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This project uses a 555 timer and a 4017 decade to produce the flash police lights.

This circuit has only 4 leds , 2 blue and 2 red.

I want to connect in 16 leds ( 8 blue and 8 red ) is that possible? and how i will do it?

My thinking is that if i repeat the same procedure from r7 and r9 and the bc547 transistors 3 more times then i will have 16 leds. but i do not know if that is correct or productive.

enter image description here

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  • \$\begingroup\$ See how each LED has a resistor, and all the LED/resistor pairs are in parallel? You can probably add more just like that. \$\endgroup\$ – user253751 Feb 5 '18 at 0:26
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The proper way is to compute total LED power the battery Wh and choose battery size to match time and power product with a suitable voltage.

For Voltage the optimal way is to make series strings, not parallel that add up to less than 1 diode drop below the minimum battery voltage.

Then use Ohm's law on the average voltage to get average current on the difference between the LED string voltage and the Battery.

When the integer number of LEDs is not a convenient value, consider a different battery or number of LED's or some other combination of series/parallel.

The switch must be able to sink all the parallel strings with a low voltage drop. Normally for BJT's this is when Ic/Ib=10 to 20.


e.g. for Vf+ R*If=9V for Red Vf=2.1 typ, Blue Vf=3.1 typ
thus If (red) = (9-2.1)/470=14.7mA thus LED Pd=Vf * If=2.1 * 14.7= 31 mW
thus If(blue) = (9-3.1)/470=12.6mA thus LED Pd=3.1*12.6=39mW
thus for 2 Red+ 2 Blue , LED power = 140mW, battery drain = V*I=9*2 * (14.7+12.6)= 491mW so efficiency is 140/491=28% with 27.3mA drain. This is because parallel instead of serial, most of the power is wasted in the 470 ohm Resistors.


Now if these are 5mm LEDs they can dissipate 65mW or so each, max.

To use 8 blue and 8 red, you could use 4 Red in series and 2 (sometimes 3) Blue for 9V. Thus the array of LEDs for Red is 2P4S and for Blue 4P2S

Let's choose 15mA for each "P" String So we have 2 Red + 4 Blue = 6 * 15mA= 90 mA which on a good 500mAh 9V battery may last 500/90 ~ 5h rounded down.

Using Rs = (9V-Vf)/If
Can you compute the If and single Rs for each colour? Then check Power dissipation of resistor.

Red 30mA Rs = (9V-8.4V)/30mA = 20 ohms >=1/16 W
Blue 90mA Rs = (9V-6.2V)/90mA = 31 Ohms = 1/2W or two 62R 1/4W for 2 strings each

This will waste less power than your 2x2 police flasher.

With almost 100mA LED drain, you would be wiser to use NFETs instead of NPN

schematic

simulate this circuit – Schematic created using CircuitLab

R1= 20 Ohms R6,R7 = 62 Ohms

Increased load demands more base R and less pull-down current.

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  • 1
    \$\begingroup\$ Teach someone to fish and they complain about not being spoonfed? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 '18 at 22:34
  • \$\begingroup\$ seemed like good advice to me... perhaps it's the understanding bit... \$\endgroup\$ – Solar Mike Feb 4 '18 at 22:48
  • \$\begingroup\$ red leds 20ma 2.1 v and blue are 20ma 3.1v. I have to use one strip red and two blue for each headlight, so i must use one resistor for each red strip and one resistor for two blue strips. The r7 , 8 ,9,10 in the circuit must stay as they are 470 or i must change them to something else? \$\endgroup\$ – panayiotis kasapis Feb 5 '18 at 18:11
  • \$\begingroup\$ thats not what I suggested \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 5 '18 at 18:29
  • \$\begingroup\$ excuse me please , i try to figure it out in the headlight of my kid car. so i use in one headlight one red strip and two blue strips and i go with wires to the other headlight and connect the remain strips. from there i use one 20ohm resistor for red leds and one 31ohms for blue leds to the circuit board. i think now i understand it. what about the resistors in the circuit i post here R7-R8-R9-R10 which are 470ohms , do they stay the same or i have to change them too? \$\endgroup\$ – panayiotis kasapis Feb 5 '18 at 20:05

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