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Qx5252f Led Driver For Mac ((INSTALL))

The fault with the above circuit lies in the fact that the output of a 555 is about 0.2v to 0.5v above the negative rail when outputting a LOW. When this is combined with a gating diode, it produces a voltage of about 0.8v on the base of the driver transistor and this will NEVER turn the transistor OFF. The remedy is to include a voltage-divider network made up of two resistors. In the following example, the 470R between base and 0v and 470R between base and gating diode, form a voltage divider. The best way to see their operation is to remove the buffer transistor. To produce 0.6v on the join of these two resistors we need to supply 1.2v at the top. This is what is now needed to turn ON the transistor and if the gating diode produces a voltage of 0.8v, the transistor will see 0.4v and it will turn OFF. This next mistake highlights the need to draw a circuit so that it is immediately obvious how the circuit works. At first impression, the first circuit appears to be scanning the LEDs in a 3x4 matrix, as the LEDs are all drawn in the same direction:However when the LEDs are re-drawn as shown in the following circuit, we can see the arrangement is known as Charlie-Plexing. Here's a circuit with a couple of mistakes:It operates as follows:The Red LED illuminates if the voltage is below about 12.5 volts.The Green LED begins to light at about 12.5v and the Red LED fades, at about 14.25v the Yellow LED begins to come on and by 14.75v the Green LED has faded out. If the pots are turned to one end; the zener, pot and transistor will be damaged. A 1k resistor needs to be placed between the base and wiper of the each pot. Here is a circuit designed by Talking Electronics to monitor a battery:The red LED turns on from 6v to below 11v.It turns off above 11v and the orange LED illuminates between 11v and 13v.It turns off above 13v and the green LED illuminates above 13v Here's another circuit from Electronics For You, and Indian publication: The first thing I have always said is this: Always draw a circuit so that it is easy to understand. I have absolutely no idea what the 4 diodes are doing and how the circuit is actually connected to the two phones. By re-drawing the circuit, this is much easier to see: The 4 diodes are a bridge and the LED is turned on when the phone is picked up. The circuit can be connected to the phone line either way as the diodes are wired so that the LED always illuminates. But the 33R has no effect at all and can be removed. Secondly, the 12p and 33p have no effect on the audio. They should be in the range 10n to 100n to have any effect at audio frequencies. And yet another circuit from an Indian website:The EHT transformer will only produce 8kv-10kv when used in flyback mode. This requires the transistor to turn off very quickly at the time when the transformer is reaching saturation. It is very important that the timing for the transistor is accurately controlled and this requires a feedback signal from the transformer. This circuit does not provide that feature. Secondly, NO TRANSFORMER is going to produce an extra high voltage when a diode is connected across the primary. This diode will completely absorb the energy from the collapsing magnetic flux and prevent a high voltage being produced. Nearly every circuit from CircuitsToday has a fault as nothing is tested before being written up on the web. More from There are many faults with this circuit. 1. The 1u tantalum can be an ordinary electrolytic. 2. The 8k2 resistor should be about 47k3. The 220n should be about 22n 4. The diode does not serve any purpose. 5. The 10k resistor should be at least 47k6. The 0.1uH should be about 0.06uH 7. The 3-10p should be at least 22p8. The 100p across the battery should be at least 22n. 9. The 220R resistor should be at least 470RSee our series on designing FM transmitters and you will learn how to design circuits like this and not make the mistakes contained in this circuit. Here we have a 4017 driving 4 buffer transistors. If you look carefully, you will see the 4 outputs of the 4017 are connected together. Each output of a 4017 is bi-polar. In other words it drives HIGH or LOW. At 12v, each output will deliver about 100mA when driving a short-circuit. When one output is HIGH, it is driving three LOW outputs and this will put considerable strain on the chip. In the following circuit, each output will rise to about 3v and that is why the designer thought the circuit would work. But this voltage is produced under enormous strain inside the chip. The outputs should have either gating diodes or resistors to prevent overload. The designer of the circuit considered the chip and stepper motor would take zero current when the chip is reset, (the 6th output - pin 1 - goes to reset) but he did not provide any feedback from the 4017 to tell the oscillator when to stop providing clock pulses. If the 4017 sits with any output HIGH, the current consumption would be considerable. Secondly, the circuit would NEVER work as ALL THE OUTPUTS are connected together and all the Darlington transistors are turned on AT THE SAME TIME!I have also asked him how the circuit guarantees to remain in reset mode when not required. In other words, what happens if the 4017 receives 3 pulses (or any of 3+5+5+5 etc). There needs to be feedback from the 4017 to tell the "clock" when to stop sending pulses. In other words, the clock needs to know where the 4017 is.I have contacted the author of the circuit: shoumik banerjee and he now claims the circuit has never been tested. Only a fool puts a project on the web without building and testing it. Not only that, but it is totally unfair to other readers, who may build the circuit and find it has no chance of working. That's why you have to be careful with everything you find on the web. I have seen some "impossible" circuits that work and some that should work, but don't. TELEPHONE BUG Look at the coil. A floppy coil like this is totally unsuitable as it will be "telephonic." This means it will vibrate when bumped and will even pick up sounds such as talking, music and footsteps, and transmit them, just like a microphone. I suggest you build this circuit and use 7 turns of thin wire on an 8-10mm pen and see how the coil picks up every sound in the room. Simply connect a 9v supply and the circuit will work. The coil should be 6t and 3mm diameter.

Qx5252f Led Driver For Mac




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