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DC vs. AC Ohmic Losses in a PV System

In a photovoltaic system, energy follows this simplified path:

PV modules (DC)DC cables → inverterAC cables → (optional AC transformer) → grid

At each stage where there are conductors (cables, transformer windings), some of the power is lost as heat: these are ohmic losses.

The basic law is the same everywhere:

\[ P_{\mathsf{losses}} = R \cdot I^{2} \]
  • \(R\): resistance of the conductors \((\mathrm{\Omega})\)
  • \(I\): current flowing \((\mathrm{A})\)
  • \(P_{\mathsf{losses}}\): power lost \((\mathrm{W})\)

What differs between DC and AC are:

  • the location where the losses occur,
  • the voltage and current levels,
  • and how they are expressed (as a % of DC losses, as a % of AC losses, etc.).

1. Ohmic losses on the DC side

On the direct current side, between the modules and the inverter, we find:

  • string cables (between modules in the same string),
  • busbar cables (junction boxes, DC cabinets),
  • DC cables leading to the inverter inputs.

This entire network can be combined into an equivalent DC resistance \(R_{\mathsf{DC}}\), as seen from the PV array.

The instantaneous losses in this section are given by:

\[ P_{\mathsf{losses,DC}} = R_{\mathsf{DC}} \cdot I_{\mathsf{DC}}^{2} \]

where \(I_{\mathsf{DC}}\) is the total current delivered by the PV array.

Important characteristics on the DC side:

  • voltages are relatively high, but so are currents (especially in direct sunlight),
  • cable lengths between PV arrays and the inverter can be significant,
  • DC losses depend very heavily on the cross-sectional area of the cables and the number of strings in parallel.

Example 1 – DC losses expressed as a percentage of the PV array’s rated power

Data:

  • PV array: \(100\ \mathrm{kWp}\)
  • Operating DC voltage: \(800\ \mathrm{V}\)
  • Rated DC power: \(100\ \mathrm{kW}\)
  • Total DC current:
\[ I = \frac{P}{U} = \frac{100\,000}{800} = 125\ \mathrm{A} \]
  • Equivalent round-trip resistance of the DC cables: \(0{,}20\ \mathrm{\Omega}\)

Calculation of DC losses

Joule losses:

\[ P_{\mathsf{losses,DC}} = R \cdot I^{2} = 0.20 \times 125^{2} = 3125\ \mathrm{W} \]

Expression as a percentage

\[ \%\ \mathsf{DC\ losses} = \frac{3125}{100\,000} \times 100 = 3{,}1\ \% \]

🧠 Interpretation

  • DC losses are expressed as a percentage of the PV array’s rated power
  • Here, 3.1% is a relatively high value
  • This may be due to:
    • excessively long cables,
    • insufficient cross-sectional area,
    • high currents in the strings.

👉 In PV design, the goal is generally ≤ 1 to 2% DC losses.


2. Ohmic losses on the AC side

After the inverter, we move to the alternating current side:

  • AC cables between the inverter and the delivery point,
  • possibly one or more AC transformers (LV/MV, MV/HV).

In this AC section, there are:

a. Ohmic losses in AC cables

  • the cables have a resistance \(R_{\mathsf{AC}}\),
  • the losses are:
\[ P_{\mathsf{losses,AC\ cables}} = R_{\mathsf{AC}} \cdot I_{\mathsf{AC}}^{2} \]

b. Copper losses in transformers

  • in the windings (wire resistance),
  • same law:
\[ P_{\mathsf{copper,transfo}} \approx R_{\mathsf{windings}} \cdot I_{\mathsf{AC}}^{2} \]

c. Iron losses in the transformer (non-ohmic, but worth noting)

  • in the magnetic core,
  • present as soon as the transformer is energized, even at low load,
  • they are often modeled as "nearly constant" power.

On the AC side, voltages are generally higher (\(400\ \mathrm{V}\), \(20\ \mathrm{kV}\), etc.) than on the DC side; therefore, for the same power, currents are lower → ohmic losses can be more limited, even over long distances.


Example 2 – AC losses expressed as a percentage of the inverter/substation’s rated power

Data:

  • Inverter: \(100\ \mathrm{kVA}\)
  • Three-phase AC voltage: \(400\ \mathrm{V}\)
  • Power factor ≈ 1
  • AC rated power: \(100\ \mathrm{kW}\)
  • AC current:
\[ I = \frac{P}{\sqrt{3}\,U} = \frac{100\,000}{\sqrt{3} \times 400} \approx 144\ \mathrm{A} \]
  • Equivalent resistance of AC cables (per phase): \(0{,}05\ \mathrm{\Omega}\)
  • Number of phases: 3

Calculation of AC losses (copper)

\[ P_{\mathsf{losses,AC}} = 3 \times R \times I^{2} \]
\[ P_{\mathsf{losses,AC}} = 3 \times 0.05 \times 144^{2} \approx 3110\ \mathrm{W} \]

Expression as a percentage

\[ \%\ \mathsf{AC\ losses} = \frac{3110}{100\,000} \times 100 = 3.1\ \% \]

🧠 Interpretation

  • AC losses are relative to the rated power of the inverter or substation
  • Even though the currents are lower than in DC,
  • the losses remain significant because:
    • the power levels are high,
    • there may be transformers (copper losses + iron losses).

👉 In practice:

  • AC cables: often < 1%
  • HV substation: an additional 1 to 2% possible

3. Comparing DC and AC losses in a PV system

We can summarize as follows:

  • DC (modules → inverter)
    • generally higher currents,
    • cables sometimes long with small cross-sections (strings),
    • significant losses if cross-sections are too small or distances are long,
    • often expressed as a % of DC losses relative to the PV array’s rated power.
  • AC (inverter → grid)
    • higher voltages, therefore lower currents for the same power,
    • AC cables + transformers (copper losses + iron losses),
    • ohmic losses sometimes lower in %, but at high power levels, so must be monitored,
    • often expressed as a % of AC losses at the inverter/substation’s rated power.

Key point: DC ohmic losses

  • Between modules and inverter (string cables, enclosures, DC connections).
  • Modeled by an equivalent DC resistance and the law:

    • \(P_{\mathsf{losses,DC}} = R_{\mathsf{DC}} \cdot I_{\mathsf{DC}}^{2}\)
  • Losses increase with the square of the current and cable lengths.

AC Ohmic Losses

  • Between inverter and grid (AC cables + transformer windings).
  • Copper losses: same ohmic laws \(R \cdot I^{2}\).
  • Iron losses in transformers: present as soon as the transformer is energized.

DC vs. AC Comparison

  • DC side: higher currents → significant losses in cables if the cross-section is undersized.
  • AC side: lower currents (higher voltage), but the presence of transformers and sometimes long cables.
  • In practice, we often refer to "% DC losses" and "% AC losses" at full power, but the annual energy lost is lower than the sum of these percentages.