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Reverse engineering the Fischertechnik blinker

By admin | July 13, 2014

The Fischertechnik blinker comes in a enclosed case with no other sensory inputs with two wires, a yellow wire and a blue wire.

The Fischertechnik blinker

The Fischertechnik blinker

As such, I reverse engineered the Fischertechnik blinker to understand how it works through desoldering, a camera, and a lot of drawing on the GIMP.

It turns out that the blinker is intended to be used like a switch in a series circuit before a load, with the yellow lead being the positive end and the blue lead being the negative end.

Series circuit with FT blinker on mySTEM board

Series circuit with FT blinker on mySTEM board

The main idea here is that the board uses a 555 timer (555C EZ606 with STMicroelectronics logo, surface-mounted) with fixed frequency (fixed R1, R2 and C) in astable mode to control a BSP 75N MOSFET transistor.

The power supply for the ICs is controlled in parallel with a diode and a 470Ω resistor to protect the ICs, while the MOSFET controls the connection to the other lead.

It features an unknown-model fuse (VW UG4 GP613) on the emitter of the transistor, presumably to protect the circuit from excess.

A diode and 100 µF capacitor cap are present, possibly to smooth out the power input for the ICs during temporary dislocations. Considering that it is a Fischertechnik component, its purpose is probably to ensure continued reliability when it is handled roughly, shaken, and dropped by students.

After complete disassembly:

Fischertechnik blinker, disassembled

Fischertechnik blinker, disassembled

With traces and PCB features:

Front view, with traces and PCB features

Front view, with traces and PCB features

Back view, with traces and PCB features

Back view, with traces and PCB features

Schematic:

Fischertechnik blinker schematic

Fischertechnik blinker schematic

Since C is a generic-looking ceramic capacitor with no markings, its capacitance was estimated by using the period.

The light was measured to blink 100 times in 41.21 seconds, giving:
\(
T = 0.4121 \text{s} \\
f = 2.426 \text{s}^{-1}
\)

Given R1 = 1 MΩ and R2 = 1.2 MΩ from above and f = 2.426, we can calculate C using the following equation:
\(
f = \dfrac{1}{\ln 2 * C * (R_1 + 2*R_2)}
\)

C then works out to about 0.175 µF.

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