Battery and resistance calculations

[mnealtx]

Greetings, all - I was asked to make a post about calculating your resistor sizing in regards to battery mods.

I'm going to lay out the basic theory behind it, then give the formulas.

You're going to need to know 2 things to do these calculations: the voltage that your batteries are supplying when freshly charged, and the resistance of your atomizer.

For demonstration purposes, we will assume a 7.5v battery (2ea. 3.7v lithiums) and a 3ohm atomizer (average for 801/901 attys, from my reading here.)

Here's how you do the calculations:

Step 1: Figure the current (amperage) through the unmodified circuit.
Theory: Current equals voltage divided by resistance.
Formula: I=V/R

Result: 7.5v / 3ohms = 2.5amps

Step 2: Determine how much you want to reduce the voltage to the atomizer.
Theory: Voltage dropped through a component is equal to the current multiplied by the resistance of the component.
Formula: Vdrop = I * R

We know we want to bring the voltage down to around the 5v-6v range, so let's try some added resistance and see what we get.

Result (0.5 ohm added): 2.5a * 0.5ohm = 1.25v projected drop
7.5v - 1.25v takes us down to 6.25v.

Step 3: Now, you have to re-calculate to see what the circuit will actually run at with the added resistance, so let's plug in the new numbers.

I = V / R 7.5v / 3.5ohm = 2.14a
Vdrop = I * R 2.14a * 0.5ohm = 1.12v
True atomizer voltage = 7.5v - 1.12v = 6.38v

Might be a bit hot, still... let's see what 1ohm added gives us.

2.5a * 1ohm = 2.5v projected drop, which takes us down to 5v.

Let's recalculate:
Current = V / R 7.5V / 4ohm = 1.9amp
Vdrop = I * R 1.9a * 1ohm = 1.9Vdrop
True atomizer voltage = 7.5v - 1.9v = 5.6v

We're right in the middle of the 5v - 6v "sweet spot", and our amperage is below 2 amps - looks like we've got a winner with the 1 ohm resistor.

Step 4: Now, let's figure out the wattage for the resistor, so it doesn't burn up.

Power = current squared times resistance
Formula: P=I^2 * R
Result: P = (1.9*1.9)*1 = 3.6watts

So, your 1 ohm resistor would need to be at least 5 watts, although stepping up to 10 watts would make sure it runs nice and cool.

See, that wasn't THAT difficult after all, now was it?

 

[mnealtx]

Another thought, and something that might make it a bit easier for some - we all know that the atomizers need at least 1 amp to work well. The upper limit seems to be somewhere around 2.5 amp from my reading here, so I'll use that for the cutoff at the high end (using 0.5 ohm steps).

I've made a list of common voltages and the TOTAL resistance needed for 1 amp to 2.5 amp through the entire circuit. Total resistance is the atomizer and any added resistors.

For 8 volts (2 ea 3.7v batteries): 3 ohm (2.67a) - 8 ohm (1a)
For 6 volts (2 ea 3v batteries): 2.5 ohm (2.4a) - 6 ohm (1 a)
For 4 volts (1 ea 3.7v battery): 1.5 ohm (2.67a) - 4 ohms (1a)

Here's the kicker - you want to try to stay near the top end (maybe around 2a current) for fully charged batteries - here's why:

8v / 4ohm = 2a
6v / 4ohm = 1.5a
4v / 4ohm = 1a

So...as your batteries fade, you're still supplying good amperage to the atomizer.

 

[mnealtx]

Here's a link to Google Docs with the spreadsheet - everyone should be able to access.

Clicky

 

[a2dcovert]

Thanks for the information. I have a question for you. Suppose I wanted to drop 5v to 4v but I also wanted to limit the current to less than 1 amp? Are voltage regulators design to limit current as well a supply a specific voltage?

K

 

[mnealtx]

I *believe* they limit amperage as well, but you might want to check with Kinabaloo, mogur, or one of the other big brains - they're definitely more up to speed on that stuff and are probably your best bet for that type of info.

 

[a2dcovert]

Thanks for the response Mike. What I want to do is to install a USB powered circuit that will charge the 14500 battery while it is in use. I'm concerned about paralleling 2 DC power systems. I also want to use it with my PC USB port without worry that it will overload the PC's motherboard. I am using a nicostick right now with a USB cable wired directly and I am not using the battery. I'd like to use it with the battery. I don't have an amp meter to check it with but I know it is drawing more than 1 amp because it overpowers my USB wallwart that is rated at 1 amp.

Kevin

 

[Ralph Hilton]

This might be a bit easier ....
1. Divide the voltage desired by atomizer resistance to give final current = 5.6/3 = 1.87
2. Calculate the needed voltage drop = 7.5 - 5.6 = 1.9
3 Divide (2) by (1) = 1.9 /1.87 = 1.02

 

[a2dcovert]

Thanks Ralph, I going to start on the re-mod tomorrow. I'm going to use a Radio Shack regulator and try to get the circuit to supply charging for the 14500 and power the atomizer. I should have enough info now to calculate the voltage needed.

Kevin

 

[mnealtx]

This might be a bit easier ....
1. Divide the voltage desired by atomizer resistance to give final current = 5.6/3 = 1.87
2. Calculate the needed voltage drop = 7.5 - 5.6 = 1.9
3 Divide (2) by (1) = 1.9 /1.87 = 1.02

Nice - thanks, Ralph.

 

[breno]

Thanks for the information. I have a question for you. Suppose I wanted to drop 5v to 4v but I also wanted to limit the current to less than 1 amp? Are voltage regulators design to limit current as well a supply a specific voltage?

K

The current is determined by the voltage and the resistance ie I=V/R so I don't see how you can control both as one determines the other. (assuming the resitance is fixed, basically the atomiser resistance is more or less fixed).


So assuming the resistance is 2 ohms, then if you set the voltage to 4v then it would seem you are defining the current I=V/R = 4/2 as 2 amps.

The only way you are going to get 1 amp though a 2 ohm resistor I=V/R
1=V/2 thus V=2 so you are saying the voltage is 2, so I don't see how you can fix both at the same time.

 

[breno]

One thing you should perhaps be aware of is that the power output is proportional to the voltage squared (assuming the atomiser resistance is fixed).
So whilst a voltage drop from 5 to 4 is (5-4)/5= a 20% drop, the power drop is (5^2 - 4^2)/5^2 = (25-16)/25 = 9/25 = a 36% drop.

So a small change in voltage produce a much larger change in power.

for example going up from 5v to 7v virtually doubles the power, and presumably nicotine inhaled!!

 

[mnealtx]

It doubles the power - but I don't think it doubles the absorption. It DOES seem to give a better 'kick', though.

 

[Dave Rickey]

You can regulate (stabilize) current by adjusting voltage, or you can stabilize voltage and let current shake out as the balance between voltage and resistance. You could theoretically adjust resistance as well as one of the other two, but in practice trying to stabilize more than one is pulling on two ends of the same string without considerably more logic than you want or need in an eCig.

In the end, what you're after is wattage applied to vaporizing eLiquid, you can't really control your resistance because the atomizer has the amount and gauge of NiChrome it has, and that's it (you can add to resistance, but not dynamically in any meaningful sense).

If you stabilize either voltage or current, you get away from the variable vapor production that plagues eCigs as the cell winds down, with the downside that as the cell voltage drops your logic is going to draw more current from it, and if the overcurrent/undervoltage protection fails (or you're using unprotected cells) you've supplied the feedback conditions for a thermal runaway (making other forms of over-current or under-voltage protection important).

Because what we're after is wattage, and load resistance is not variable in any meaningful sense, current regulation is our best route to stabilizing vapor output. Combine that with user-controlled airflow (through restriction or how hard they suck) and you can get a much more consistent result, at least within the limitations of atomizer reliability.

--Dave

 

[Robert]

If your using a resistor- then your battery is heating a resistor.

I use an inline voltage regulator.

 

[breno]

If your using a resistor- then your battery is heating a resistor.

I use an inline voltage regulator.

I think a voltage regulator will still produce heat.
I was think they would be a good idea too because I realised a resistor
would heat up, now I am thinking how do those regulators work?

I think the basicially do the same thing as a resistor in a round about way.


Anyway here is a picture of one:-

Note it has aheat sink on it or where it can be attached to a bigger one.

Here is a much more sophistated one, notice the whopping great heat sinks!!

I think the first one is a simple transistor, and the name comes from shortening "transfer resistance" and that is what they basically do, they transfer resistance in to the circuit.

So I am not to sure if is any better.

 

[mnealtx]

If your using a resistor- then your battery is heating a resistor.

I use an inline voltage regulator.

And you (and I, since I use one too) are heating them up just like a resistor would.

 

[mnealtx]

I think the first one is a simple transistor, and the name comes from shortening "transfer resistance" and that is what they basically do, they transfer resistance in to the circuit.

So I am not to sure if is any better.

The name 'transistor' came from a combination of transmitter and resistor. It's more like a rheostat - you can go from a virtual open to a virtual short (minus bias losses).

 

[Dave Rickey]

A linear regulator is electrically just a self-adjusting resistor, you're still shedding the excess power as heat and your efficiency is no better than the ratio between the input voltage and output. A switching buck regulator can be much more efficient (90-95% being pretty typical), it works by turning the current off part of the time, then smoothing the output in an inductor. But you need at least two more components; a capacitor and an inductor, the inductor being the physically large part of that, the more you're stepping the voltage down, the bigger it has to be to turn the sawtooth output into a steady voltage, feeding a sawtooth into an atomizer would probably be bad (repeatedly cause the physical join between the NiChrome and the leads to flex, that being the usual source of total failure).

--Dave

 

[quovadis]

This is a naive question.

If what you guys say is true, then we should get the Chinese to build an atomizer with a perfect ohm resistance amount to satisfy different vapers using different voltages?

 

[Þornbjörg]

Related... I found a nice bit of testing done on various high capacity 18650's, with actual capacity at various rates... might be worth including for knowledge.

CandlePowerForums - View Single Post - New 18650 with 3000mAh from KD

If you're customizing atty resistance, you probably would consider a custom battery, so now you can intelligently decide on best battery performance too!