[TriEmbed] Power advice

Brian triembed at undecidedgames.net
Wed May 9 09:39:35 CDT 2018


Hi Craig,

Not knowing your background, here is a big fat dump of knowledge.

A cutting blade should have either a voltage or a wattage rating with 
it.  If it's a voltage rating, your job is easy; you just need a power 
supply capable of delivering $beaucoup amps at that voltage.

If it's a wattage rating, you'll need to do some measurements and math 
to figure out the correct voltage for your power supply.  Here are the 
equations you'll mainly be using:

E = IR (Ohm's law, potential (Volts) equals current (Amps) times 
resistance (Ohms))
P = EI (Power (Watts) equals potential times current)

Say your blade is rated 100 Watts.  Then you know that E * I must equal 
100.  So, you could build a power supply capable of delivering 1 Amp at 
100 Volts, or 100 Amps at 1 Volt, and have the correct amount of power. 
  But which one is right?  That's where the resistance of the blade 
comes into play.

Technically you need to know the at-temperature resistance, but the cold 
resistance will do and is a lot safer to measure!

Let's say your blade measures 1.5 Ohms cold.  Back to the equations:

E = I * 1.5
and
100 = E * I

Now we do some algebra:

100 = (I * 1.5) * I
100 = 1.5 * I^2
I^2 = 100 / 1.5
I = sqrt(66.67)
I = 8.16 Amps

So we have a current; to get that current through a 1.5 Ohm resistor, we 
need E = (8.16) * (1.5) = 12.24 Volts.

Checking our math, we do P = EI = 8.16 Amps * 12.24 Volts = 99.9 Watts.

So for our example, you need a 12-volt power supply capable of 
delivering 8 Amps, which will run your knife just slightly cooler than 
its rating.

Some loose ends:

- The knife probably has a positive thermal coefficient (PTC), meaning 
its resistance increases as it gets hot.  The calculations done using 
cold resistance mean that it will run slightly cooler than designed.  To 
eke out every last watt, you'll need the hot resistance in the equations 
above.

- The simplest approach is an unregulated linear power supply, but that 
means a pretty heavy chunk of iron.

- Depending on the voltage requirements, a castoff computer power supply 
may be able to suit your needs.

- The power supply's ratings affect the device this way:

   Volts - how hot does it get
   Watts - how fast does it get hot (and how well does it stay hot)

- The safest way to determine hot resistance is to measure the current 
through, and voltage across, a running blade.  Note that "safest" here 
means "marginally less likely to lead to severe burns."

- You may find that the best convergence of ease vs cost is to buy a 
cutter off the shelf and build a jig to hold the blade how you need it held.

- The wires connecting power supply to blade need to be thick and short 
as possible.  Since the blade's resistance is (probably) quite small, 
the resistance of the wires start to become a significant factor.  Wire 
resistance is proportional to length and inversely proportional to 
cross-sectional area (i.e. shorter and thicker means less resistance).

- AC/DC is nearly* irrelevant for this application.  Use whatever is 
handy.

- If you use AC, note that the voltage figures in the equations should 
be interpreted as RMS (root-mean-square) voltage; multiply RMS by 
sqrt(2) (approximately 1.4) to get peak-to-peak voltage, and make sure 
you know which one is which when you're measuring and reading labels.

Hope at least some of this was helpful!

-Brian


On 05/08/2018 07:57 PM, Craig Cook via TriEmbed wrote:
> Lets say I want to make a custom one of these:
> https://smile.amazon.com/gp/product/B00CP70IDU/ref=ox_sc_act_title_1?smid=A2IGZ09LPUKLVH&psc=1
>
> The blade gets up to 1200 F in a few seconds.
>
> What do I need to investigate to heat a blade safely like that? (I can
> buy that blade for ~$23, or open to other blade ideas)
>
> I know I could buy one of these machines, I am interested in changing
> the form factor though.  Maybe add a motion sensor, when no motion, shut
> the system off.
>
> I can borrow one to get measurements if that helps.
>
> Thanks
>
> Craig
>
>
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