A close friend asks me if I can help in the creation of an automated mechanism for his plastic. Must replace the motorized part with the gear ratio of a beacon light with something without moving parts because of vibrations and noises, in the long run damage the same structure in which they are accommodated. Problem not easily solved. The first thing that comes to mind is use led in circles that light up in cyclic sequence to give the impression of motion and you're done. Made all in flying shape to have an opinion of the person concerned does not meet expectations, the transition from one LED is very visible, also the observation point is such that just a single LED to give the desired result. Murphy's law makes no exceptions, "The first solution that comes to mind is always the wrong one!”. Practically must vary the brightness from a minimum to an average value, at that point a flash light of the flash of light should overlap to the normal brightness for then once exhausted to fall back down to the minimum.

Having rejected the idea of ​​using two separate LEDs to have a single point light must devise a circuit that makes that function.

By analyzing the waveform must first create a triangular signal and then synchronize it to a peak but starting from the power 5V which originally piloting the motor with gear ratio of revolutions.

The most simple circuit to generate triangular wave is this

In Vo is has a square-wave voltage V1 while I have the triangular waveform. A simple explanation of the circuit is a must, Also to explain the choices made circuit. R1 - R2 - R3 determine the value of hysteresis and V2 formulas are

for all three identical values ​​of the resistors has for V2 equal 1/3 Vcc e 2/3Vcc, R1 and R2 determine the symmetry while R3 determines the excursion.

Playing with the values ​​one can determine every aspect of the triangular wave.

These formulas, however, may apply only if you have an ideal operational or one rai to rail, with excursion of the output that comes to feeds. In my case in the drawer I have available a paltry LM358 which has as positive excursion Vcc 0.1V-1.4V and negative in the case of single supply. The two values ​​then can be deduced with a simplified circuit, assuming that the non-inverting input of the operational amplifier does not absorb a current that affects significantly on the calculations we will have for R1 and R2 = 100K, and R3 = 22K

before continuing it is best to post the final schematic and then reasoning about what

R4 and C1 are the components that determine the frequency and the data components have a period of 3,3 seconds.

R5 and R6 form a voltage divider of voltage high enough total value do not significantly affect the charging of C1. This choice was made necessary because of the low LED run 5V supply voltage and the voltage of at least 3V there would have been sufficient voltage for driving the next stage, a voltage-current converter.

As shown in the figure above VR6 now it oscillates between 145mV and 549mV, this voltage will be compared by the operational U1B with the voltage that develops in R12, the transistor Q3 will be piloted to maintain the voltages to the operational amplifier inputs identical, therefore the current will be directly proportional to the incoming voltage non-inverting. The same current will pass through the LED which is known to be piloted in current and not in tension.

The maximum current will be I = 0.549 / R12 = 8,1mA the voltage drop on R11 will VR11 = 47 x = 8,1mA 380mV which, added to give VR12 549 + 380 = 929mV this voltage is such as to allow the full functionality of the LEDs with the supply date, in fact 5V – 3V – 0,929V = 1,07V fall on the transistor Q3.

The Q1 and Q2 function is to synchronize the peak current in correspondence with the change of slope of the triangular wave. It works like this:

We start from the output of U1A high, C2 will discharge because both armatures are at the same potential forced by R7 and consequently the high gate Q1 that will take him condition, Q2 is cut off because the gate is located in potential mass forced to zero by Q1.

Output of U1A goes to zero now for a brief moment the capacitor behaves as a short circuit to ground up to that R7 does not load again the plates of C2 to the supply voltage, this brief moment with the data components has a duration of 5mS , during this period will be Q1 and Q2 conductive interdict, conducting it short-circuits to ground Q3 and R12 leaving the only limiting resistor R11 plus the resistance of Q2 condition ( 5 ohm) to limit the current in the LED.

With the data components will have a peak of 30mA in white LED, this value having regard to the very short duration does not damage the LED.

This produces a high current pulse through the LED, with the right timing to produce a strobe effect. The result is an LED, that gradually becomes brighter, then flashes even brighter before subside again. This produces a light for the viewer, that simulates what you would see viewing function in a lighthouse.

Given the limited space I created a base adapted to contain all components without the need for jumpers and quite spaced tracks it can also be done by hand.

2 replies
  1. doctor
    doctor says:

    Bravo Amilcare Strong and exhaustive as always by Doktor . I take the occasion to make you know that the famous power supply 0 – 500 V 5 The is almost ready for those vuple dtiver try to scr . The only way to drive with power like that after so many failures around 2/ 300 skipped components and takeoffs of continuous counter and fibrillations of national network


Leave a Reply

Want to join the discussion?
Feel free to contribute!

Leave a Reply