# Is it possible to increase the amperage of a DC circuit while keeping voltage the same?

#### Skeeterweazel

Joined Oct 28, 2021
1
Hello. I have a potential project that i have a question about: Is it possible to increase the amperage of a DC circuit while keeping voltage the same?

I'm looking at converting a model train locomotive to DCC (computer controlled). The problem is the available decoders that go in the locomotives aren't made to supply enough amperage for the large loco I'm planning to convert.

I'll give a quick explanation of how DCC works as i understand it; i'm new to this and learning as i go. The base station supplies power and instructions to the track. It is a square wave of roughly +15 to - 15 volts DC. Picture below shows form. The signal is used by all the locos on the track to tell them how fast to go, direction, turning on/off lights and sound, as well as supply power for the motors.

So my question is can i use an existing decoder, but boost the amperage that it supplies to run the motor? An average decoder may put on +-18 volts DC @ 1 amp. I may need up to 15 amps at same voltage. How to add additional power from the track and get it to the motor?

I realize this may not be possible, but i wanted to ask to see if possibly it is.
.
Thanks
Marty

#### dl324

Joined Mar 30, 2015
15,460
Welcome to AAC!
I realize this may not be possible, but i wanted to ask to see if possibly it is.
If I understand your question correctly, no. You can't get more power out than you put in.

#### Papabravo

Joined Feb 24, 2006
19,611
In general the answer is no. Whatever the electrical power is used for has a certain load impedance. In most cases this load impedance is fixed or varies by small amounts around an operating point. Ohms Law which is fundamental to all things electrical says that among the three quantities of voltage, current, and impedance, if two of them are fixed, then the third has a predetermined value which must be maintained unless something else changes.

So if you are able to raise the voltage by a small amount, the current will increase by a small amount. If you continue this process so that small amounts become large enough, the load will dissipate enough power to destroy itself. Power dissipation in a fixed impedance goes up proportional to the square of the current.

As I understand what you are describing you are distributing both power and data on the same pair of conductors. With a bipolar signal you have to consider the average value for voltage and current. Each '1' bit has an average value of 0V and each '0' bit has an average value of 0V, but the stretched 0 ha a high part which is longer than the low part. I imagine what they are doing is rectifying the data/power waveform and charging a capacitor which is actually used to run the motor. I further imagine that the capacitor has enough energy to run the motor under all conditions of dat transmission.

I don't have any idea if you can make the leap from 1 Ampere to 15 Amperes. I depends on the losses from the length of track and how hot you want things to get.

Back of the envelope calculation. You have a track layout with a DC resistance of 10 Ω, and you want to push 1 Ampere through the tracks. The power required for the tracks is 10 Watts = 1 x 1 x 10. If we now try to run 15 Amperes through the same track we have 2.25 kilowatts - 15 x 15 x 10. do you see the problem? Also the voltage drop along the track would be 150 volts = 15 amperes x 10 Ω and your ± 15 Volt or ± 18 Volt supplies will fall to their knees and not bother to stop there. Actually they wouldn't actually be able to push 15 amperes through 10 Ω.

#### crutschow

Joined Mar 14, 2008
31,128
Do you have a track supply that can provide 15A to your train?

If so you may be able to add a transistor boost circuit to the decoder output to increase the current to the motor.

#### AlbertHall

Joined Jun 4, 2014
12,190
Will the track be able to cope with 15A?

#### BobTPH

Joined Jun 5, 2013
6,080
Are you saying that the power supplied to the track is capable of > 15A but the decoders are only able to supply 1A each? If so it might be easy to do what you want.

If the output of the decoder is a PWM, you could simply use that signal to drive a beefier MOSFET.

Bob

#### djsfantasi

Joined Apr 11, 2010
8,676
A DCC receiver extracts commands from the data stream to control the motors spreed and directions as well as accessories (sound and lights for example)

The controller uses PWM to change the motors speed. Power is sourced by the DC component. It also uses the DC power in the tracks to power itself (and the sound generator, if that feature is included.)

Increasing the voltage would likely destroy the controller. So, no, there is no way to increase the amperage.

#### Alec_t

Joined Sep 17, 2013
13,232
I agree with Bob (post #6). Providing your supply, the DCC controller booster and the track are all capable of handling 15A then the loco's decoder output could drive a power-MOSFET bridge feeding the loco motor.
I'm a bit concerned that 8-10kHz rectangular pulses of 10A through considerable lengths of track would generate a lot of RF noise.

#### djsfantasi

Joined Apr 11, 2010
8,676
I agree with Bob (post #6). Providing your supply, the DCC controller booster and the track are all capable of handling 15A then the loco's decoder output could drive a power-MOSFET bridge feeding the loco motor.
I'm a bit concerned that 8-10kHz rectangular pulses of 10A through considerable lengths of track would generate a lot of RF noise.
I agree, too. That’s a good description of what is needed in each DCC component…

#### ElectricSpidey

Joined Dec 2, 2017
2,182
I would also be concerned about the RF from the motor PWM.

And I would also be concerned about filtering out those negative parts of the control from the motor, this is probably pretty easy with a motor that has a current draw of under 1 amp, but maybe not so easy for a motor that draws 15 amps.

#### MisterBill2

Joined Jan 23, 2018
13,804
I see a reference to "the data stream" but where is this data stem connected? That matters a bit.
As long as the various connections can handle the current, and the load will accept the current, that should work. 15 or 18 volts and ten amps will be a lot of power, and so there will be a need to reduce the resistance at every point. I am not aware of small train tracks that could handle ten amps without overheating, and so you may need to do a lot of work. And what I have seen of the pickup brushes on model train engines, they also will not survive ten amps.
OR is this something else, similar to a model train but quite different in scale? Really, I can not imagine any of the model trains that I have seen needing anywhere near that much current. Of course, this may be one of those larger "sit-on" trains that I have seen a very few times.

#### GetDeviceInfo

Joined Jun 7, 2009
2,125
I had played with this years ago, and I asked anyone who would listen, why continue with an aging protocol. Why not bluetooth or even wireless ethernet. No good responses, other than those that say yea, why not, that way my fancy web based track management system could directly tie in without a converting system.

I agree though, your converter should simply drive a power element, assuming your supply can deliver. Brushing that current into the loco could require advanced techniques.

#### milo0291

Joined Oct 29, 2021
2
Hello, unfortunately, you can not increase it myself, I tried in various ways, but it did not work, unless I can not, but I think it can not be done at all

#### AlbertHall

Joined Jun 4, 2014
12,190
I had played with this years ago, and I asked anyone who would listen, why continue with an aging protocol. Why not bluetooth or even wireless ethernet. No good responses, other than those that say yea, why not, that way my fancy web based track management system could directly tie in without a converting system.

I agree though, your converter should simply drive a power element, assuming your supply can deliver. Brushing that current into the loco could require advanced techniques.
Like this (but it won't do 15A): http://bluerailtrains.com/

#### GetDeviceInfo

Joined Jun 7, 2009
2,125
Like this (but it won't do 15A): http://bluerailtrains.com/
I would omit DCC altogether and go straight to each device, Mesh or whatever. Enumerate the device, request its functionality, map it into the layout/controls. Small stuff like landscape lighting might be better handled with a localized controller.

#### MisterBill2

Joined Jan 23, 2018
13,804
Like this (but it won't do 15A): http://bluerailtrains.com/
There is a VERY GOOD reason to not go with the latest fad in control communications, and that is that it changes so fast that new products would need to be created every tear or so, and the various devices that were used one year would be obsolete the next year, and totally unavailable 3 years later. But control systems using some older technologies and PARTS THAT ARE STILL AVAILABLE keep on working. And that is another thing. Model RR is not a ten--week hobby, where folks get bored and dump it in short order. So some layouts are 2 or more years old, and some parts of some layouts much older than that. So to start adding in that rapidly changing stuff makes not much sense. And besides all of that, the older stuff is much easier to understand, and the parts big enough to not need a 10X magnifier just to see them.

#### ElectricSpidey

Joined Dec 2, 2017
2,182
Back in the day I did a lot of slot car racing with modified motors that needed a lot of juice (not 15 amps) on very large layouts, we would solve the voltage drop problem by using power distribution that consisted of heavy cables that were tapped into the track every X feet or so.

I think if the same solution was used here there would be "some" RF mitigation (using shielded cables) and any voltage drop issues would be solved.

Would still need to work out some other issues...that's for sure.

#### djsfantasi

Joined Apr 11, 2010
8,676
There is a VERY GOOD reason to not go with the latest fad in control communications, and that is that it changes so fast that new products would need to be created every tear or so, and the various devices that were used one year would be obsolete the next year, and totally unavailable 3 years later. But control systems using some older technologies and PARTS THAT ARE STILL AVAILABLE keep on working. And that is another thing. Model RR is not a ten--week hobby, where folks get bored and dump it in short order. So some layouts are 2 or more years old, and some parts of some layouts much older than that. So to start adding in that rapidly changing stuff makes not much sense. And besides all of that, the older stuff is much easier to understand, and the parts big enough to not need a 10X magnifier just to see them.
DCC was introduced in 1989. It’s been a standard for 30 years. Model RR controls do not change rapidly. The last change to model RR controls took place in the 1940s.

#### MisterBill2

Joined Jan 23, 2018
13,804
DCC was introduced in 1989. It’s been a standard for 30 years. Model RR controls do not change rapidly. The last change to model RR controls took place in the 1940s.
I was referencing the suggestions in posts #14 and #15, that suggested a new scheme that will be obsolete in a year or so. And not have parts available shortly after that.

#### GetDeviceInfo

Joined Jun 7, 2009
2,125
I was referencing the suggestions in posts #14 and #15, that suggested a new scheme that will be obsolete in a year or so. And not have parts available shortly after that.
if you check, you'll find a number of protocols that will likely disappear in the hobby. Wifi will be a major player and is well on its way. Be there or be square, unless you like nostalgia, as that can be fun too.