2. INTERFERENCE
β’ Many devices operate in the 2.4GHz range. Have
you ever lost your Wi-Fi connection while
cooking something in the microwave?
Interference is a common problem which was
solved by allowing routers to emit 2.4GHz on
different channels. If your router detects channel
1 to contain too much noise, it will move to an
open channel.
β’ The Wi-Fi jammer was designed to emit large
amounts of 1.0 Vp-p 1KHz and 1Vp-p 400KHz
sine waves across the entire 2.4GHz band
causing a direct denial of service.
3. MATERIALS USED
β’ TP4056 Micro USB 1A Charging Board
β’ TX6729 2.4 GHz 4 CH A/V Transmitter
β’ 470 Ohm Resistor
β’ 3.7v 2000mAH Lithium Ion Battery
β’ Hammond 1591ESBK Project Box
β’ Dremel and Power Drill
β’ 24 AWG conductor Wire
β’ 110V Soldering Iron
β’ SMA Female to PCB Solder Pigtail cable
β’ Router Dipoles
β’ On/Off Rocker Switch
5. WI-FI JAMMER SCHEMATIC
β’ The jammer is a multi-loop
circuit. When the switch is closed,
and the charging board is connected
to a standard 120V outlet, current
flows from the 4.2V/1A charging
board, through a 470 Ohm resistor,
into the A/V transmitters and
Lithium Ion Battery. This current
charges the battery while providing
energy to the transmitters.
β’ When the switched is closed
without being connected to an
outlet, the battery powers the
transmitters with 3.7V / 0.5A.
6. CALCULATIONS
This shows that the resistor is only going to allow around 1mA to flow from
the 4.2V USB source to the 3.7V battery and load. This was designed so the
battery was never given too much current.
We can find π2 by Applying KVL across the second lo
3.7 β 7.73 β π2 = 0. π2 = 0.479mA.
We can find π1 by Applying KVL across the first loop.
4.2 β 470 β π1 β 3.7 = 0. π1 = 1.06mA.
In order to analyze the circuit further, we need to find the resistance of the A/V
Transmitters. We will use π = πΌ2
π =
π2
π
= πΌπ. From the spec sheet the transmitters
use 90mA at 5V. When all four circuits are connected, a voltmeter reads 3.73 at
the copper terminals.
P=(90mA)(5V)=0.45W.
0.45W*4 =
3.932 π
π
R=7.73 Ohms
We can also use Ohms law to calculate current
3.73π
7.73 πβππ
= 0.483ππ΄
