# Parasitic Inductance, rough estimate -Single cable through steel

• Windadct
The conversation revolves around a DC power distribution system in which the customer has run their cable through metal boxes, which would not be allowed in an AC system due to eddy currents. This has caused potential control issues, and the speaker is looking to measure the inductance of each box entry and exit to justify potential problems with the installation. There is also discussion about potential solutions and the issue of incorporating AC control on a DC system.

#### Windadct

TL;DR Summary
Looking to estimate the inductance of a 262kcmil entering a steel, NEMA type box, 14 Ga. Ideally with fittings ( lol)
DC case - but customer ran their cable run through a couple boxes, this would not be allowed in AC due to the eddy current created in fittings and the enclosure steel. ( I do not even know if code would allow this for DC) - but anyway. This is part of a high power dc system and I suspect the series inductance may be causing a control issue.

Any idea how to calculate the inductance of each box entry ( and exit of course)

Can you just measure it? (or is it not made yet?)

Also, the two wires are the power out and return, and you are wanting to calculate the increase in inductance of that loop based on the close metal fittings?

Windadct
It is all built - and this will be down in the mH range for sure. Yes this is 2 Pos and 2 Neg conductors through 4 enclosures, most with fittings.

Hmmmm - but not a bad idea to connect the ends at the load and measure the entire loop.

We do not have people onsite so all of this is guidance / or requests to the client. I am still surprised the electricians that did the install did it this way.

berkeman
Last i knew this sort of installation is allowed for DC but is a big no-no for AC. A thin sawcut between the holes where the conductors go through the box will likely take care of any choking effect.

Windadct
Ignoring eddy currents for the moment, model the circuit as two conductors of a return circuit passing through separate holes in a magnetic lamination. The return circuit will double the flux in the sheet material between the conductors. Will material in that region be saturated by the magnitude of the DC current? Any saturation would modulate the AC impedance of the return circuit, dependent on DC current.

hutchphd
It seems the easiest way to avoid excess induction to the boxes is to route a power and its return in each conduit. The external fields tend to cancel that way.

For AC power wiring this is a code requirement anyhow, to have all conductors of a circuit in the same conduit/raceway.

Tom.G said:
It seems the easiest way to avoid excess induction to the boxes is to route a power and its return in each conduit. The external fields tend to cancel that way.

For AC power wiring this is a code requirement anyhow, to have all conductors of a circuit in the same conduit/raceway.
Exactly - that is the standard way - and how I would have done it. But the system is built. I was looking for an actual inductance figure to justify to our and the customer engineering teams that this may be part of the problem.

Baluncore said:
Ignoring eddy currents for the moment, model the circuit as two conductors of a return circuit passing through separate holes in a magnetic lamination. The return circuit will double the flux in the sheet material between the conductors. Will material in that region be saturated by the magnitude of the DC current? Any saturation would modulate the AC impedance of the return circuit, dependent on DC current.

Basically what I was thinking, then halve the value and apply it to every case in the power run.

Windadct said:
Exactly - that is the standard way - and how I would have done it. But the system is built. I was looking for an actual inductance figure to justify to our and the customer engineering teams that this may be part of the problem.
Beware; we routinely reject posts on PF that appear to be seeking to gather information potentially used in litigation or disputes. Your question sounds borderline.

Windadct said:
Summary: Looking to estimate the inductance of a 262kcmil entering a steel, NEMA type box, 14 Ga. Ideally with fittings ( lol)

DC case - but customer ran their cable run through a couple boxes, this would not be allowed in AC due to the eddy current created in fittings and the enclosure steel. ( I do not even know if code would allow this for DC) - but anyway. This is part of a high power dc system and I suspect the series inductance may be causing a control issue.
Can you please clarify this? Inductance is not an issue for DC, but you mention AC control issues. Is there an AC control communication waveform superimposed on this DC power distribution system? If so, how do they deal with the low impedance power source and low impedance power loads on the system? I work with Link Power systems a lot, and the issues of superimposing communications waveforms on high power distribution wiring are non-trivial.

Also, what percentage of the linear run of this DC power distribution cabling system is through these metal boxes? If it is a low percentage, then the inductance of the separated power wires in the overall system will dominate the total inductance of the loop, IMO.

DaveE and Windadct
This is a large test system, with 4 wire (remote V sense), we have seen long and "sloppy" cable runs introduce enough inductance to be an issue, and cause an oscillation of the DC. I do not think this is a passive component oscillation (LC)- but a control system response to a far more complex load than it was set up for. I am sorry if I said "AC control"? - it is a DC system - but the wiring they used would not be allowed in an AC system.

For reference -- we have about 2000 of these systems between 100-2.4MW and normally can control 650V to better than 0.5 V - even when under inverter as a load. Here with no real load we have a 40 V P-P.

anorlunda said:
Beware; we routinely reject posts on PF that appear to be seeking to gather information potentially used in litigation or disputes. Your question sounds borderline.

Understood - I have been on her for a while and I assure you this is not the case. Before we ask our client to de-construct an installed system I am trying to justify it - even if with a napkin calculations.

berkeman
Windadct said:
This is a large test system, with 4 wire (remote V sense), we have seen long and "sloppy" cable runs introduce enough inductance to be an issue, and cause an oscillation of the DC. I do not think this is a passive component oscillation (LC)- but a control system response to a far more complex load than it was set up for.
Oh, that's different. So the feedback system in this voltage delivery system has an oscillation and you are trying to understand the source. Got it.

How much does the load current vary, and at what bandwidth? What control do you have over the feedback system's transfer function? If you need to run a test with a transfer function test instrument (like an HP 4194 Gain/Phase instrument), could you get access to the system briefly, or does this have to get diagnosed/fixed with all systems running?

hutchphd and dlgoff
Also, what is the power source like? Is it a switching power conversion system? If the load varies near the switching frequency (or harmonics) of the power source, that can cause large instabilities. (Please do not ask me how I know this...)

The power source is the high power test instrument, as I said mature and a large population at large, used in the leading power research labs in the country - this is a 1 in >100 application case problem. We are not going to diagnose the total system here. I was just hoping to be able to sketch up an approximate schema with the inductances to bring to the team.

Tom.G
Perhaps it is not Inductance per se that is a concern, you essentially have a shorted turn surrounding the conductor.

The usual problem with unbalanced magnetic fields is magnetic attraction. In my experience, a few hundred amps in 3-phase circuits can easily break a pencil inserted between conductors,

My biggest concern would be your conductors moving around at each pulse and destroying the wire insulation; a bit more problematic than an unstable power supply.

See section 3.2 of:
https://electrical-engineering-portal.com/wiring-cables-conductors-low-voltage-switchboard-dos-donts-precautions#protection-against-magnetic-effects

(above found, on pg. 3, of:
https://www.google.com/search?q=effect+of+conductor+surrounded+by+metal+ring)

Cheers,
Tom

hutchphd
If the outlet conduit it is a magnetic conduit of 4" inner diameter and 4.5” outer diameter [the μrfe≈1000] .

Consider the magnetic flux through the conduit only and the conductor

in the center then H=I/(2*pi()*r) ; r=2 inches ; I=100 A [for instance].

r=2*25.4/1000=0.0508 m H=I/(2*pi()*r)=313.3 A/m. Bfe= μrfe*μo*H=0.394 Wb/m^2

μo=4*pi()/10^7 H/m

Magnetic flux per meter [approx.] Φ = Bfe*wd*lng wd=conduit thickness=0.25”= =0.25*25.4/1000=0.00635 m ; lng=1m Φ=0.0025 Wb L=Φ/I=0.0025/100=2.5E-05 H/m

Calculating induction in the open space between conductor and conduit will be approx.1.67E-07 H/m so it could be neglected.

Windadct
Babadag - thanks for looking into this. A single conductor in conduit is a pretty common method for creating a heat trace, with an AC Loop

In the case I am looking at there is no conduit - a single insulated cable ( 6262MCM) enters a metal NEMA box via a metallic fitting ( gland) .

In my opinion we can separate the inductance of the gland from the inductance of the enclosure with respect to cable.
For an approximate solution we may consider the enclosure as a cylinder of the same circumference.
Let say the gland it is a cylinder of 2" outside and 1.5" inside and 2" long.
Calculated the L1=1E-7 H [non-saturated steel, centred conductor].
For a steel box of 10*10" and 1/10" thickness the equivalent cylinder diameter is
40/pi()=12.73” , the length of 10” then Lencl=4.35965E-05 H

Last edited:
Thanks again -

I think I did a mistake when I translate from Imperial to MKSA units.
If the outside surface of the junction box it is 4*10*10 then the H run on this surface it is 40"=40*25.4/1000=1.016 m ; Let's say I=100 A
Then on a cylinder H=I/1.016=100/1.016=98.425 A/m
Bfe=1000*4*pi()/10^7*98.42=0.1237 Wb/m^2
Magnetic flux through 1/10"-2.54/1000=0.00254 m and 10" length=0.254 m
The area A=0.254*0.00254=0.000645 m^2
The magnetic flux then it is Bfe*A=0.1237*6.45e-04=7.98E-05 Wb
L=7.98e-05/100=7.98e-07 H