistances:
R Total = (R1 × R2) ÷ (R1 + R2)
0.4 × 0.05 ÷ (0.4 + 0.05) = 0.02 ÷ 0.45
= 0.044Ω. (Always remember the Order of Operations!)
Voltage Drop (DC Resistance)
- Rail = 0.4Ω X 2.5A = 1V
- Copper Wire 0.05Ω X 2.5A = 125mV
Rail plus Copper wire in parallel: 0.044Ω × 2.5A = 0.111V (111mV)
Exchanging the C100 rail for C70, the resistance increases to ~ 1Ω, so the voltage drop on the rail alone is 2.5V
Effect Without a DCC Power Bus
Relying exclusively on the rails to carry the power will introduce a significant amount of voltage drop.
Using the table below, the impedance of C100 rail is 0.076Ω per metre. If the track runs 10 metres, the impedance is 0.76Ω X 2, or 1.52Ω. A one amp current will result in a 1.5V drop across the circuit. Placing a bus with an impedance of 0.55Ω in parallel with the track results in a total impedance of 0.4Ω
For easier math, without the power bus, at 1A Vdrop is 1.5V, with the power bus, VDrop is 0.4V. Without the Power Bus arrangement, the voltage drop is 3.75 times more.
Apply that to Code 80 rail: Z80 is 10 X 2 X 0.108Ω, or 20 X 0.108 = 2.16Ω.
At one amp, VDrop80 = 1A X 2.16Ω = 2.16V.
Multiply that by 5A and VDrop80 is now 10.8V.
Voltage Drop - AC Measurements
Since the rail impedance is an AC measurement, it would be better to use AC equivalents for the power bus too. An Ohmmeter only measures the DC resistance, it doesn't see any inductance or capacitance present. Due to the nature of the DCC Waveform, the booster sees the track and power bus as an impedance, which is much larger than the DC resistance of the circuit. See Track Bus Impedance for more details on the power bus measurements.
The power bus is 12AWG and has an impedance of 0.55Ω/10m loop, or 20m total. (0.55-1 = 1.8.)
|DCCWiki.com Voltage Drop Table|
|Code||Rail Impedance||Bus Impedance||Total Z Bus // Rail||Both Legs||Voltage Drop at 1 Amp|
- Calculated as follows:
- Length of power district is 10 metres
- Bus is estimated at 0.4 Ω per line
- Rail and Bus are connected in parallel, with the bus value stated above
- The result is doubled to represent the sum of both legs of the circuit
- The Voltage Drop is simply the total impedance multiplied by 1A. Multiply by 5 to see a value at 5A.
- For 1A to flow in the circuit, the sum of the loads across the booster outputs would be equal to Volts ÷ Amps
- These are worst case scenarios. Most of the time your current draw will not be 5A in a given power district.
- Recommended maximum VDrop for DCC Wiring should be 5%, not exceeding 10%.
- Planning for the worst case is the best course of action. At some point in the future it is possible to draw enough current to create a 10% voltage drop.
Assuming HO scale with 15V on the rails, the maximum permissible voltage drop is 10%, or 1.5V. At 5% this value is 0.75V
Using the results found above, the C100 results at 5A exceed VDropmax of 1.5V. The same applies to C83 and C70 rail. At a lower current, such as 2A, the voltage drop is within the permissible maximum of 1.5V at V. It is also close to the value for VDrop5% at 0.8V.
The Vdrop for Code 83 at 5A exceeds VDropmax, but it is acceptable at 2A with a Vdrop of 1.0V.
Notes on Voltage Drop
If you were to place a load at the end of the track (the 10m point) which causes a draw of 5A total, measuring across that load:
VLoad = Vsource − (VDropA + VDropB)
- Vsource =15V
- VDropA = 1V
- VDropB = 1V
VLoad = 15V − (1V + 1V) = 15 − 2 = 13V
Using this example, while you may measure 15V at the booster output, there will be 13V at the load. As the power (volts X amps) must stay constant, as the voltage drops the current increases.
(The total voltage must equal the sum of all the voltage drops.)
As seen through the various calculated examples, the lighter the rail, the heavier the power bus must be to counteract the increased impedance of the rail. Doing so reduces the Voltage Drop to a value within acceptable limits. This becomes more important in smaller scales with lower track voltages.
|Recommended Track Voltages by Scale|
|Scale||Recommended Track Voltage||VDrop5%||VDrop10%|
See page Rail Size for more details on the impedance of Nickel Silver Rail.
|DCCWiki Rail Impedance Table|
|Code of Rail||Impedance per metre, mΩ|
Rail impedance at 1 amp, 60Hz, measurement accuracy is less than 50ppm. Different alloys and rail profiles will have different values.