Large array of 5mm LEDs (555 timer,LED flasher)

[The_Sinpy]

Referring back to older post of mine about flashing high power LEDs
https://www.electronicspoint.com/th...-555-timer-highpower-led.269840/#post-1620622
Would the same thing work for what i'm doing or would I just need to use a transistor like I have in the photo listed below?

Power supply Car/12.5~14.5v
LED 3.2~3.4v 20mA
LED being powered at 3.0v 17mA
Also should I base my resistors for LED's from the 12.5v or the 14.5v

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Schematic from
http://www.instructables.com/id/41-LED-Flasher-Circuit-using-555-IC/

 

[KrisBlueNZ]

That design in "60 led flasher.png" will work except for the transistor drive. You have omitted the base current limiting resistors that are in the design on instructables.com.

In your design, if you follow the current path from the positive supply, through the base-emitter junction of the PNP transistor, through the base-emitter junction of the NPN transistor, to the 0V rail, you will se that you have two P-N junctions connected in the forward direction in series across your power supply. A P-N junction starts conducting heavily above about 0.8V and if you try to apply 12V across one, it will lose its magic smoke extremely quickly, especially if you use an automotive supply which can provide hundreds of amps!

This problem can be avoided by using MOSFETs instead of transistors; this will also be more efficient and will provide slightly more of your supply voltage across the LED strings. Replace the PNP with a P-channel MOSFET with its source to V+ and replace the NPN with an N-channel MOSFET with its source to 0V.

I notice you still haven't added load dump protection to your circuit. Perhaps one day when you turn off the igniton and a big puff of smoke comes from your circuit board, you will reconsider that decision.

As for calculating the resistors, to be safe you should first calculate a minimum value for them to ensure that the maximum continuous current specification for the LEDs is not exceeded under worst case conditions. Worst case conditions would be maximum supply voltage and minimum forward voltage across all LEDs.

Then calculate a nominal value assuming a typical forward voltage (3.3V, unless there's a typical characteristics graph in the data sheet that shows you the mean forward voltage) and a typical supply voltage, say 13.5V I guess, to give the desired forward current (not sure if that's 20 mA or 17 mA). As long as the result of that calculation is higher than the minimum safe value, I would use it.


I just noticed that you've written on your schematic
"3.2~3.4V 20 mA max"
"Going run 3.0V 17 mA"

If you mean that you want to run them at 3.0V and 17 mA, you can't do that - that's not how it works. Each LED has a characteristic graph that plots forward voltage against current. There will be some variation from one LED to another, even in the same batch, and temperature has an effect too, but the point is that you can't set the voltage and the current independently - they are interdependent, and the relationship is a characteristic of the LED itself.

If you used a constant current driver, you could set the current, and the voltage would depend on the LED characteristic. Then you just need to ensure that there's enough voltage available to cover the maximum voltage needed by all of the LEDs that are in series, for that amount of current. But constant current drivers are more complicated than the simple current limiting resistor method.

Using a series resistor for current limiting, you can't set either the voltage or the current exactly, because the current depends on how much voltage is left over when all of the series forward voltages are subtracted from supply voltage (and a little bit more is subtracted for voltage drop in the MOSFET).

It's reasonable to want to run the LEDs at around 17 mA, but it's meaningless to say that you want to run them at 3.0V. If you feed 17 mA through an LED, the voltage across it is determined by the LED's characteristics, and if the voltage is 3.2~3.4V at 20 mA, it won't be very much less than that at 17 mA.

You need to go to the paragraph that starts with "As for calculating the resistors," and work out the minimum safe resistance, and the nominal resistance. If the LEDs are rated for a maximum of 20 mA, you'll probably find that the minimum safe resistance will be higher than the nominal resistance you calculate; if you want to respect that maximum current limit, you'll have to use the minimum safe resistance, which will work out to a nominal current that's somewhat less than the nominal current you want.

 

[The_Sinpy]

Sorry took so long reply, My post never posted

By using mosfes are you talking about this thing or the photo beside it? (Photo bottom of post)

I wouldn't use both N channel and P channel. Just the N channel mosfet. Reason being...I was playing around with few LEDs to see how it would look with the alternating flash pattern. But didn't like the results. I found using two of the 555 timers is getting the flash pattern I wanted. Just have to change one of the resistor values to one of timer.

I'm running leds at what you said.
Also the reason I didn't have anything in there about load dump protection is because bit lost in the mist about it.
Added few new things to the schematic

 

[KrisBlueNZ]

Both of those circuits in post #3 will work. The N-channel MOSFET can be fed directly from the output (pin 3) of a 555 without a resistor, though a low-value resistor (e.g. 33Ω or so) is recommended.

The second circuit uses the MOSFET in an arrangement called a current limiter. R4 isn't needed, and a resistor (around 100Ω) in series with the transistor's base is recommended, but the circuit will work as shown. The MOSFET limits its drain current to about 200 mA (calculated as roughly 0.7V / R5). This current will be shared by all of the LEDs, as long as the resistors in series with the LED strings are high enough to allow good current sharing.

This is a good way to control the LED current when your supply voltage is not constant. But it adds a voltage drop of around 0.8V minimum in the driver circuit.

So you want to run each LED at 20 mA. Ten strings of three LEDs each, for a total of 200 mA through the MOSFET.

So the voltage across each string will be:
V_F min = 3.2V × 3 = 9.6V
V_F nom = 3.3V × 3 = 9.9V
V_F max = 3.4V × 3 = 10.2V

The supply voltage is 12.5~14.5V. Subtract 1V for the MOSFET and current setting resistor leaves 11.5~13.5V.

So there will be at least (11.5 - 10.2V) = 1.3V across each current sharing resistor. That's good. Around 0.8V is plenty, so each current sharing resistor will be R = V / I = 0.8 / 0.02 = 40Ω; use 39Ω.


So that right hand circuit will work fine. Use a resistor of 33~39Ω for each string of three LEDs. Maximum dissipation in the MOSFET will be about P = V × I = 3.2 × 0.2 = 0.64W so you may need a small heatsink - something like http://www.digikey.com/product-detail/en/507302B00000G/HS115-ND/5849.

Re load dump protection, a large TVS such as http://www.digikey.com/product-detail/en/V18ZA40P/F3078-ND/1009327 from the positive input to 0V, then a series resistor of about 1.5Ω, then a 15V 5W zener (http://www.digikey.com/product-detail/en/1N5352BG/1N5352BGOS-ND/1474096) to ground (anode to ground), should do the trick.

 

[The_Sinpy]

Ok Ill give it try. Thank you very much

Im just having hard time finding that large TVS.
I would get it off the Digikey web site but looks like they upped their shipping price :\

I've scrap few circuit boards and found a few. But part number of them showed they were in the 240v range

-------Edit---------
Attached a file

 

[KrisBlueNZ]

1. The 15V 5W zener should be connected across the incoming supply, not in series with it. (Shown above.)

2. You had the incoming supply connected to pin 7 of the 555, not to the circuit's positive rail. (Shown above.)

3. You could reduce R6 to 10k, the same as R4. I recommend any value between 3k3 and 10k. That may give slightly better current regulation.

4. Have you checked the duty cycle of the flashing at both positions of the switch? At both positions, the LEDs will be ON most of the time, with a short interruption. This is most noticeable when the switch selects R2. If you want roughly 50% duty cycle in both positions, you need to make some changes. (a) Connect a fixed resistor between pin 7 and pins 6+2, instead of two resistors with a switch; (b) This resistor must be a lot higher in value than R4 - you can reduce R4 to, say, 3k3 and use, say, 47k between pin 7 and pins 6+2; (c) change C1 to a lower value, to set the fast flash rate; (d) add another capacitor with an SPST switch in series with it, across C1, so that when you close the switch, it will be in parallel with C1, and choose the value of that new capacitor to set the slow flash rate.

 

[The_Sinpy]

R4 should been 7.5k Sorry I didn't notice that and the power going to pin 7 when I made the changes to it to the photo

I've used few leds and it seems be working fine for the type of flash I wanted. A fast strobe and flash when I flip the switch.
Don't mind the video title

 

[KrisBlueNZ]

OK, if the flash rate and the duty cycle are what you want, no problem. I was just checking that you were happy with them.

 

[The_Sinpy]

Well guess I'm not happy with them now....lol

When I added the mosfet part of the circuit it made a change in the flash rate.
Is this normal?

 

[KrisBlueNZ]

Hi again and sorry for the long delay.

I have been reconsidering my advice on protecting circuits from the automotive electrical environment.

I've discovered that automotive load dump is not a normal occurrence; it only happens if the battery becomes disconnected while the engine is running - for example if you go over a big bump and your battery terminals are not properly tightened. In this circumstance I think it's acceptable for a component to be destroyed, as long as the rest of the circuit is protected from damage.

There are many other sources of disturbances on the automotive power supply but they are of much shorter duration than a load dump surge.

The V18ZA40 varistor that I used to recommend is no longer stocked by Digi-Key. Littelfuse have an improved range called the V18AUML series which is specifically designed for automotive use. Unfortunately, they are only available in SMT (surface-mount technology).

I've drawn up a complete design for you.



The supply protection circuitry is at the left. There are two resistors in series, RF1 and RF2, both 1Ω 2W wirewound fusible resistors. These will contribute a total voltage loss of 0.4V when the LEDs are ON and drawing 200 mA. This is still just within the voltage budget for the circuit.

DR protects the circuit against reverse voltage, and RV is a Littelfuse V18AUMLA2220 varistor that clamps its voltage at 40V maximum. RF1 is the weak point in this part of the circuit and it will be the one that fuses on reverse power supply or very high input voltage.

The 555 is only supposed to operate at up to 15V. Quite a weak point for a device that's regularly used in automotive circuits! So DP is a 15V, 5W zener diode, and RF2 provides the weak point between varistor RV and zener diode DP.

CP smooths the power supply rail. There is also a decoupling capacitor on the 555 (CD1, which should be connected as directly as possible between pins 1 and 8).

For the oscillator I have specified a 10 μF timing capacitor (CT) instead of 100 μF, so the timing resistors (RA and RB) can be higher values. With R1 between pin 7 and VCC, and the timing resistors between pin 7 and pins 2 and 6, the duty cycle will be fairly close to 50%, i.e. the LEDs will be ON and OFF for roughly equal lengths of time. You will need to determine the values for RA and RB by trial and error or by using a potentiometer or trimpot to get the flash rate you want, then measuring it.

The output circuit (current sink) is as before, though I've specified a lower gate resistor to make the MOSFET switch slightly faster and tighten up the current regulation a bit, and I've noted tht the MOSFET must be heatsinked. It will dissipate a bit of power while the LEDs are ON, especially when the supply voltage is near its maximum limit. Dissipation will normally be less than a watt, so a small clip-on heatsink should be enough

 

[The_Sinpy]

Thank you for all the the nice information you have gave and for explaining it out for me. I also like the schematics you drew up for me. I really appreciate all of your help you gave me.

I've changed my 100uF to a 10uF and changed all of my resistors to the 555 timer. I've got the flash speeds I want.Nothing in the circuit is getting hot.

I'm sorry about the late reply I gave.

 

[horah]

I'd like to make something similar, but my leds are 4 serial conected (together have 5W). 20 pieces of this kind I'd like to connect in parallel, what means all of them would be about 100W, all supplied with 12V dc. What modification do you suggest? Thank you very much.