Sorry, Adafruit. My Bad

Gary Schafer, 26 February 2015

I bought a 20x4 LCD from Adafruit the other day. It arrived three nights ago. I plugged it into a circuit that I've used on several occasions to power a 16x2 LCD. Well, when I plugged this LCD into the circuit, I discovered that the third row would only show the top-half of the pixels. Yes, I was annoyed.

Then, when I contacted Adafruit to tell them I had a defective product, I got even more annoyed. Here's why. I assumed that I would have to send the LCD back. I tell them "defective part", they give me an RMA (return merchandise authorization?), I send it back, and they send me a new one. Pretty straightforward. I've had to do this with several items I've ordered over the Internet. The product arrives, it's broken or defective, then I have to take the time to re-pack and then pay to send it back. Typically, the cost of sending it back is much less than buying a new one. It's also the risk I assume when I buy stuff from far away and have it shipped to my house.

So, with my assumption that I had to send it back, I was a bit... miffed... when I sent my e-mail requesting instructions on returning the LCD, and they responded that I had to post my problem on their tech forum. WTF?!?! The LCD is defective. How hard is this to understand? You'll find it out for yourself when I return it. Just let me exchange this for a working unit. C'mon!

I posted a short, curt message on the Adafruit forum. The response was also brief, and perhaps curt.

Could you post clear, detailed pictures of your board and the connections to it?

By now, I've almost got steam coming out of my ears. I have a defective LCD! Just let me exchange it! Still, I took some pics showing the defective LCD, then posted them on the forum.

This morning, I saw another post on the forum stating,

Looks defective to me. Email with a link to this thread for a replacement display.

That's when it hit me. My initial assumption was wrong. Adafruit is not expecting me to send the defective LCD back; they're simply going to send me a working unit. No need for any more of my time. No need for me to pack and ship this thing back. They simply wanted confirmation that I did, indeed, have a defective display because they now knew I wasn't simply trying to get free stuff out of them.

Yeah, I felt like a dick. And an idiotic one, at that.

So, Adafruit, if you sensed annoyance on my part during our exchanges, that's because I can be a flaming asshole when I don't understand things. I didn't understand how this exchange was going to be. Sorry about that.

Calculating Radiation Exposure - The Other Way

Gary Schafer, 4 February 2015

I think I now understand how the different values of radiation exposure are calculated. That's the interesting part. They're calculated, not measured. For example, last month, I posted a link to a material data safety sheet for cobalt-60. That sheet stated that the dose rate was 370 uSv/h/GBq, or "microsieverts per hour per gigabecquerel". I first assumed that this was a measured value. I thought that they'd taken a known quantity of cobalt-60, set a radiation meter near it, and measured it. Come to find out, probably not.

It seems that this value was actually calculated, but not using the same equation as from last month's post. Instead, it uses a different algorithm. It's not something that is easily made into a single equation. It requires a couple of steps to arrive at the proper answer.

The basic equation used for the Oak Ridge tabulation is as follows:

ln D(E) = A + B(ln(E)) + C(ln(E)2) + F(ln(E)3)

The values of the constants A, B, C and F above are shown in Table 1, below.

Table 1: Gamma-Ray Flux to Dose Rate Conversion Factors
Polynomial Coefficients in Analytic Form
Photon Energy (MeV)ABCF
0.01 - 0.03-20.477-1.7454
0.03 - 0.5-13.626-0.57117-1.0954-0.24897
0.5 - 5.0-13.1330.72008-0.033603
5.0 - 15.0-12.7910.283090.10873

According to the worksheet, cobalt-60 expels three gamma rays with each decay. These gamma rays have energies of 1333, 1173 and 693.8 keV. The probability of each a decay generating these different gamma rays is 1.0, 1.0, and 0.0002, respectively. For simplicity sake, I'm going to not use the 693.8 keV gamma ray for the remaining calculations.

Using Table 1, both the 1333 and 1173 keV gamma rays use the third line, "0.5 - 5.0". This means that the constants are as follows:

  • A = -13.133
  • B = 0.72008
  • C = -0.033603
  • F = not used

This equation has to be calculated for each, individual gamma ray. For the first, higher energy gamma ray from cobalt-60 (1.333 MeV), the overall equation becomes:

ln D(1.333) = -13.133 + (0.72008)(ln(1.333)) + (-0.033603)(ln(1.333)2)
= -13.133 + 0.206974 - 2.776185e-3
= -12.9288

Then, for the other gamma ray (1.173 MeV):

ln D(1.173) = -13.133 + (0.72008)(ln(1.173)) + (-0.033603)(ln(1.173)2)
= -13.133 + 0.114899 - 8.55561e-4
= -13.01896

The overall exposure rate is given by:

Γ = [ Σ Si D(Ei) ] * 1013 / (4 π R2)


  • Γ = the total exposure rate, in uSv/h/GBq.
  • Si = the probability of each gamma ray being released per decay.
  • Ei = the energy of each gamma ray, MeV.
  • D(Ei) = the exposure rate, calculated from the polynomial coefficients table.
  • R = distance to point of exposure from sample, in cm.

Using the above equation:

D(1.333) = exp(ln(D(1.333))) = 2.427133e-6
D(1.173) = exp(ln(D(1.173))) = 2.217885e-6
R = 100 cm (1 meter)
S1,2 = 1
Γ = [ Σ Si D(Ei) ] * 1013 / (4 π R2)
= [(1)(2.427133e-6)+(1)(2.217885e-6)] * 1013 / (4 π (100)2)
= 369.6 uSv/h/GBq

In case you're curious where the "1013" came from in the equation above, that's the conversion factor to go from rem/h to uSv/h/GBq.

Note that this final answer is pretty close to the "370" from the MSDS, and it actually equals it if you round to the nearest one (369.6 ~= 370). So there you have it. This appears to be how they calculate the exposure rates for different radioactive elements.

Here's a Random Fact...