When Rogers high-frequency circuit boards, such as RT-duroid laminates, carry direct current (DC) or radio frequency (RF) current, the temperature rise depends on factors such as conductor resistance, material properties, current flow, and thermal management.

1. Factors Influencing Temperature Rise

1. Conductor Properties:
   • Thickness and Width: Wider and thicker traces reduce resistance.
   • Material Conductivity: High-conductivity materials like copper lower resistive heating.
2. Current Characteristics:
   • Direct Current (DC): Temperature rise is primarily due to I2RI^2RI2R (Joule heating).
   • Radio Frequency (RF): Skin effect and dielectric losses contribute to heating.
3. Substrate Thermal Properties:
   • Thermal Conductivity: Rogers materials like RT-duroid 5870 (~0.2 W/m·K) and 5880 (~0.2 W/m·K) impact heat dissipation.
   • Dielectric Loss Tangent: High tan⁡δ\tan \deltatanδ materials produce more dielectric heating at RF frequencies.
4. Environmental and Design Factors:
   • Ambient Temperature: Higher ambient temperatures reduce thermal headroom.
   • Thermal Management: Heat sinks, thermal vias, and air circulation mitigate temperature rise.

2. Temperature Rise Calculation

For DC Currents:

The temperature rise due to resistive heating can be estimated using the formula:P=I2RP = I^2RP=I2R

Where:

• PPP = Power dissipated (W)
• III = Current (A)
• RRR = Resistance of the trace (Ω\OmegaΩ)

The resistance of a trace is given by:R=ρ⋅LAR = \frac{\rho \cdot L}{A}R=Aρ⋅L​

Where:

• ρ\rhoρ = Resistivity of copper (1.68×10−8 Ω⋅m1.68 \times 10^{-8} \, \Omega \cdot \text{m}1.68×10−8Ω⋅m)
• LLL = Trace length (m)
• AAA = Cross-sectional area of the trace (width×thickness\text{width} \times \text{thickness}width×thickness, m²)

The temperature rise is then:ΔT=Ph⋅As\Delta T = \frac{P}{h \cdot A_s}ΔT=h⋅As​P​

Where:

• hhh = Heat transfer coefficient (W/m²·K)
• AsA_sAs​ = Surface area for heat dissipation (m²)

For RF Currents:

The total power loss includes resistive and dielectric losses:

1. Resistive Loss (Skin Effect):

At RF frequencies, current flows on the surface of the conductor. The effective resistance increases with frequency:RRF=1σ⋅δ⋅WR_\text{RF} = \frac{1}{\sigma \cdot \delta \cdot W}RRF​=σ⋅δ⋅W1​

Where:

• σ\sigmaσ = Conductivity of copper
• δ\deltaδ = Skin depth (2ωμσ\sqrt{\frac{2}{\omega \mu \sigma}}ωμσ2​​)
• WWW = Trace width

2. Dielectric Loss:

Dielectric heating is given by:Pd=12⋅ω⋅ϵr⋅tan⁡δ⋅E2P_d = \frac{1}{2} \cdot \omega \cdot \epsilon_r \cdot \tan \delta \cdot E^2Pd​=21​⋅ω⋅ϵr​⋅tanδ⋅E2

Where:

• ω\omegaω = Angular frequency (2πf2 \pi f2πf)
• ϵr\epsilon_rϵr​ = Relative permittivity
• tan⁡δ\tan \deltatanδ = Loss tangent
• EEE = Electric field strength

3. Practical Design Considerations

1. Thermal Vias and Ground Planes:
   Use thermal vias to transfer heat to a ground plane or heat sink.
2. Copper Thickness and Trace Width:
   Increase trace dimensions to reduce resistive losses. Common copper thicknesses are 1 oz/ft² (35 µm) or 2 oz/ft² (70 µm).
3. High Thermal Conductivity Materials:
   For demanding applications, select Rogers laminates with higher thermal conductivities.
4. Impedance Control for RF Traces:
   Ensure proper characteristic impedance to minimize signal reflection and losses.
5. Simulation Tools:
   Use thermal and RF simulation tools (e.g., Ansys HFSS, COMSOL Multiphysics) to model temperature rise accurately.

4. Example Estimation:

For a 10 cm copper trace carrying 2 A DC:

• Copper thickness = 35 µm
• Width = 1 mm
• Ambient temperature = 25°C

1. Calculate Resistance:R=ρ⋅LA=1.68×10−8⋅0.10.001⋅0.000035=0.048 ΩR = \frac{\rho \cdot L}{A} = \frac{1.68 \times 10^{-8} \cdot 0.1}{0.001 \cdot 0.000035} = 0.048 \, \OmegaR=Aρ⋅L​=0.001⋅0.0000351.68×10−8⋅0.1​=0.048Ω
2. Power Dissipated:P=I2R=22⋅0.048=0.192 WP = I^2R = 2^2 \cdot 0.048 = 0.192 \, \text{W}P=I2R=22⋅0.048=0.192W
3. Temperature Rise: Assuming h=10 W/m2⋅Kh = 10 \, \text{W/m}^2\cdot\text{K}h=10W/m2⋅K,ΔT=0.19210⋅(0.1×0.001)=19.2 °C\Delta T = \frac{0.192}{10 \cdot (0.1 \times 0.001)} = 19.2 \, \text{°C}ΔT=10⋅(0.1×0.001)0.192​=19.2°C

Final board temperature = 25+19.2=44.2 °C25 + 19.2 = 44.2 \, \text{°C}25+19.2=44.2°C.

Would you like assistance with more detailed simulations or specific RF current scenarios?