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Typical Bus Bars Design Example

Electrical Bus Bar Requirements:
Current Carrying : 300 Amps operating current @ 30°C max temp rise.
Application Dependent Parameters: Minimum Voltage drop, Max Capacitance, Minimum inductance.

Mechanical and Physical Requirements:
Product Configuration: Two Layer, Rigid Epoxy Glass Board, Edge Potting;
Shape: Planar
Dimensions: 24" long by 1.5" wide max
Materials: Copper alloy 110, Mylar Tedlar Inner Insulation
Termination Method: Threaded Fastener
Mounting Method: Insulated thru holes
Environment: High humidity environment, minimum vibration
Design Parameter
Design: Formulas and Tables Used
Results
Cross Sectional Area A = 300 x l x [1 + 0.75(N-1)]
formula (2.7)
A = 0.097 sq in
   l = 300 amps
 N = 2 layers
A = 300 x 300 x [1 + 0.75(2-1)]
Conductor Width (w) & Thickness (t)

w = A / t
formula (2.8)

t = 0.093"
w = 1.032"
 

t = selected thickness values from the available std thickness to get the maximum w / t ratio and practical to the application
A = 0.097 sq in

Thickness (t) 0.125" 0.093" 0.062"
Width (w) 0.776" 1.043" 1.564"
w /t Ratio 6.20 11.21 25.23

The width requirement is 1.5" max therefore 11.21 (0.93"/1.043") is the max w / t ratio practical to the application

(Optional method)
Use the ampacity table and select the combination of w & t practical to the application and which will yield the lowest inductance (max w / t ratio)

t = 0.093"
w = 1.040"
Resistance
R = ρ / A ohms/foot
formula (2.1)
R = 0.084 Milli Ω / foot @ 20°C
  ρ = 8.1 (Ω • sq mil/ft) at Ambient Temp. 20°C
table 3
A = 96,750 sq mil
R = 8.1 / 96,750 Ω/foot
R2 = R1 [1 + α (T2-T1)] ohms/foot
formula (2.2)
R2 = 1.074 Milli Ω / foot @ 50°C
  R1 = 0.084 Milli Ohms as calculated above
α = .393 from table 3
(T1-T2) = 30°C
R2 = 0.084 [(1 + 0.393(30)]
Voltage Drop ΔV = R x l x l
formula (2.3)
ΔV = 48 MIlli volts @ 20°C
  l (Conductor length) = 2 ft
R = 0.084 Milli Ohm / foot at ambient temperature
l = 300 Amps
ΔV = 0.084 * 2 * 300 = 50.4 Milli Volts at ambient temperature
  R2 = 1.074 Mili Ohm / foot at the 50°C (the max allowed temperature) ΔV = 0.644 Volts @ 50°C
ΔV = 1.074 * 2 * 300
(if this voltage drop is too large, increase cross sectional area)
Capacitance C = 0.224 (k)(w)(l)/d picofards
formula (2.4)
C = 0.0095 microfarads
  K (Dielectric constant Mylar tedlar) = 8.5 from table 4
w (width) = 1.040"
l (length) = 24"
d (dielectric thickness) = 0.005"
C = (0.224)(8.5)(1.040)(24)/.005
Inductance L = 31.9 (l) (d/w) nano Henrys
formula (2.5)
3.68 nano Henrys
  l = 24"
d = .005
w = 1.040
L = 31.9 (24) (.005/1.040)