Solar battery sizing for a PLC and screen

What do I need before I can size the solar and battery?

You need four numbers before anything else: each load's measured watts and hours per day, the modem's supply voltage and daily transmit time, the site's worst-month peak sun hours, and the battery's usable fraction from its datasheet.

Measure before you buy. A DC clamp meter or a shunt-based battery monitor on each load gives watts you can defend. Nameplate ratings are often worst-case figures; backlight level, standby mode and polling behavior move them by tens of percent. Alongside the measurements, gather:

• each device's input voltage range, so the bank's full and discharged voltages both sit inside it
• the modem's transmit schedule, from the polling logic or the report interval
• the monthly peak-sun-hour table for the site's latitude and tilt
• the battery datasheet: usable fraction, temperature table, discharge rating
• the autonomy target in days, chosen by what a dropout costs

Three days of autonomy is the common remote-site target; a site that loses product or security on a dropout justifies five.

How many watt-hours per day does a 20 W PLC and a 10 W screen draw?

A 20 W PLC and a 10 W screen powered for all 24 h of the day use 30 W × 24 h = 720 Wh per day. That energy figure, not the wattage, is what the battery has to store.

Watts measure the rate; the battery stores energy, so multiply by the hours each load is actually powered. The PLC contributes 20 W × 24 h = 480 Wh and the screen 10 W × 24 h = 240 Wh. Take the reading over a full shift with a clamp meter where you can. Screens that idle down and PLCs that only communicate in bursts read well under their plates, and the measured figure replaces the nameplate in every later step. If part of the site load is a standalone sign instead of the operator screen, battery-powered digital signage: size the power first works the same problem from the screen's side.

How do I count a radio modem that draws 500 mA only while transmitting?

Multiply volts by amps: on a 12 V feed, 0.5 A is 6 W while transmitting, so a modem that transmits 48 minutes a day adds 4.8 Wh, and its idle draw often adds more than the bursts do.

Current alone is not power. The watt figure needs the modem's supply voltage: 0.5 A × 12 V = 6 W, and at a 24 V feed the same current is 12 W. The transmit time comes from the polling schedule: a 10-second report every 5 minutes is 10 ÷ 300 = 3.3% duty, which is 48 minutes a day, so 6 W × 0.8 h = 4.8 Wh. Idle and receive states usually outweigh the bursts. A modem idling at a measured 30 mA takes 0.36 W on that same feed, which is 8.6 Wh over a day, close to twice the transmit energy. Take the current profile from the maker's documentation or measure it at the supply terminals, and do the multiplication at whatever voltage the datasheet's figure assumes.

Which losses go into the budget before the battery?

Divide the load's watt-hours by the converter efficiency, so 733 Wh of daily load at 90% becomes about 815 Wh a day that the battery must deliver.

Every DC-DC converter between the bank and a device takes its cut. Read the efficiency at your actual output power from the datasheet curve; most converters land between 85% and 95%, and 90% is a fair planning figure when the curve is missing. The load list so far totals 720 + 13 = 733 Wh, and 733 ÷ 0.90 = 815 Wh a day at the battery terminals. Converter application reports from semiconductor makers show how to read that efficiency curve and how much heat the missing percentage dumps into the enclosure, which matters for a sealed cabinet in summer. Charge-controller and battery-charging losses belong to the array derate below, so count them once. Energy and peak current stay separate problems; the amps come in the controller section.

How many battery amp-hours do 3 cloudy days take?

Multiply the daily battery energy by the autonomy days and divide by nominal volts times the usable fraction: 815 Wh × 3 ÷ (12 V × 0.5) = 407.5 Ah, so specify about 410 Ah of nameplate capacity at that nominal voltage.

The usable fraction is the depth of discharge the chemistry tolerates for the cycle life you want. Flooded and AGM lead-acid live longest near 50%, so 0.5 is the planning figure; lithium makers commonly allow 80%, which turns the same 2445 Wh draw into 2445 ÷ (12 × 0.8) = 255 Ah. Temperature moves the number too. Ratings assume about 25 °C, at freezing a lead-acid battery delivers roughly a fifth less, and near -18 °C roughly half, so substitute the maker's cold-temperature figure for an unheated cabinet. Autonomy is the biggest lever: the same budget at 5 days is 815 × 5 ÷ 6 = 680 Ah. Choose the days by what a dropout costs, then stop arguing with the result.

How many panel watts does the worst month of sun need?

Divide the daily battery energy by the worst month's peak sun hours times a 0.7 derate, then add the energy needed to refill the bank after your autonomy event; for this budget the steps land at 390 W break-even and about 620 W with a 5-day refill.

Peak sun hours are the day's solar energy expressed as equivalent hours of full noon sun, published month by month for a location and tilt. Size from the lowest month of the year: mid-latitude winters often sit at 2 to 4 h a day, and the annual average lands a third or more under the worst-month array. The 0.7 derate bundles the measured losses: an MPPT controller near 95%, lead-acid charge efficiency near 80 to 85%, and soiling, wiring and temperature around 90%. Multiplied out that is close to 0.7, and it is why a nameplate watt never arrives as a stored watt-hour.

The array also has to put back what the autonomy event drained. The 3-day drawdown here is 2445 Wh, and refilling it over 5 days adds 2445 ÷ 5 = 489 Wh a day, so the array sees 815 + 489 = 1304 Wh and needs 1304 ÷ (3 × 0.7) = 621 W, which rounds to three 210 W modules. Skip the refill term and the 390 W array barely feeds the load, leaving the bank drained until the days lengthen. Sun hours move the answer fast: at 4 peak sun hours the same 1304 Wh needs 466 W, and at 2 it needs 931 W.

Worked example: what does a 30 W site with a 500 mA modem come to?

The full budget for a 20 W PLC, a 10 W screen and a 500 mA modem is 815 Wh a day, a 440 Ah battery bank at the nominal voltage and about 620 W of array with a 70 A class controller.

Run the steps in order and check each against the last:

1. Loads. (20 W + 10 W) × 24 h = 720 Wh a day.
2. Modem. Transmit: 6 W × 0.8 h = 4.8 Wh. Idle: 0.36 W on the same feed, × 24 h = 8.6 Wh. Modem total 13 Wh.
3. Converter. (720 + 13) ÷ 0.90 = 815 Wh a day at the battery.
4. Battery. 815 × 3 ÷ (12 × 0.5) = 407.5 Ah; build 440 Ah as two 220 Ah batteries in parallel at the same nominal volts, about 8% of margin.
5. Array. Refill adder 2445 ÷ 5 = 489 Wh a day; (815 + 489) ÷ (3 × 0.7) = 621 W; three 210 W modules give 630 W.
6. Controller. 630 W ÷ 12 V = 52.5 A, × 1.25 = 65.6 A, so a 70 A class MPPT.

Recommendation: budget from measured watt-hours and buy the next standard size up, never the calculated exact. For a 30 W PLC-and-screen load with a lightly used 500 mA modem, that is about 815 Wh a day, a 440 Ah lead-acid bank for 3 days of autonomy, 630 W of panel so the bank refills within 5 worst-month days, and a 70 A controller. Drop the refill allowance only where a slow recovery costs nothing, and move the bank to 24 V if 52 A of charge current troubles you.

What size charge controller, fuse and wire does that array take?

The 630 W array pushes about 52.5 A into the bank; with the 125% margin the code applies to loads that run 3 h or more, that is 65.6 A, so a 70 A controller, an 80 A battery fuse and 4 AWG copper.

Amp-hours store energy; this check keeps the peaks alive. With everything running and the modem transmitting, the load side draws 30 + 6 = 36 W, which is 36 ÷ 0.90 = 40 W and 3.3 A at the bank. That is trivial for 440 Ah but worth redoing if the load list grows a heater or a bigger radio. The same peak-and-inrush logic is what sizing a separate 24 V DC supply for HMI inrush current works through for cabinet screens.

On the charge path, 630 W at the bank's nominal 12 V is 52.5 A and 65.6 A with the margin; at a 24 V bank the same array is 26.3 A and a 40 A class controller carries it. 4 AWG copper holds the 65.6 A design current with headroom under standard ampacity tables. Run the wire further than a few metres and voltage drop, checked to stay under 2%, starts moving the size up before ampacity does. Fuse the bank at the terminal before its first energize and vent the battery box: a lead-acid bank of this size can push thousands of amps through a dropped wrench, and flooded cells collect hydrogen under a lid.

How do I build and check the installation against the budget?

Commission against the budget, not against the panel's glow: measure a full day's watt-hours at the battery, check voltage at the loads during a transmit burst, and test the low-voltage cutoff by the controller's documented method.

1. Wire the array, controller and bank with the fuse at the battery terminal, and energize with the loads off to confirm polarity before anything draws.
2. Run a full day with every load enabled and read watt-hours at the bank with a shunt meter. Compare against the 815 Wh budget; more than about 10% over means remeasure the loads before buying panels.
3. Watch the voltage at each device during a transmit burst. A dip below the device's minimum input points at wiring, connections or the fuse, and it will be worse on a cold morning when the bank voltage sits low.
4. On a clear day, confirm the controller reaches the absorb and float targets the battery maker prints and the bank actually reaches full by early afternoon.
5. Test the low-voltage cutoff using the controller's documented method, and set it above both the battery's damage threshold and the PLC's minimum input.
6. Keep the day-one watt-hour reading and repeat it at the end of the worst month; those two numbers are the baseline every later check compares against.

To trend the load over months, log run hours in the PLC so the count survives the outages you are designing for; keeping PLC run hours through a power loss shows the counter setup. And if the cabinet also holds a separate notice screen fed from the same low-voltage bus, a DIN-rail 24 V DC supply keeps that wiring tidy in the enclosure.

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