Calculating Current on a DC Low Voltage Circuit

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I have a 50VCD class 2 power supply with a 2A breaker on each output. The devices that I am powering with this supply operate between 44VDC - 54VDC.

My question is on a 200' run of 18AWG wire and 40W load, my voltage at the end is approx. 47.96VDC. When figuring my current, do I use my source voltage for the calculation or the voltage at the device that I am powering.
For example, at my max class 2 100W load, will an end voltage of 48VDC trip my 2A breaker since 100W/48V would exceed 2A or do I utilize the 50V source voltage in the calculation.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160423-1301 EDT

As a Systems Engineer I might assume you have an EE degree and if this is the case, then you should be able to answer your own question.

What is a 100 W or a 40 W load? You need to know how its current varies with applied voltage to the load. As examples: does the load look like a resistance (not constant power with voltage), or a constant power device (current increases as voltage goes down), or something else.

More later.

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160423-1352 EDT

Continuing.

Most ordinary circuit breakers are not very accurate devices and are not expected to carry their current rating as a continuous load. 80% of the breaker rating would be a normal maximum load.

I believe you used approximately 2.04 ohms as your 200 ft loop resistance.

Suppose your ideal voltage source is 50 V, the total source resistace as seen at the load is 2 ohms, and the load is a constant power load of 100 W, then 50 = 2*I + Vload. The current is about 2.2 A. This will trip an ideal 2 A breaker.

Now assume the load is resistive, then 100 W at 50 V is 2 A or a resistance of 25 ohms. Assume 2 ohms for the source resistance, then current is 50/27 = 1.85 A and an ideal breaker won't trip, but would be above 80% of 2.0 A or 1.6 A.

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I guess I should clarify, I am a systems engineer (Data Networking Systems and Low Voltage Tech)

The device I am powering is the equivalent to a cable modem. This device operates at 20W base but also provided POE power to externally connected devices that could increase the total Watts to 100W.

My power supply provides multiple 50VDC/100W outputs with a 2A push button breaker for each output.

I have had 2 different explanations on the max available load that I can put on one of these devices that is located at a distance at which the calculated voltage at the end is less than the 50VDC source (47.96VDC). Thus my question, is the current on the circuit 40/47.96 = .83A or 40/50= .8A
 

Smart $

Esteemed Member
Location
Ohio
I guess I should clarify, I am a systems engineer (Data Networking Systems and Low Voltage Tech)

The device I am powering is the equivalent to a cable modem. This device operates at 20W base but also provided POE power to externally connected devices that could increase the total Watts to 100W.

My power supply provides multiple 50VDC/100W outputs with a 2A push button breaker for each output.

I have had 2 different explanations on the max available load that I can put on one of these devices that is located at a distance at which the calculated voltage at the end is less than the 50VDC source (47.96VDC). Thus my question, is the current on the circuit 40/47.96 = .83A or 40/50= .8A
Under the NEC, you use nominal system voltage and nameplate current (some exceptions exist, such as motors). With a nominal voltage of 50VDC and a nameplate rating of 100W, that's 2A. If your load is comparable to or essentially equivalent to a cable modem, the usage will not be a consistent 100W. Most 2A breakers will not trip under nominal operating conditions no matter what the voltage is at the end of the run.

Given the preceding evaluation, calculating the current at 40W consumption is pointless in regards to tripping the breaker.

As to accurately calculating the current, you use the consumption of the device and the voltage at the end of the run... or the consumption of the device plus the I²R loss of the conductors and the voltage at the source.

PS: Welcome.... :thumbsup:
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160423-2344 EDT

Jeromyk:

You are asking for a simple answer to a question, and you have not yet figured our how to ask the correct question.

If I have a power source with a rather typical circuit breaker on its output, and I want reliable operation, then I don't want to load that breaker to its full rating. If I have an electronic current limiter that can be precisely set to some value, then I also don't want to load it to that limit, but I might go closer to the limit than on a thermal type breaker.

Now to what is the question to ask. You must first determine the characterustics of your load. Likely you have electronic loads with some sort of internal regulator to provide constant voltage to the electronics. This would make the load a constant power load when that load is at a steady state condition. This means the load current as seen at the input terminals of the load is inversely related to the voltage across those input terminals. Lower the input voltage and the current increases. This would be the case with a switching regulator in the load. The modern way of doing circuits.

If the regulator in the load is a series pass type, then the input current would be approximately constant.

You have to know your load characteristics. You can run experiments to determine how your load works.

The device I am powering is the equivalent to a cable modem. This device operates at 20W base but also provided POE power to externally connected devices that could increase the total Watts to 100W.
This statement is your critical ctiteria. If you have a 100 W load at 50 V and this is a constant power load, then at 50 V at the load input terminals the input current is 2 A. If that input voltage is 45 V, then its input current is 2.22 A. Basic electrical circuit analysis tells you where to measure voltages and currents. You need to find a book on this subject.

For a pure stable resistor in a DC circuit V = I*R, P = V*I or V^2/R or I^2*R. Where V is the voltage across the resistor (load), not somewhere else in the circuit, and I is the current thru the load.

Around any closed loop the sum of the voltages is 0. At any point in a circuit the sum of the currents is 0.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160425-0850 EDT

Jeromyk:

You also need to determine the internal impedance (resistance in this case) of your DC supply. This needs to be added to your wiring resistance.

I have an 8 VDC supply with a transformer capable of about 20 A. This supply has an apparent internal resistance of about 0.1 ohm when viewed from the output terminals.

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