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# I don't get it. ... am I doing wrong?

Discussion in 'Modding Forum' started by MidnighToker, Aug 9, 2009.

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1. ### MidnighToker Super MemberVerified MemberECF Veteran

Jun 27, 2009
NC
OK, so I posted the result of my first box mod. Was trying to run 2 x 14500 in series, and end up around the 5v mark. A few people here helped me out with the calculations and I wound up with a 10W 1ohm resistor. End result...atty fried in a split second from the 8.3v that it was getting.

Scrapped that box and built a parallel one until I could regroup. Parallel box lasted all of two days before the Radiosh*t switch failed. Of course I had epoxied the switch in....another box to the scrap pile.

Ordered a bunch of different resistors after seeing a thread on here about a mod running the same setup with a 2W 2ohm resistor and ending up with 4.7v. Just built mine and I get 8.1v (I remembered to check before I connected an atty) both before and after the resistor .

Aren't I breaking some law of physics here?!?!

2. ### kinabaloo Vaping MasterECF Veteran

1 ohm is far too little.

The current through the atty will be at least 1A at the desired 5V.

So you need to drop about 3v at 1A = 3 ohms. Betwee 2 and 3 ohms is what you need.

~~

With no atty and only the meter, the current will be tiny so virtually no voltage drop; the resistor will do nothing on it's own.

Another way to think about it is that the resistor and atty (about 3 ohms) divide the voltage. If the resistor value is equal to that of the atty, the voltage across the atty will be half that of the battery/ies.

3. ### MidnighToker Super MemberVerified MemberECF Veteran

Jun 27, 2009
NC
So I should be OK to attach the atty with a 2 ohm?...assuming that the calculations were correct (atty is an 801 and measuring 3.7 ohms).

4. 8.1v @ 3.7 ohms gives you 2.2a to the atomizer (I=V/R) - I would *think* that it should handle that.

Lets run some numbers and see what we can figure out - don't forget, we have to calculate everything once the resistor is in place, because it affects amperage through the circuit as well as voltage drop across the resistor.

A 2 ohm resistor before the atomizer makes the total resistance 5.7ohms. Amperage would be voltage divided by resistance (8.1v / 5.7o), which equals 1.4a.

Voltage dropped by a component is calculated by multiplying the current (I) by the resistance (R) of the component. In a series circuit, current stays the same through all components, so your formula is I * R(component) = Vdrop, or voltage dropped through the component. Power dissipation is current squared (I^2) times resistance, to get watts.

So, we calculate for the full circuit with the resistor in place:
Amperage: 8.1v / 5.7o = 1.4a.
Vdrop (resistor): 1.4a*2 = 2.8v dropped (8.1v - 2.2v = 5.9v at atomizer)
Power dissipation (resistor): 1.4a^2*2 = 4w

1 ohm resistor would make the total resistance 4.7ohms. Amperage would be (8.1v / 4.7o) 1.7a.

Calculate for full circuit:
Amperage: 8.1v / 4.7o = 1.7a
Vdrop (resistor): 1.7a*1o = 1.7v dropped (8.1v - 1.7v = 6.4v at atomizer)
Power dissipation (resistor): 1.7a^2*1 = 3w

Looks like either one will get you 'in range'...but 2ohms looks to be 'safer' for the atty.

5. ### MidnighToker Super MemberVerified MemberECF Veteran

Jun 27, 2009
NC
Thanks guys. Worked like a charm. Haven't had a chance to measure with full batts yet, but slightly drained ones are giving me 4.1v.

6. The acid test is going to be with fully charged batteries. Did you use the 2ohm or 1ohm?

That's going to be my next mod - 2x 14500 in series. Just have to get some resistors.

7. ### SmokinScott Super MemberECF Veteran

Apr 21, 2009
Acton MA, USA
What's the advantage to using resistors over the 5V voltage regulator?

Is the heat from the regulator too much for battery box plastic (or epoxy glue)?

8. ### kinabaloo Vaping MasterECF Veteran

If the batteries upply 8v, then with a 2ohm resistor the voltage across the atty would be:

8 * (3.7 / (3.7 + 2))

9. ??? I'm not sure where you're at with this, kina - can you lay it out for me?

10. ### kinabaloo Vaping MasterECF Veteran

The two resistors divide the voltage.

The atty will have 8 * (3.7 / (3.7 + 2)) volts (about 65% of the voltage)

and the resistor 8 * (2 / (3.7 + 2)) volts (about 35% of the voltage)

~~

Atty v : 8 * (3.7 / (3.7 + 2)) = 8 * 3.7 / 5.7 = 8 * 0.65 = 5.2v

Or to dip below 5v across atty, a 2.2 ohm res would be good.

Do the same thing for voltages other than 8v. This method involves no current values.

However, the current would be 8/5.7 which is same as 5.2/3.7 which is the same as (8-5.2)/2 = 2.8 / 2

= 1.4 A

About the most one can get away with.

~~

The power consumption would be 1.4 x 1.4 x (3.7 + 2) = 1.96 x 5.7
which is the same as 8 x 1.4
= 11.2 W

The useful power is 0.65 x 11.2 = 7.3 W

With just one lithium and no resistor, the useful power would be 4 x (4 / 3.7) = 4.3 W

~~

In short, although atty res veries a bit, if use two lithiums in series, a resistor of 2 ohms or 2.2 ohms (a standard value) will run the atty at about 5v, similar to a USB passthrough.

~~

A more detailed explanation is here: http://www.tji-java-ide.com/e-cigarette/index.php?option=com_content&view=article&id=169:the-resistor-to-use-with-2-lithiums-in-series&catid=42:mods&Itemid=59

11. Ok...but he's using a single atomizer now and running a resistor to cut the voltage - that's why I was confused.

12. ### kinabaloo Vaping MasterECF Veteran

Yes, one atomiser and one resistor. The atomiser is a resistor too!

So one can model the system as two series connected resistors.

13. Ok, with the amplification, I see where you were coming from - a different way of deriving the voltage at the atomizer.

14. ### kinabaloo Vaping MasterECF Veteran

"The atty will have 8 * (3.7 / (3.7 + 2)) volts (about 65% of the voltage)
and the resistor 8 * (2 / (3.7 + 2)) volts (about 35% of the voltage)"

The ratios of the resistances translate to the split in the voltage.

~~

Alternatively: calculate current fom the total resistance:
current will be V/R (from V=IxR) = 8/(2+3.7) = 1.4A

So the voltage across the atty will be 1.4x3.7=5.2v
and across the resistor will be 1.4x2=2.8v

5.2+2.8=8v

The reality is that we need a certain power to generate heat. I still don't know what this exactly is for each atty and the normal air flow, but lets go with the common assumption of 5V and 1A. So that is about a 5W heater (coil).

We don't know the resistance of a hot coil. It is probably 10% higher then at room temp. (Mogul also has some measurements)

We need about 0.5A to heat a 38ga NiCr to about 300degC with no airflow, so that seems about right to bump it up to 1A with airflow.

The voltage is totally meaningless, except that we are directly using batteries so that means the battery power can provide with a direct connection is P=V^2/R so for 5W it seems like 5V is the sweet spot.
If you directly apply a 3.7V battery you will only get half the heat at the coil.

It seems crazy to use a 1W resistor just to turn your batteries into wasted heat.

If you have 7.4V of supply, you can put in the same space as a giant power resistor a 555 timer and a mosfet and duty cycle, or a simple DC-DC converter like a LM317, or a simple current limiting circuit would be best. If you limited the current to 1A, you could plug it straight into your car lighter at 12V. A LM317 works up to 1.5A so it is almost build in overheat control... It will work with battery voltage 3V or greater above your desired regulated voltage. (so 7.4V becomes 3.4V.. so the LM317 may not be the best choice, but there are other DC-DC) Look up the "Current Limited 6-V Charger Circuit" That's what you want.

16. My point to the above is that a 2ohm 1W or 1ohm 10W power resistor is more expensive and larger then just using a LM317 and three resistors. You can even solder it directly to the LM317 legs like they do in chipamp designs. (google chipamp)

17. I can get a 1 ohm 10w resistor \$1 per at RS. I also don't have to do the other calculations necessary to figure resistor values for the chipamp you mention.

18. ### jonnyb Unregistered Supplier

Jun 12, 2009
redford,michigan
my friends and I use a 1ohm 2watt resistor and it limits the voltage to5.4v using 2 spiderfire 3.6v batts.

19. The calculation is
Vo= Vin (1 + R2/R1)
for 5.04V ==> R1 = 330, R2 = 1000

reuk.co.uk/LM317-Voltage-Calculator .htm

A LM317 costs \$0.50 to \$0.80 and also is at The Shack for \$2.30

The only problem is they aren't very efficient so about 50%, so the extra power wasted as heat is about the same as a power resistor. The benefit of using a LM317 is:
* less space (but it might need a heat sink)
* use any DC power supply (6.5V and above)
* same exact heating of atty even as battery voltage drops
* R2 can be adjustable resistor, you can tune how hot to make atty

-----
Of course if we are pretty sure 5.0V is the way to go, a 5V regulator would be perfect, you could throw any voltage at is and they have 0.5V dropout and 1A output current limited.

So a LM2940-5.0 would be perfect. You could use it with USB or a car battery (in the USB case, the atty would only see 4.5V, everything else is 5V).

Power wasted then is (Vout - Vin)*Iload or 0.5W for USB but 7W for a 12V battery (might need a heat sink)

They are \$1.83 at digikey.

20. So assuming about 1A in the heater.

1.8V IR drop with a 1ohm power resistor means your coil is using 1.8A.
(Also 5.4V/3ohm = 1.8A)

For the burn time of 5s, you are putting 3.2Watts through the resistor and 9.7Watts are used in the heater coil. A quarter of the battery life is wasted. (A voltage regulator gets down to 1.8Watts or 1/10 of battery life.)

(18mAh used per 5s burst which translates to 100-200 uses, does this seem close to your experience??)

Also, it seems like 1.8A is way too much. I'll bet it is glowing yellow/orange, unless the NiCr is thicker wire.