Simple Peltier Refrigerator Circuit

In this post I have explained a straightforward procedure for building a simple refrigerator using Peltier device for generating the required cooling effect inside the fridge.

How Peltier Device Works

We are all familiar with a Peltier device and know how it functions.

A Peltier device is a 2-wire semiconductor device having two surfaces that generate hot and cold temperatures across them in response to electricity supplied on its wire terminals.

Basically it works on the principle of thermo-electric effect (opposite of Seebeck Effect) where a potential difference is used for making or producing hot and cold temperatures over the two ends of a dissimilar metal assembly.

A Peltier device has two terminals in the form of wire ends which requires to be connected across a voltage source rich in current content.

The application of voltage instantly starts transforming one surface of the unit hot and the reverse surface cool very fast.

However, the hot end must be quickly managed so that the heat does not reach higher levels, which can completely hamper the heating and cooling process and ruin the device itself.

Therefore the hot surface must be attached with heavy heatsinking materials like aluminum or copper metal of suitable sizes.

How to Build a Simple Fridge using Peltier Device

The simple construction of a simple peltier refrigerator circuit shown in the figure demonstrates the above discussed set up where two such devices are appropriately fixed with aluminum plates for radiating different degrees of temperatures from their relevant sides.

The plates responsible for generating the cooling effects must be trapped inside a well insulated enclosure made up of thermocole or polyurethane foam etc.

The inside chamber may be used for storing water bottles or water packets as desired.

The hot heatsinked surfaces must be exposed in the outside air for radiations and for controlling the temperatures of "hot" ends of the unit, see figure.

Complete diagram for understanding How to Make a Simple Peltier Refrigerator at Home.

Peltier Performance Specifications

• Hot Side Temperature (ºC) 25ºC / 50ºC
• Qmax (Watts) = 50 / 57
• Delta Tmax (ºC) = 66 / 75
• Imax (Amps) = 6.4 / 6.4
• Vmax (Volts) = 14.4 / 16.4
• Module Resistance (Ohms) = 1.98 / 2.30

Video Demo

Formulas for Calculating Peltier Parameters

Peltier Heat (Q_p):

Q_p = Pi * I

Where:

• Q_p = heat absorbed or released at the junction (in watts, W)
• Pi = Peltier coefficient (in volts, V)
• I = electric current passing through the element (in amperes, A)

Thermal Resistance (R_th):

R_th = ΔT / Q

Where:

• R_th = thermal resistance (in °C/W or K/W)
• ΔT = temperature difference between the hot and cold sides (in °C or K)
• Q = heat transfer rate (in watts, W)

Temperature Difference (ΔT):

ΔT = Q_p / R_th

Where:

• ΔT = temperature difference across the Peltier element (in °C or K)
• Q_p = heat pumped by the element (in watts, W)
• R_th = thermal resistance (in °C/W or K/W)

Joule Heating (Q_j):

Q_j = I² * Re

Where:

• Q_j = Joule heat (in watts, W)
• I = electric current (in amperes, A)
• Re = electrical resistance of the Peltier element (in ohms, Ω)

Total Heat (Q_total):

Q_total = Q_p + Q_j

Where:

• Q_total = total heat to be removed from the hot side (in watts, W)
• Q_p = heat pumped by the Peltier effect (in watts, W)
• Q_j = heat generated by Joule heating (in watts, W)

Solving an Example Peltier Problem

Given:

• Peltier coefficient, Pi = 0.5 V
• Electric current, I = 3 A
• Thermal resistance, R_th = 0.1 °C/W
• Electrical resistance of the element, Re = 1 Ω
• Calculate the Peltier heat, Joule heating, total heat, temperature difference.

Step 1: Calculate the Peltier heat (Q_p):

Q_p = Pi * I
Q_p = 0.5 V * 3 A
Q_p = 1.5 W

Step 2: Calculate Joule heating (Q_j):

Q_j = I² * Re
Q_j = (3 A)² * 1 Ω
Q_j = 9 W

Step 3: Calculate the total heat to be dissipated (Q_total):

Q_total = Q_p + Q_j
Q_total = 1.5 W + 9 W
Q_total = 10.5 W

Step 4: Calculate the temperature difference (ΔT):

ΔT = Q_p / R_th
ΔT = 1.5 W / 0.1 °C/W
ΔT = 15 °C

• The Peltier element produces a 15°C temperature differential between the hot and cold surfaces.
• The total heat that must be dissipated from the hot side is 10.5 watts.