Brass loco current draw : LUSENET : Wiring for DCC : One Thread

I have a numer of brass loco's all but one are can motors. I want to be sure I am correctly measuring the DC current. I put the loco on a track, connected an amp meter to the engine and slowly turned on the power. I held the tender as the driving wheels slipped and increased power until the amp meter no longer increased. I got from 250ma up to 900ma(for the open frame motor) on the various motors. Is this the correct method?

-- Chris Dante (, September 28, 1999


Yes, and No. You've used a correct method but have only found the maximum current demand when connected to your existing power source, which may be limiting the current that can be supplied. What you really want to know is what the current could be when connected to a new DCC power source. In other words, you really need to know the motor Resistance. Measure it this way... Provide an ammeter in series with the test track. Provide a voltmeter connected to the test track. Turn lights and aux equip (sound/smoke generators) OFF. Run the loco, at 6-8 volts or so. (Smooth DC but not critical) Hold the loco so that the drive wheels slip.(Do not press down) Record the amperes. [Irun] This is the maximum average current demand while running. A DCC decoder must be size slected to provide these amperes continuously. Press down on the loco to temporarily lock the drive wheels. Record both the volts and amps readings.(Quickly to avoid damage) Calculate motor resistance as Resistance(ohms)=Volts/Amps. (Do several times and average the data) Calculate Peak current on a DCC supply as Ipeak=12Volts/R(ohms). (12V is for HO/N scale may be 14 or 16 for S/O/G scales)

A DCC decoder needs to be able to supply these peak amperes without damage. Most DCC decoder manufacturers will provide a peak current rating. Certain 'high performance' motors will have a very low resistance, and subsequent high peak current demand. [Early P2K PA's for example] If the calculated peak current is too high, consider using a decoder with build-in short circuit protection (usually more costly) or adding a discrete resistor of several ohms in series with the motor leads to increase its effective resistance. [Such a resistor will get warm. It needs to have a watt rating of at least Irun*Irun/R(resistor ohms).

-- Don Vollrath (, September 29, 1999.

Opp's, Sorry. There is a typo mistake in my last message. The watt rating of a resistor in series with the motor needs to be Irun*Irun*R (ohms)! [All multiplication, not division]

-- Don Vollrath (, September 29, 1999.

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