# How to calculate the voltage drop in the PCB over time

The voltage drop in your PCB can be calculated using the following formula:

**Voltage drop = Current * Time / Battery capacity**

Where:

**Current:**The current drawn by the controller in sleep mode (135 µA)**Time:**The time for which the controller is in sleep mode (in seconds)**Battery capacity:**The capacity of the battery (in mAh)

In your case, the battery capacity is 200 mAh. Let's assume the controller is in sleep mode for 24 hours:

**Time = 24 hours 60 minutes/hour 60 seconds/minute **

** = 86400 seconds**

Now, let's calculate the voltage drop:

**V= 135 µA * 86400 seconds / 200 mAh **

** = 0.0596 volts**

Therefore, the voltage drop in your PCB would be approximately 0.06 volts when the controller is in sleep mode for 24 hours. This is a very small voltage drop and is well within the acceptable range for a 3.7V battery.

However, it is important to note that this calculation only considers the current drawn by the controller in sleep mode. Other components on your PCB might also draw current, which could further increase the voltage drop. Additionally, the actual voltage drop may vary depending on the specific battery you are using and its internal resistance.

For a more accurate assessment of the voltage drop in your PCB, you can measure it directly using a multimeter. This will give you a better understanding of how much voltage the PCB drops under actual operating conditions.

## Derive Voltage Drop Formula:

The formula for voltage drop, V_drop = I * t / C, can be derived from the fundamental concepts of electricity and conservation of energy. Here's a breakdown of the proof:

**1. Ohm's Law:**

Ohm's law states that the voltage drop across a conductor is directly proportional to the current flowing through it and the resistance of the conductor:

**V_drop = I * R**

where:

V_drop = voltage drop across the conductor (volts)

I = current flowing through the conductor (amperes)

R = resistance of the conductor (ohms)

**2. Energy Conservation:**

In a closed system, energy cannot be created or destroyed, only converted from one form to another. In this case, the electrical energy supplied to the battery is converted to various forms of energy, including thermal energy due to the resistance of the conductor.

**3. Power Dissipation:**

The rate at which energy is converted from electrical energy to thermal energy is called power dissipation. The power dissipated in a resistor is given by:

**P = I^2 * R**

where:

P = power dissipated (watts)

I = current flowing through the resistor (amperes)

R = resistance of the resistor (ohms)

**4. Relating Power and Energy:**

Energy is the integral of power over time:

E = ∫ P dt

where:

E = energy (joules)

P = power (watts)

t = time (seconds)

**5. Applying the Concepts:**

Combining Ohm's law, energy conservation, and power dissipation, we can relate the voltage drop to the energy stored in the battery:

V_drop = I * R = P / I = (I^2 * R) / I = I * t / C

where:

C = battery capacity (ampere-hours)

t = time (hours)

**6. Converting Units:**

Note that the units of time in the formula need to be consistent. If "t" is given in seconds, the battery capacity "C" should be in ampere-seconds.

Therefore, the formula V_drop = I * t / C is a valid and proven formula for calculating the voltage drop across a conductor due to the internal resistance of the battery and the current drawn from it over a given time period.