OCPD tripping on 24V DC circuit

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philly

Senior Member
I am troubleshooting a 24V DC purely resistive circuit consisting of light bulbs and resistors. This circuit is protected by a 20A circuit breaker. The circuit has been operating fine for some time, however recently the 20A breaker has been tripping. When measuring the current on this circuit we measure about 23A, obviously enough to trip the breaker after some time. The 20A breaker was specified and installed by manufacturer.

We have toubleshot circuit and not found anything obvious. We were thinking that maybe the current draw has been close to this tolerence for the last 6 months, and just now we are seeing it high enough to trip breker.

We have discussed upsizing the breaker to a 25A breaker however the wire in the circuit is only 14AWG. The wire is run in a bundle with other wire both in free air, and through small enclosures.

Do you think we can go ahead and upsize this breaker even though we have 14AWG wire? Does the way its routed help reduce heating, and allow more current capacity? Does the fact that the current is DC produce less heating and allow for a greater current capacity?

Do you think we should look to increase the size of this breaker?
 

philly

Senior Member
Is there a transformer associated with this circuit?

No this circuit is purely resistive with (4) light bulbs and (2) resistors. Since this problem is similar to another current post on resistance in series I'll give a rough sketch of the circuit as follows:

(24V)------------------------------------ 2ohm ------------(200W)---N
| |
|------| |---------|-----(200W)---N
|
|-------2ohm-------------(200W)---N
| |
|------| |--------|------(200W)---N

The (4) lights are 200W lights rated at 30V and the 2ohm resistors in series with each light pair is used to dim the lights. The (2) contacts are used to put lights at full power. Only one contact is closed at any given time.

I calculate the resistance of each bulb based on 200W at 30V as follows:

R = V^2 / P = 30^2 / 200 = 4.5 ohms (at rated temperature)

So if for some reason both contacts were closed I would expect to see (4) lights in parallel I get a combined resistance of 1.12 ohms. 24V/1.12ohms = 21.4 A which is close to the 23A which we are seeing. However we cannot see that both contacts are closed at the same time applying full voltage to lights.

If the circuit operates as designed with only one contact closed at a time I would expect to see 10.6A total going to the (2) lights on the closed contact at full power and some lesser current on the other two lights in series with the 2ohm resistor.

Looking at the (2) lights in series with the resistor the (2) lights would have a combined resistance of 2.25 in series with the 2 ohm resistor. With this only about half the voltage or 12.7V across the lights and thus only 35.8 watts. I would also expect to only see about 5.6A of current through this branch based on 24V / 2.25ohms = 5.6A. The part that I'm not sure about however is how to model the resistance of these two lights that will only be at half power due to the fact that they will not be at full temperature and thus not be at full resistance. With the lights only at 1/5 their power is it safe to assume that resistance of lights will only be 1/5 of 4.5ohms?

The circuit did not post as intended but basically each contact is in parallel with each 2ohm resistor and there are (2) lights off off each contactor/resistor combo.
 
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philly

Senior Member
I'm not sure weather this is true or not but I'm for modeling the resistance change in the lights I'm going to assume that the resistance change is linearly proportionally to the temperature change.

So assuming that the cold temp of the bulbs is about 1/10 of 4.5 ohms (measured actually at .5ohms) then the initial current through the bulb would be 24V/.0 = 48A. However since the highest the initial current can possibly go is is 12A limited by the 2ohm resistor then I'm assuming that the highest the bulb temp can rise is only 1/4 of rated current and therefore the bulb resistance would only raise to 1/4 of rated or 1/4(4.5) = 1.125ohms.

Therefore the (2) bulbs in in series with each resistor will draw a combined current of 9.37A. So with both contacts open and both sets of bulbs in series with each resistor we would have 9.37 * 2 = 18.75 which is very close to the 18.5A we see when measuring circuit with both contacts open.

Does this approximation seem sound or am I way off?
 

hurk27

Senior Member
200w @ 30 volt lamps is telling me aircraft landing floods or spots?
Is this a stage lighting setup?

sure doesn't sound like it's NEC compliant at least for a dwelling install.

I done many lighting designs for stages, and using 5 24 volt aircraft spots in series allowed me to control them right off my dimmer pack, or 10 12 volt, same thing. but to control them individually or in pairs, would take some circuiting.

I would check the resistors to see if any breakdown in the insulation has happened, 400 watt resistors would be hard to find unless wire wound.

Other then raising the resistance to provide more current limiting, I just don't know.
 

SG-1

Senior Member
Can you measure current through each branch of the circuit ? Are they equal ?

With the power off you might try some resistance readings & compare the two circuits.

To see if only one contact is closed you only need to measure voltage across it. Zero volts would indicate it is closed. Full supply voltage would indicate it is open.

Have the bulbs been replaced lately ?

Has the ambient temperature at the breaker increased for some reason ? Connections tight at the breaker ?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
091028-0725 EST

philly:

Create an X-Y plot of the lamp current vs voltage. I extrapolated from a 25 W 120 V bulb and did not make accurate measurements. The results are
30.0 --- 6.67 A
27.5 --- 6.31
25.0 --- 5.89
22.5 --- 5.60
20.0 --- 5.25
17.5 --- 4.83
15.0 --- 4.38
12.5 --- 3.99
10.0 --- 3.51
07.5 --- 2.80
05.0 --- 2.25
02.5 --- 1.61
00.0 --- 0


Next on an X-Y plot draw a straight line from 0 V 12 A to 24 V 0 A. This is your 2 ohm resistor. The intersection of these two curves is the operating point of the 2 ohm resistor in series with the 200 W lamp. I get about 15.2 V at 4.4 A.

At 24 V on your lamp I get about 5.8 A.

Two full brightness lamps 11.6 A.
Two full plus one partial 11.6 + 4.4 = 16 A
Three full brightness 11.6 + 5.8 = 17.4
Two full plus two partial 11.6 + 8.8 = 20.4 A
Three full plus one partial 17.4 + 4.4 = 21.8 A
Four full 23.2 A

I could not follow the description of your switching logic.

.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
As a _rough_ approximation, the power consumed by an incandescent lamp scales as the V^(1.5) Since power is simply voltage * current, this approximation can be restated as the current consumed by an incandescent lamp scales as V^(0.5) You can calculate the resistance from the 're-rated' voltage and current if you wish.

Your lamps are rated as 200W at 30V, or 6.67A at 30V. At 24V the approximation suggests 5.96A.

If you want to solve the series circuit of a resistor and a lamp, I'd suggest writing the equation in terms of voltage and current rather (in a resistor I scales as V^1 rather than converting lamp voltage/current into resistance.

-Jon4
 

broadgage

Senior Member
Location
London, England
Have you checked the actuall wattage of the lamps ?
Have perhaps 250 watt lamps been fitted instead of 200 watt ?

A 25 amp breaker might be considered acceptable on #14 wire in some circumstances.
It would be a code violation for any common type of permanent install, but this sounds as though it might be an "appliance" and not an "installation"
If however it is an installation then I believe that at least #10 will be required.

A DC current will have exactly the same heating effect as an AC current.
Unless stated otherwise, AC currents are allways RMS currents, and one definition of an RMS current is "that AC current which has the same heating effect as a corresponding DC current"
 

templdl

Senior Member
Location
Wisconsin
If the breaker can be reset immediatel after a trip it is with most certainty is tripping instantaneously (magnetically) as a result of a fault or a high magnitude of instantaneous current.
If the breaker tripped thermally the thermal element must be given time to cool befor the breaker can be reset latched and then closed. If the thermal element is not left to cool the breaker can not be latched.
There could be a possibility that the breaker may be out of calibration is the current that is causing the breaker to trip is determined to be within or less than the thermal magnetic trip curve. If that trip curve has moved to the left then you could have an issue.
 

Power Tech

Senior Member
Is it possible to run another conduit or wire and split the circuit?

20+ amps on a 14?

Lighting loads are ruthless.

I I have found it a good practice to always oversize wire on lighting.

I have serviced many factories with hot pipes.
 

philly

Senior Member
091028-0725 EST

philly:

Create an X-Y plot of the lamp current vs voltage. I extrapolated from a 25 W 120 V bulb and did not make accurate measurements. The results are
30.0 --- 6.67 A
27.5 --- 6.31
25.0 --- 5.89
22.5 --- 5.60
20.0 --- 5.25
17.5 --- 4.83
15.0 --- 4.38
12.5 --- 3.99
10.0 --- 3.51
07.5 --- 2.80
05.0 --- 2.25
02.5 --- 1.61
00.0 --- 0


Next on an X-Y plot draw a straight line from 0 V 12 A to 24 V 0 A. This is your 2 ohm resistor. The intersection of these two curves is the operating point of the 2 ohm resistor in series with the 200 W lamp. I get about 15.2 V at 4.4 A.

At 24 V on your lamp I get about 5.8 A.

Two full brightness lamps 11.6 A.
Two full plus one partial 11.6 + 4.4 = 16 A
Three full brightness 11.6 + 5.8 = 17.4
Two full plus two partial 11.6 + 8.8 = 20.4 A
Three full plus one partial 17.4 + 4.4 = 21.8 A
Four full 23.2 A

I could not follow the description of your switching logic.

.


I could not determine where you were getting the current values from listed above. I tried several different calculations but could not come up with the same Voltage and current relationship that you did?

I have attachd a better schematic of the circuit. Basically these lights are on a locomotive, and one set is for the front and the other for the rear. While the locomotive is idling both contacts are open, and both set of lights are through the resistor and thus dim. When the unit is put into gear on of the contacts closes and supplies the full 24V to the lights thus making them brigher as the others stay dim. Both contacts are never closed at the same time to make both sets bright.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
091029-1715 EST

philly:

Your new schematic is whole different story.

For my estimates of lamp current I will double the values in my table. Now the intersection point is approximately 10 V for the lamp voltage with 2 ohms in series with a pair of lamps.

You can determine the actual value by measuring the lamp voltage.

Both switches open. 2 ohms in series with a pair of lamps paralleled has a current of about 7 A. So with both switches open total is 14 A.

Close one switch and the current will be about 5.8 + 5.8 + 7 = 18.6 A. The 5.8 is from my earlier post.

You need to use a clamp-on DC ammeter to measure the actual currents. Measure voltage as well.

.
 

nakulak

Senior Member
the 14 wire is a violation.

as others noted, this breaker is overloaded from the getgo. you should not be putting more that 16 on it continuously.


is the 20 a bolt in breaker ? if so, I would check to see that it is tight. this might allow you to continue to overload the ckt and the wire until it eventually burns up.
 
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