A game of skill and patience
This project combines some of the components and building blocks we have covered in the courses in this e-magazine. It's a game of skill that tests your reaction-time and patience.
Patience, because a certain condition appears on the indicator LEDs, only every few seconds and it corresponds to them both being OFF.
When they are both off, you must press the control button and this supplies a little energy to a reservoir electrolytic. The voltage on this electro is detected by a high impedance amplifier and the result is shown on a "LED-Pair.' As the voltage rises, one of the LEDs in this pair gradually dims and its brightness is transferred to a LED of the opposite colour.
This process is very gradual and may take up to 20 to 30 pushes. But if your timing is out, all the progress of 20 pushes will be lost to the discharge circuit.
The time when it is safe to press and NOT press is very abrupt and you have to be very careful not to extend into the “forbidden" zone.
The first building block is the low frequency oscillator, of which there are two. Each oscillator operates independently and its flash rate is indicated on a LED. The LED is buffered by an inverter to use the gates in the chip.
This arrangement also allows the output of the oscillator to rise to full voltage and allow the capacitor to charge. Any loading of the output of the oscillator does affect charging and if it does not rise to 66% of rail voltage, the oscillator will not operate.
The value of the timing components in each oscillator is not very important and we have chosen 47k and 22u for one set and 100k and 47u for the other. This will give a flash rate of about 4:1 and means the two output LEDs are both OFF for a very small fraction of the time.
The next block is the GATE, made up of two signal diodes and a 10k resistor. This forms an AND gate and operates as follows: The circuit requires a HIGH on the end of the 10k resistor to charge the reservoir electrolytic and this is available ONLY when the cathode of both diodes is HIGH.
This condition corresponds to both LEDs being OFF and thus the circuit is designed around the OFF state.
When the output of both oscillators is HIGH, the charging path for the electrolytic is via the 10k and 3k3 resistors and during the short period of time when this occurs, the button should be pressed. When either oscillator goes LOW, it brings the discharge path into operation.
The discharge path is made up of a 470R and signal diode. This path has a much lower impedance than the charge path (the 3k3 resistor) and thus most of the good work done during the charging will be lost by one false move.
During charging, the diode (in series with the 470R) forms a 'blocking' effect and does not play any part in the charging.
Next we come to the high impedance DETECTOR circuit. It is actually an emitter-follower arrangement, designed to turn ON one of the indicator LEDs. The question we ask is "Why use two piggybacked transistors?" In simple terms, the answer is to keep the LED illuminated between presses of the switch. This block is a good example of how a transistor operates. It shows that a current is required by the base, for it to supply collector-emitter current.
If we remove the top transistor and connect the base of the lower to the electrolytic, the LED would gradually dim between pushes of the switch. This is because the base requires current for the transistor to drive the LED and it draws on the electrolytic for this energy. The voltage across the electrolytic would gradually fall and the effect of the game would be lost.
By placing another transistor across the first, the current drawn from the electrolytic is reduced by much as 1/200th and the brightness is maintained.
This arrangement is called a SUPER ALPHA pair or SUPER TRANSISTOR as the gain of the transistors is multiplied together to produce an arrangement similar to a single transistor with enormous gain.
The output of this stage is taken to a common-emitter arrangement (for a PNP transistor) in which the load is in the collector. This stage turns off when the previous stage turns ON.
This is the entire circuit. It consists of one chip, 3 transistors, 4 indicating LEDs and a few resistors, diodes and capacitors. The circuit is supplied by 9v and is designed to accept a tri-coloured LED or a red and green LED in the output.
Building the project will help you understand the circuit a lot more and bring together the theory we have covered.
Begin construction with the smallest components. They are easier to fit when no other components are on the board. Begin at one end and fit the 3 diodes and 9 resistors. The 3 diodes must be fitted so that the cathode end (as indicated by a black band around one end) goes over the 'bar or line' shown on the overlay.
Next fit the 3 transistors. Two are NPN and will be BC 548 or 2N 2222. The other transistor is PNP and will be BC 558 or 2N 3906.
After this we fit the 4 LEDs. The two mini LEDs are 3mm and the cathode lead
must be fitted near the edge of the board. You have a choice of output LED. It can
be either a 5mm red LED and green LED or a single tri-coloured LED. Either of
these arrangements can be fitted to the board. But NOT both.