Quote and layout for a light bar, fridge socket and auxiliary battery
A workshop is preparing a 2019 dual-cab for touring. The customer wants a 240 W LED bar on the roof and a 15 A fridge socket in the tub, and supplies this layout: main battery positive → 40 A fuse → 8 mm² cable, 5.5 m one way with the chassis as the return → 100 Ah AGM auxiliary battery in the tub. From the auxiliary battery positive → 30 A fuse block → relay terminal 30 and the fridge socket. Relay terminal 85 to the tub body, 86 to a dash switch fed from a 10 A ignition-switched fuse, 87 to the light bar, light bar earth to the tub body. The tub is bonded to the chassis by a strap measuring 0.08 Ω. The alternator regulates at 14.4 V and the AGM maker specifies 14.4–14.7 V charging. Copper resistivity is 0.0175 Ω·mm²/m. The fridge socket must be dead with the ignition off.
1. Calculate the light bar current and choose a fuse and cable for the bar run (3.0 m each way) allowing no more than 3 % drop. Check whether the 30 A fuse block and the relay are suitably rated.
Read the worked answer
Current
240 W ÷ 12 V = 20 A. At a charging 13.2 V it is about 18 A, but design for 20 A so the installation is safe at low voltage and the driver is not run at its limit.
Cable for 3 % drop
- 3 % of 14.4 V = 0.43 V
- R max = 0.43 ÷ 20 = 0.0215 Ω over 6.0 m of loop
- A = 0.0175 × 6.0 ÷ 0.0215 = 4.9 mm²
So 6 mm² is the minimum for the bar run, with a 25 A fuse at the source end (or 20 A, which still carries the load and protects 6 mm² comfortably).
The 30 A fuse block
The block feeds the 20 A bar and the 15 A fridge socket from one 30 A fuse. Combined potential is 35 A, so the fuse will nuisance-blow the first time the fridge compressor starts with the bar on, and even if it holds it is protecting the block at its limit. Split it: a 25 A fuse for the bar and a separate 20 A fuse for the socket.
The relay
A 30 A relay is only just adequate. A 240 W LED bar has a capacitive driver with a high inrush, which pits contacts over time. Fit a 40 A relay with a suppression diode across the coil and crimp the terminals - do not solder heavy lugs that will be subject to vibration.
2. The fridge socket must be live only with the ignition on. Explain how the layout must change and calculate the load seen by the dash ignition fuse if it were wired as drawn.
Read the worked answer
As drawn
The socket is fed straight from the auxiliary battery through the 30 A fuse block, so it is permanently live. The fridge will flatten the auxiliary battery over a weekend and the customer will blame the wiring.
The change
Add a 40 A normally-open relay for the socket. Contacts take the heavy feed from the auxiliary battery through a dedicated 20 A fuse to the socket; the coil is fed from an ignition-switched circuit through the dash switch. That way the switch only carries coil current, about 0.15 A, and the ignition circuit never sees the fridge load.
If it were wired through the ignition fuse as drawn
- The fridge socket is rated 15 A and a compressor fridge can pull 5–8 A running, with a start surge well above that.
- The dash circuit is protected by a 10 A fuse. Two outcomes: the fuse blows on the first compressor start, or the fuse holds and the undersized dash wiring and ignition switch contacts carry the current and heat up.
Neither is acceptable. Keep the load on a dedicated feed, switch only the relay coil, and fuse the socket at 20 A close to the source. Also check the socket's own rating: a 15 A cigar-style socket is not a 15 A continuous connector, and a fridge that runs for hours will warm it.
3. Evaluate the charging path to the auxiliary battery. Calculate the voltage drop at 40 A including the earth strap, and state the effect on the AGM battery.
Read the worked answer
Positive run
- Loop length 11 m of 8 mm²
- R = 0.0175 × 11 ÷ 8 = 0.0241 Ω
- At 40 A: 0.0241 × 40 = 0.96 V
- At 60 A (a deeply discharged AGM will draw hard): 1.45 V
Earth return
The auxiliary battery negative is bonded to the tub, and the tub to the chassis through a 0.08 Ω strap.
- 0.08 × 40 = 3.2 V
That strap is the killer. Total drop could exceed 4 V, so with 14.4 V at the alternator the AGM may never see more than about 10 V.
Consequences
An AGM held below its specified 14.4–14.7 V never reaches full charge. It sulfates, loses capacity, and the 40 A fuse may blow anyway from the inrush when it is deeply flat. It will also fail early and be blamed on the battery.
Fix
- Run a dedicated earth of the same size as the positive back to the main battery or a clean chassis point, not through the tub strap. Replace the 0.08 Ω strap and clean its mounting points.
- Upsize the positive to 16 mm² (R = 0.012 Ω, 0.48 V at 40 A).
- Better still, fit a DC-DC charger at the auxiliary battery: it compensates for the drop, gives the AGM its correct multi-stage profile, and protects the start battery. Size its input fuse to the cable and its output to the AGM's acceptance rate.