# Regulated Box Mod Mathematics

#### starrvenom

Joined Apr 12, 2016
2
Hi Everyone! I'm pretty new to the forums here but after taking a look around I thought this was the best place to seek an answer to a question that has been irritating me to no end. I'm a vaper (holla to the others who might be in here), and a pretty advanced one at that. I am however just now beginning to desire a deeper understanding of the mechanics of these devices that i use every day, so that, hopefully, one day soon i can build them. So i have a series box mod that is regulated (power mode in W). My question is why does my mod only fire, if regulated at a certain wattage, down to a certain percentage, say 70%, of my remaining battery power? I have a vague understanding of why, but I'd really like to have a concrete answer, and to see the math behind it, and I haven't been able to find that anywhere else on the web.

If anyone here is kind enough to show the math for this problem, the details of the particular circuit i have set up would be a series of two batteries with capacities of 3000 mAh and a max of 35A discharge connected to a parallel set of .36 ohm resistors, and I have the wattage set to 120, which by the way is well within the parameters of the device that they are housed in.

#### ronv

Joined Nov 12, 2008
3,770
Hi Everyone! I'm pretty new to the forums here but after taking a look around I thought this was the best place to seek an answer to a question that has been irritating me to no end. I'm a vaper (holla to the others who might be in here), and a pretty advanced one at that. I am however just now beginning to desire a deeper understanding of the mechanics of these devices that i use every day, so that, hopefully, one day soon i can build them. So i have a series box mod that is regulated (power mode in W). My question is why does my mod only fire, if regulated at a certain wattage, down to a certain percentage, say 70%, of my remaining battery power? I have a vague understanding of why, but I'd really like to have a concrete answer, and to see the math behind it, and I haven't been able to find that anywhere else on the web.

If anyone here is kind enough to show the math for this problem, the details of the particular circuit i have set up would be a series of two batteries with capacities of 3000 mAh and a max of 35A discharge connected to a parallel set of .36 ohm resistors, and I have the wattage set to 120, which by the way is well within the parameters of the device that they are housed in.
To answer your question we will need to know the battery voltage and the circuit that provides constant power/current. But I would suspect that the battery voltage falls below the point that the circuit can provide the 25.8 amps into .18 ohms.

#### starrvenom

Joined Apr 12, 2016
2
To answer your question we will need to know the battery voltage and the circuit that provides constant power/current. But I would suspect that the battery voltage falls below the point that the circuit can provide the 25.8 amps into .18 ohms.
The Voltage of the batteries is 3.7, as for the voltage output falling once they reach a certain level of charge, that much i assumed, i was just wondering if there might be a formula or a general mathematical principle to determine at what level of charge the circuit would fail.

#### ronv

Joined Nov 12, 2008
3,770
The Voltage of the batteries is 3.7, as for the voltage output falling once they reach a certain level of charge, that much i assumed, i was just wondering if there might be a formula or a general mathematical principle to determine at what level of charge the circuit would fail.
What is inside this box?

So i have a series box mod that is regulated

#### OBW0549

Joined Mar 2, 2015
3,566
The Voltage of the batteries is 3.7, as for the voltage output falling once they reach a certain level of charge, that much i assumed, i was just wondering if there might be a formula or a general mathematical principle to determine at what level of charge the circuit would fail.
It depends entirely on the characteristics of the particular voltage regulator chip used.

#### Papabravo

Joined Feb 24, 2006
16,509
There are two relationships at work:
1. Ohm's Law, which establishes the relationship between voltage, current, and resistance. Voltage(E in volts) = Current (I in Amperes) × Resistance (R in Ohms)
2. Power equation which relates power to voltage and current. Power(P in Watts) = Voltage(E in Volts) × Current(I in Amperes)
Starting with a fresh battery of 3.7 Volts and a load of 0.18 Ohms, I would expect that the current in the load would be:

$$\frac{3.7\; \text Volts \math}{0.18\; \text Ohms \math} \approx 20.55\; \text Amperes$$

The power consumed by the setup would be about 76 Watts. As the battery discharges the voltage starts to drop slowly at first and then quit rapidly. When the voltage is at 3.2 Volts for example the current will drop to about 17.75 Amperes and the power will also drop to about 56.89 Watts and it contiues from there until the battery can no longer supply any current at any voltage.

BTW -- what is a series box mod?

#### ronv

Joined Nov 12, 2008
3,770
There are two relationships at work:
1. Ohm's Law, which establishes the relationship between voltage, current, and resistance. Voltage(E in volts) = Current (I in Amperes) × Resistance (R in Ohms)
2. Power equation which relates power to voltage and current. Power(P in Watts) = Voltage(E in Volts) × Current(I in Amperes)
Starting with a fresh battery of 3.7 Volts and a load of 0.18 Ohms, I would expect that the current in the load would be:

$$\frac{3.7\; \text Volts \math}{0.18\; \text Ohms \math} \approx 20.55\; \text Amperes$$

The power consumed by the setup would be about 76 Watts. As the battery discharges the voltage starts to drop slowly at first and then quit rapidly. When the voltage is at 3.2 Volts for example the current will drop to about 17.75 Amperes and the power will also drop to about 56.89 Watts and it contiues from there until the battery can no longer supply any current at any voltage.

BTW -- what is a series box mod?
a series of two batteries with capacities of 3000 mAh

#### Papabravo

Joined Feb 24, 2006
16,509
a series of two batteries with capacities of 3000 mAh
Thanks -- I thought it was something complicated and/or innovative.

#### paulktreg

Joined Jun 2, 2008
802
Thanks -- I thought it was something complicated and/or innovative.
New member + Sarcasm. Hello and welcome to AAC!

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#### Papabravo

Joined Feb 24, 2006
16,509
New member + Sarcasm. Hello and welcome to ACC!
No trigger warning either. I'm feeling uncomfortable.

#### ronv

Joined Nov 12, 2008
3,770
No trigger warning either. I'm feeling uncomfortable.
Sorry, I was just trying to point out that it was 2 batteries and not one. So his 120 watts is possible.