I have a microphone cable with 2 conductors in shielded twisted pair in the center. Material is copper. Each conductor is made of 28 strands x 0.10 mm thick.

They're covered in braided shield with 84 x 0.10 mm.

Outer diameter of the cable is 6 mm.

What is the AWG of the cable?

(28*0.10)*2+84*0.10 OD:6 mm

  • 1
    \$\begingroup\$ It doesn't work that way. Those are separate conductors, each will have its own rating. \$\endgroup\$
    – JRE
    Nov 15, 2016 at 15:26
  • \$\begingroup\$ There is no "AWG number" for cable. AWG is a scale for thickness of wire, and you have 28 strands of 38AWG wire in each conductor. What are you actually trying to calculate, that you think you need the AWG of this cable for? \$\endgroup\$
    – Jack B
    Nov 15, 2016 at 15:28
  • \$\begingroup\$ because this one says it's 24 gauge. sweetwater.com/store/detail/2condshldm \$\endgroup\$
    – Moe Sweet
    Nov 15, 2016 at 15:42
  • \$\begingroup\$ and what does 24 gauge mean in that cable? \$\endgroup\$
    – Moe Sweet
    Nov 15, 2016 at 15:58
  • \$\begingroup\$ ... It means that each conductor is 24AWG... \$\endgroup\$ Nov 15, 2016 at 15:58

2 Answers 2


From your comment:

How do you calculate 28*0.10mm into 38?

You don't. A single 0.10mm strand is AWG38.

The formula is based on AWG36 wire being exactly 0.005" (5 mils) in diameter. All other values are based on an exponential curve:

$$diam = 5\text{ mils} \cdot 92^{\frac{36 - AWG}{39}}$$

Solving for AWG as a function of diameter gives:

$$AWG = 36 - 39\cdot log_{92} \frac{diam}{5\text{ mils}}$$

28 strands of AWG38 has the same cross-sectional area as AWG24.


In order to determine the effective AWG of the conductor you must first determine the cross-sectional area of a single strand, then multiply that by 28 (since there are 28 strands in there), and then consult a copper wire table where the total area can be converted to AWG.

Here's how to do it:

For the cross-sectional are of a single strand,

$$ A = \pi\ r^2 = 3.14\times 0.05mm^2 = 0.00785 \text{ square millimeter}$$

For the bundle,

$$ A = 28\times 0.00785 \text{ square millimeter}= 0.2198 \text{ square millimeter.} $$

Then, after consulting a copper wire table:

enter image description here It appears that the conductors are closer to AWG 24 than to AWG 23, which should make very little difference for feed from a microphone.

  • \$\begingroup\$ Actually, your table contains an error. The "Area" column figures are actually in units of "circular mm" (diameter squared), rather than true square mm (pi * radius squared). Note that the value you calculated for 0.1mm (approximately AWG38) does not correspond to the value in the table -- this should have been a clue! \$\endgroup\$
    – Dave Tweed
    Nov 15, 2016 at 22:07
  • \$\begingroup\$ @ Dave Tweed: Thanks for the reality check, and I'll update the table with something more reliable, but I don't understand why you chose to include the pejorative: " this should have been a clue!" in your comment. Can you elaborate? \$\endgroup\$
    – EM Fields
    Nov 16, 2016 at 1:00
  • \$\begingroup\$ An engineer always cross-checks his calculations. Didn't you glance down the table to see whether the entry closest to 0.1mm DIA had the same area you just calculated? \$\endgroup\$
    – Dave Tweed
    Nov 16, 2016 at 4:55
  • \$\begingroup\$ I didn't check the area of the 0.1mm diameter entry in the table because it was irrelevant since I calculated it. However, I should have checked the correspondence between AWG and the target area. I still don't understand the reason for your patronizing remark, which is what I asked you to elaborate on. \$\endgroup\$
    – EM Fields
    Nov 16, 2016 at 9:22
  • \$\begingroup\$ I was just trying to help you avoid looking foolish. Your credibility here is based on the accuracy of the information you present, and using data from some random table you found on the Internet without checking it first is a rookie mistake that an experienced engineer such as yourself shouldn't be making. So yes, I'm going to "tweak" you a bit about it. But this conversation is not helping your cause at all. Saying that the simple check that I proposed is "irrelevant" hints at a level of overconfidence that calls into question the reliability of the data provided in your other answers, too. \$\endgroup\$
    – Dave Tweed
    Nov 16, 2016 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.