How much array can one inverter actually swallow? This gives you the DC/AC ratio, the exact
ratio at which clipping begins for your site conditions, the annual kWh it
costs you, and the nameplate size that is worth buying. The clipping model is derived on this
page, not fitted to a chart.
Where the clipping number comes from
Two things decide whether an array clips. First, how much DC power the array actually makes
when the sun is strongest — which is not its nameplate, because hot cells make less.
Second, how often the sun is strong enough to reach the ceiling.
T_cell = T_amb + (NOCT − 20) ÷ 800 × G_peak
k = (G_peak ÷ 1000) × [1 + γ × (T_cell − 25)]
P_dc,peak = P_dc,STC × k
R_th = 1 ÷ (k × η_inv) ← clipping starts above this DC/AC
x_th = R_th ÷ R_actual (capped at 1)
loss_clear-sky day = (1 − x_th)^1.5
loss_annual = f_clear × loss_clear-sky day
The (1 − x_th)^1.5 term is the part worth explaining. Model the clear-sky day as a
parabola, G(t) = G_peak (1 − t²). The time spent above any irradiance
G is then H_day × √(1 − G/G_peak). Integrating the excess —
everything above the inverter ceiling — over that profile gives
(1 − x_th)^1.5 of the day's energy. It is a closed form, not a curve fitted to
someone's simulation, and you can reproduce it with the formula above.
The f_clear term scales it to a year: only energy produced under near-clear skies
can reach the ceiling at all, and that share is site-dependent. It is an input for exactly
that reason — if your site is cloudier or clearer than the default, change it.
Worked example
60 modules × 550 W = 33.0 kWp on a 25 kW inverter,
97.5 % weighted efficiency, γ = −0.35 %/°C, NOCT 45 °C, 30 °C ambient at peak sun.
- Cell temperature at peak: 30 + 25 ÷ 800 × 1000 = 61.3 °C
- k = 1 × [1 − 0.0035 × (61.3 − 25)] = 0.873 — the array makes
87.3 % of nameplate when it matters most
- P_dc,peak = 33.0 × 0.873 = 28.81 kW → 28.09 kW AC, above the 25 kW ceiling
- R_th = 1 ÷ (0.873 × 0.975) = 1.175 — below this ratio, no clipping at all
- Actual ratio 1.32 → x_th = 0.890 → clear-sky-day loss (1 − 0.890)^1.5 = 3.65 %
- At 55 % clear-sky share: 2.01 % of the year = 924 kWh = about
$111/yr at $0.12/kWh
Is the bigger inverter worth it?
This is the question the tool exists to answer, and the answer is usually no. Using the array
above, moving from 25 kW to 30 kW:
| 25 kW unit | 30 kW unit |
|---|
| DC/AC ratio | 1.32 | 1.10 |
| Annual clipping loss | 924 kWh | 0 kWh |
| Extra capital | — | ≈ $1,100 (at $0.22/W × 5 kW) |
| Part-load efficiency penalty | — | ≈ −138 kWh/yr (≈ 0.3 % poorer) |
| Net recovered | ≈ $94/yr → payback ≈ 11.7 years |
Eleven years is longer than most inverter warranties and close to the service life of the unit
itself. Keep the 25 kW. The premium only pays when the tariff is high
(above roughly $0.25/kWh), or when the larger unit costs under about $0.10/W more. Substitute
your own numbers — the two that decide it are the price per watt and the tariff, and both are
inputs here.
Limitations
- The clipping model assumes a parabolic clear-sky day and a single clear-sky energy share.
Treat the annual loss as ±30 % in relative terms — good enough to choose
between two inverter sizes, not good enough for a bankable energy assessment.
- Inverter efficiency is a single weighted constant. Real efficiency sags at light load, so
a badly oversized inverter loses a little more than this tool shows.
- Only cell temperature derates the DC power. Soiling, mismatch, wiring and orientation
losses belong to the annual energy figure you supply, not to this model.
- DC input current and MPPT allocation are not checked here. Use the
string configurator for the voltage and current windows of each MPPT input.
- No ampacity, overcurrent protection, DC isolator rating or grid-code check. This tool
matches array power to inverter power and stops there.
What DC/AC ratio should I use?
1.10–1.35 for most fixed-tilt systems. Below 1.10 you pay for capacity the array rarely uses; above 1.35 clipping usually costs more than the smaller inverter saved. Run the numbers for your tariff rather than copying the band.
Does clipping damage the inverter?
No. It is the inverter holding output at its rating by moving off the maximum power point — a design loss, not a fault. What does matter is the DC input side: array STC power and string voltage must stay inside the datasheet limits, which this tool flags separately.
Why does my array never reach its STC rating?
The nameplate is measured at 1000 W/m² with the cell at 25 °C. On a real roof at peak sun the cell runs 55–70 °C, and silicon loses 0.35–0.45 % per degree above 25. Typical real peak: 80–92 % of STC. That is exactly why a ratio above 1.0 does not automatically clip.
Is a bigger inverter worth it?
Only if the recovered energy pays the premium before the unit needs replacing. In the worked example, 25 → 30 kW recovers $94/yr against $1,100 extra — an 11.7-year payback, so no. It flips at high tariffs or when the larger unit costs under about $0.10/W more.
Does this check ampacity or code compliance?
No. It models array-to-inverter power matching only. Ampacity, overcurrent protection, DC isolator ratings and grid code are out of scope, and no wiring code is read.