Two 200 W drivers wired in parallel do not give you 400 W of usable capacity. In the worst case you get barely more than one unit can deliver, and what you actually get is a cascade failure: one unit carries nearly all the current, hits its overcurrent protection, drops out, the second inherits the full load, trips in turn, and the whole display goes dark. That is strictly worse than the single 400 W unit you were trying to avoid buying.
Paralleling has exactly one legitimate reason — redundancy. If your reason is capacity, buy the bigger unit.
Constant-voltage drivers regulate to a set voltage with a tolerance, typically ±1% to ±3%. On a 24 V nominal output:
Now connect their outputs in parallel. Driver A, being higher, tries to hold the bus at 24.24 V while driver B sees a voltage above its set point and backs off to zero current. In practice A supplies the load until it reaches its current limit, the bus voltage sags to B’s set point, and B picks up the remainder.
The result is neither current sharing nor clean handover. Whichever unit has the higher set point runs hot and carries most of the load; its lifespan drops accordingly. And the load is not the only current flowing — the mismatch also drives circulating current between the two outputs through the paralleling impedance.
Worse, many CV drivers sense their output through the same terminals they deliver through. Two drivers fighting over one node can push each other into hiccup mode, where both cycle on and off. The visible symptom is a display that pulses.
Not really paralleling, and the correct answer in most cases.
A 300 W load split across two zones does not need two 200 W drivers — it needs one 400 W driver, or two 150 W drivers on separate zones. Two separate zones give you genuine partial-failure behaviour (one zone dark, one lit) with zero efficiency penalty and no extra components.
The real benefit shows up on large installations: four 300 W drivers on four zones cost less than one 1,200 W unit, and a single failure costs you 25% of the display instead of 100%. The load calculator worksheet walks through the split.
Put a diode in series with each output before the common bus. The higher-voltage unit conducts; if it fails, the diode reverse-blocks and the second unit takes over with no interruption.
Two design consequences:
Voltage drop. A Schottky diode drops 0.3–0.5 V at rated current. On a 24 V bus that is 1.2–2%, which shows up as slightly reduced output at the far end of the cable run. Compensate by trimming the driver’s output adjust up by the diode drop — most industrial units have a ±10% adjustment range — or accept the loss.
Heat. At 10 A, a 0.4 V drop is 4 W per diode. That needs a heatsink or a PCB-mounted diode with adequate copper. Undersized OR-ing diodes are a common fire-starting detail in field-built redundancy schemes.
A MOSFET plus a controller IC replaces the diode, dropping 20–50 mV instead of 400 mV. Controllers in this class switch in microseconds and also block reverse current if a driver fails short.
The added cost per channel is roughly 3–8 in components — small next to the driver cost, and it removes both the voltage-drop problem and the diode-heat problem. If you are building redundancy deliberately, this is the option to specify.
3–
Two 12 V drivers in series will produce 24 V, and this works only if the outputs are galvanically isolated from mains and from each other. Most are, but verify it:
Do not series-connect to reach a higher voltage unless the datasheet explicitly states the output is floating. If it does not say so, assume it is not safe.
Build N+1 when the cost of darkness exceeds the cost of the second driver:
Skip it when the failure is visible and cheap to fix: a shop-front sign above a serviceable door with a spare on the shelf does not need redundancy.
The arithmetic: N+1 with four units means each runs at 75% of its rating — 3 units of load across 4 drivers. That light loading is a bonus, not a waste: 75% sits inside the efficiency sweet spot and reduces thermal stress. See 500W Switching Power Supply: Sizing & Derating Guide.
Load: ____ W at ____ V → ____ A Zones: ____ (prefer ≥2) Redundancy required? Y / N (if Y: what does one hour of darkness cost?) If Y: Topology: separate zones / Schottky OR-ing / ideal-diode OR-ing Driver rating: ____ W each (must be ≥ 100% of load) Diode drop budget: ____ V → output trim: +____ V Diode heat at ____ A: ____ W → heatsink required? Y / N Per-output fusing: ____ A DC OK signalling: Y / N
If the conclusion is “buy one bigger driver instead,” the range is:
Send your load, zone layout and uptime requirement — we will tell you whether you need redundancy at all, and if so which topology. Sometimes the honest answer is one bigger driver. → Request a quote · 500W sizing guide
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