When selecting wire diameter we usually lok for cross reference tables where you can find recommended wire diameter for maximum current drive. But sometimes may be more useful to calculate by formula than look in to the tables. This way you can have more accurate results. There is nothing new just simple physics.

Wire resistance (Ω)is calculated as follows:

R = ρ·l/S or R = (1.27·ρ·l )/d²;

Where ρ – resistivity of material (Ω·m), found in tables, l – wire length (m), S – cross selection area(m²), d - wire diameter (m). According to this we can calculate wire length, when other parameters are known:

l = R·S/ρ or l = (0.785·R·d²)/ρ

Cross selection area can by found as follows: S=0.785·d^2.

Wire resistance depends on temperature. Lets say R₂ is resistance for temperature t₂ and R₁ is for t₁(usually t1 = 18°C), then:

R₂=R₁·(1+α·(t₂-t₁))

Where α – temperature coefficient (K^-1).

Lets say that maximum current for cross selection area is marked as Δ (A/mm²), then maximal current can be found I=0.785·Δ·d². Needed diameter can be found by formula(√ - square root symbol):

d=√((1.27·I)/Δ)

So if allowed load is Δ=2A/mm², then d=0.8√I.

If diameter is les than 0.2mm, then melting current can be calculated by formula:

I_m=(d-0.005)/k

where k – coefficient (k_copper=0.034; k_nickel=0.07; k_iron=0.127). Then diameter would be : d=k·I_m+0.005. This formula may be used when calculating fuses.

For example, if we need fuse that melts at 5A, then copper wire diameter would be d_5A=0.034·5+0.005=0.175mm.

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This entry was posted on Monday, February 5th, 2007 at 4:59 pm and is filed under Electronics Tutorial. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.