Smooth headlight or LED ignition on a microcontroller

Smooth ignition of LEDs

Good night all. With radio technology and circuits, I’m on you, but I wanted to solder a soft-lighting LEDs. The scheme took this (http://c.a.d-cd.net/99ebbu-960.jpg). Why are there two + (constant 12V and control) can we do with one control? I want to install it in the dashboard and the + and – will be taken from the contacts of light bulbs (to leave the possibility to adjust the brightness). And the most important question, when you put +12V is delayed (5 seconds) before starting the lighting LEDs. How to get rid of this? I would like to turn on the lights and the LEDs began to flare, without a pause.

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This is due to the non-linearity of the IAC of LEDs: at voltages below 1.5 to 2 volts, they don’t light up. Try putting a resistor between the gate and the ground somewhere between 150-200 kohms and match it to the delay time.

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Oh no! Then the first recommendation won’t work.

This is due to the non-linearity of the IAC of LEDs: at voltages below 1.5 to 2 volts, they don’t light up. Try putting a resistor between the gate and the ground somewhere between 150-200 kohms and match it to the delay time.

Maybe it is difficult for TC, but if you connect LEDs to op-amp circuit, included the repeater, and open the field transistor by output of op-amp signal, then the nonlinearity of the LED (SAGE) will be compensated and LEDs at the source of the op-amp will be lit and shine with linear dependence on the voltage at the input of Op-amp. You can turn them on even from a cassette – a kind of color music.

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Maxidroomys, all that is left is your head and hands. Take a part of the circuit, where the LEDs are included in the source, attach this circuit that is in the footnote and go ahead. You can use any opener powerful enough to open the field effect transistor. To do that, look at the datasheet for DTs, for example tda2030, datasheet for poles and go. Yes throw out the capacitor from your circuit, and instead of diodes include Op-Amps. If you use Op-Amps like 2030, then you can experiment with my circuit. The idea is this: when you put + at the input of Op-Amp (or you adjust the trimmer), in short, change the potential at the input of Op-Amp, the potential at the output also changes, goes to the LED VD and lights it. Since the LED is in the op-amp circuit, the potential at the output changes mirror-like – inversely to the LED’s SAC. The LEDs included in the drain (your case) or source circuits will shine directly proportional to the applied voltage from some initial voltage level and light up as smoothly as you apply the voltage and/or sound. If you put a 100 mcf capacitor at 16 volt with two resistors on the Op-Amp input and put + on the capacitor from something, you will get what you want. The resistor in series with the capacitor will give the charging time and thus the LEDs lighting up and the parallel resistor will give the LEDs extinguishing time. The circuit in SPLAN.

Smooth headlight or LED ignition on a microcontroller

I have already made some prototypes and it works amazingly. But recently ran into a problem – when installing the circuit headlights are lit regardless of whether the light is on or not. as it turned out the Asians control is made minus. What can be finalized in the circuit that it would not be fundamentally important what polarity occurs control

That is, if you put a transistor at the input, it will not matter which pole to turn off?

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The original circuit was designed for figure 1. You’re looking for Fig.2. And the answer is about the output transistor. And not just replace it, but put it on the ground wire.

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The original circuit was designed for figure 1. You’re looking for Fig.2. And the answer is about the output transistor. And not just replace it, but put it on the ground wire.

The problem is not in the headlight wiring diagram, but in the control scheme. The problem is that if you disconnect the minus on the input, the poleswitch is permanently open. On the table the same effect is observed.

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Pull up the input (after the diode) to the common wire with a 1K-10K resistor. And what does minus control even mean? Minus control is when the input voltage range is below the common wire. Maybe you mean an inverted input?

_________________ only pure true norwegian blackx

Pull up the input (after the diode) to the common wire with a 1K-10K resistor. And what does minus control even mean? Minus control is when the input voltage range is below the common wire. Maybe you mean an inverted input?

I mean that the common ground is disconnected at the input, not the plus wire

Okay, got it. Then C5 should be shunted with a 5M-10M resistor.

And in general imho here you can put R6 – 100K, C5 – 20uF, and shunt then 510K – 1M.

In any case this variant gives a smooth shutdown.

PS. I’m dumb though and I still don’t understand what the Asians are guilty of. How do you wire the circuit in the first place? Why the input (85)?

_________________ only pure true norwegian blackx

blackx I will try it tomorrow. Thanks for the tip.

Generally speaking, the soft shutoff of the lamp is evil. It makes the coil sag and it’s going to be a quick kill.

If you cut the common wire from the circuit (which in principle is wrong in itself), it does not help a damn thing. Draw a picture of how you hook it up. The circuit must be grounded __always_. The toggle switch should control __only__ the input – pin (85). The headlights should also be connected to common ground at all times, and the output of the circuit should be connected to “+”.

_________________ only pure true norwegian blackx

There you go, now you can see what’s wrong. Connect the common wire of the circuit directly. Take the control input (85) away from the +Power. Now you need to figure out how to control this input. When you turn on the toggle switch it can pull our input to ground. But we need to make sure that the control input gets the +Power level. So what do we do? Build a small inverter:

(Googled the picture, you will have your own voltages).

Output of the inverter (point A) to (85). The input is on + Power supply (!).

And you connect the transistor base to the tumbler. You can do it directly without the resistor.

Rb and Rc take 10K each. If you don’t have enough control power you can reduce it to 1K.

The culprit here is not Asians, but those who wired the button this way. Of course I don’t know anything about cars, but you should always cut off +power supply, not the ground.

_________________ only pure true norwegian blackx

blackx

If I understand correctly, as long as there is no mass on the base and on the emitter, the plus will not go to the circuit? And the poleswitch won’t be open with just the plus?

Yes, you seem to get the point right.

Инвертор инвертирует входной сигнал – 12 В -> 0, 0 ->12 В. When the key (toggle switch) is open, the +Power current flows through the base, the transistor is open, and “presses” the inverter output to ground with its small resistance. The control input in this case gives a low level signal – the headlights do not light up. As soon as the key closes, the current stops flowing through the base – it is easier for it to go to ground through the key itself. The transistor closes, “cutting off” Rc from ground. The inverter output is set high. The headlights come on.

PS. And don’t confuse the transistors – bipolar is used here. A field effect transistor has no base and no emitter at all.

_________________ only pure true norwegian blackx

blackx So the problem is that you can’t press it to the ground. The device is plugged into the relay socket in the pad and there is no minus on it when the toggle switch is off.

So you need to forcibly connect the device as it should. +Power and ground _always_ have to be on. At least when you turn the key in the ignition. This is a good place to start. It’s easiest to ground in a car – the whole body is ground

PS. If you can’t, you can try to screw with a big capacitor, which will take the charge of the fuse to itself.

_________________ only pure true norwegian blackx

Here’s another twist that came to mind. For experimentation. I don’t know how it will work exactly, but it should.

UPD. !@#$%, you have a stoned 24V step-up there. Then you need a 2-3 meg resistor. Otherwise it will always be on at half strength. But try 1M for now as an experiment.

You should connect the minus on the emitter and the minus from the “controlling” minus contact of the relay to the base.

If there is no minus at the base of the transistor, then the + is pulled up to the ground. If the base is minus (switch is on), then the plus goes to the circuit

I don’t get it, sorry I have already explained how the inverter works, let’s not discuss it. For it to work, it must be connected to +Power and to the common wire (ground).

I have given you some kind of solution above, try this first.

_________________ only pure true norwegian blackx

I tried both inverter and resistor, no effect.

In place of A1 you need a switch that will supply “plus” to “85” when the “minus” ground is applied to it

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