[Do Dual Coils Affect TC Accuracy?]

Jalcide said:

I guess it's all moot now. I just ran the math. 

I was wrong in a key assumption.

The rise and fall of twice the ohms in the parallel circuit does happen, like I suspected, but the math of the TCR formula, when paired with parallel resistor math, works out to keep the temperature the same in each coil, between dual and single.

Here's the math for a .15-ohm Ni200 build, with a .1 ohm rise and drop in resistance:

TCR formula:
(Rref + rD) = Rref [1 + a(T - Tref)]

 

R = resistance at temp T

Rref = resistance at temp Tref

a = TCR

T = temp

Tref = ref temp for the TCR

rD = resistance delta (rise)

Single coil build TCR formula:
(.15 + .1) = .15[1 + .006(T - 20)]
A .1-ohm rise and fall equates to 131.1C, or 267.9F.

For the dual build to read at .15-ohm, each coil needs to be .3-ohm.
So the same .1-ohm rise of the circuit, means each resistor inside it must rise .2 ohm.
Dual coil build TCR formula:
(.3 + .2) = .3[1 + .006(T - 20)]
A .2-ohm rise and fall required for each coil equates to 131.1C, or 267.9F


I still don't fully have my head wrapped around why this is true. But I'm going to accept the math, for now.

I think the things are simpler than that. If you apply voltage and current to a resistor the latter will heat up. This heat depends on the power dissipated by the resistance in Watts and the time in seconds during which you hold the trigger fire.

For a resistance of 1 Ohm and traversed by a current of 5 A the power dissipated will be 25 W (P = R * i ^ 2). If I hold the trigger for 2 sec, the dissipated energy will be 50 Joule (E = W * t). It is the energy applied to the wick (without taking into account the losses).

Now, if I apply a current of 5 A on two resistors of a value of 2 Ohm each, I will have the same power on the whole circuit that is to say 25 W (2 Ohm in parallel = 1 Ohm) . The question is: what is the power dissipated by one resistance and therefore applied on a wick?

In a circuit in parallel and with two equal resistances the current will be divided by two. We thus have 2.5 A by resistance and therefore a power P = 2 * 2.5 ^ 2 = 12.5 W. In conclusion the dissipated power is divided in two as well. Each resistance therefore delivers only half of the total power. For a trigger held for 2 sec, the energy applied on each wick is 25 Joule! The Half !

It is for this reason that for a double coil installation it is necessary to double the power or the temperature to maintain the same energy applied per wick.

It will be interesting, if Evolv, adds an option in the Escribe software that allows the user to specify the number of parallel coil used.

Sorry for my english ! Google Translate.

[dna 250 low usb current]

Hi,

I’ve got the same problem on my DNA. So, I plugged my DNA on a laboratory power supply with a fixed voltage and current. (voltage = 5v, current = 2A). Guess what? Same problem, my DNA only consume 0.31 A (Confirmed by the PSU)! So I've tested with all my usb cable. The result: my Triade 250 consumes 1.98 A (confirmed by PSU), only with the little tiny usb cable suppled with it !

 

So, my advise is to use the little cable with a tablet charger (often supply a current of 2.1 A). I hop this helps you.

 

Sorry for my english (it’s note my native language).