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]

Originally Posted by 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

Nice - thanks, Ralph.

[breno]

Originally Posted by 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

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.