Whats the theory behind this circuit? - Page 1
Forum section: electronics / beginners

author: yada
author_rank: Frequent Contributor
timestamp: April 27, 2017, 03:41:11 am
content: http://www.555-timer-circuits.com/bike-turning-signal.htmlI realize it makes a cascade effect but could you add more LED to it? That last resistor in the cascade is confusing me. So why does this work? Also at lower voltages it will only light up three LED but still at the same rate, not like all five are on but dimmer.

author: ironmonkey
author_rank: Contributor
timestamp: April 27, 2017, 03:50:56 am
content: It seems a kind of voltage divider using each pn junction voltage drop as a level. First, the group 1 lights up, and as the voltage at the rc tank increases, the following groups. The first group will turn on arround 1.2V and the following will do in 0.6 V increments, and thats why at lower voltages you only get the first onesEnviado desde mi MotoG3 mediante Tapatalk

author: ebclr
author_rank: Super Contributor
timestamp: April 27, 2017, 03:54:24 am
content: On the very philosophic concept, the trick is that each base-emitter needs 0,6 V to turn the transistor since you have a 555 oscillator and an RC circuit, the transistor will be on when vbe is +-0,6V and since the transistor are cascading, each one will turn on at a different level, 1st 0,6 2nd 1.2V 3rd 1.8 and so on

author: Brumby
author_rank: Unknown
timestamp: April 27, 2017, 07:13:20 am
content: Quote from: yada on April 27, 2017, 03:41:11 amI have moved those resistors around ... does this seem clearer?

author: ebclr
author_rank: Super Contributor
timestamp: April 27, 2017, 09:20:36 am
content: It's all the same thing, the unique difference is that each one will need 0,6V more to turn on the transistor since the every transistor will "take" 0,6V for herself, also consider another 0,6V for the end diode, means +- 1,2V will be the minimum for 1st LED and additional 0,6 V for every transistor attached is series ( on the point of view of base emiter). This circuit have a little error, to have all Led light near the same bright the resistors 10K must have different value for each transistor in a way that VBE is near the same

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 27, 2017, 09:35:12 am
content: The first one seems to be wrong.The second one seems to be less wrong... but that doesn't quite make it right.But what should you expect, it's got a 555Tim

author: ebclr
author_rank: Super Contributor
timestamp: April 27, 2017, 10:16:11 am
content: Another way to make thishttp://www.ti.com/lit/ds/symlink/lm3914.pdf

author: BrianHG
author_rank: Super Contributor
timestamp: April 27, 2017, 10:31:53 am
content: An even better way to make this, though you need to write a little MCU code:Any cheap 65 cent PIC MCU would do as well. You can have perfect programmed animated speed, no transistors needed, just the 8 pin PIC + 10uf VCC-GND cap + the 5x string of LEDs on 5 of the IOs, or, a 16 PIC pic using 10 IOs, 5 for left, 5 for right, 2 inputs for left/right indicator inputs, and, 1 more for solid on brake light mode. Also add an optional photocell to 1 ADC input, you can have auto brightness saving battery life during the evening by running the LEDs at a lower brightness. Maybe add 3 protection resistors for the switch input.For the 16 pin pic solution, the IOs are also low current making your battery contacts to the leds are short, continuously on, you wont get the mechanical switch failure like in flashlights since the switches will be super low current off of a really thin wire..Your battery supply would operate between 5v through 3v and the red or white LEDs would be wired in parallel instead of serial.

author: Zero999
author_rank: Super Contributor
timestamp: April 27, 2017, 10:52:20 am
content: Quote from: BrianHG on April 27, 2017, 10:31:53 amI agree, an MCU is a sensible solution but you still may need the transistors. The micro-controller will need to run off 5V or less, so it will need a voltage regulator. It will work with red LEDs in series but blue or white LEDs have a higher voltage drop, so will need to be connected in parallel (a resistor for each LED of course) or will need a transistor to switch the two LEDs in series to +12V.

author: kosine
author_rank: Regular Contributor
timestamp: April 27, 2017, 12:29:28 pm
content: The attached circuit is an interesting variation. It's a passive 9V battery tester which is a nice way of demonstrating that bipolar transistors don't actually need a 0.6V base-emitter voltage to turn on.If you build the circuit (or simulate it in Spice), a 9V supply will light all five LEDs with about 6-7mA (the differing resistor values balance the currents). The base input to Q5 will be about 3.9V, so each transistor has about 0.6V Vbe as expected.However, if you attach a flat battery (about 5V), the last LED (D1) will still light up with about 3mA of current, and D2 may be just visible with about 0.3mA. The base input to Q5 will only be about 2.2V, yet for Q1 to be on, all the other transistors must also be "on".If you measure them, Q5 will have a Vbe of about 0.2V, Q4 about 0.3V, Q3 about 0.45V. Q2 about 0.6V and Q1 about 0.65V, giving a total of about 2.2V.Technically, all the LEDs are actually "on" as well, but the currents are too small to make them visible.If you want to build the circuit for real, the capacitor just creates a small delay so the LEDs light up one-by-one. It makes a really nice battery tester, especially with green, yellow, orange and red LEDs.

author: Brumby
author_rank: Unknown
timestamp: April 27, 2017, 12:49:10 pm
content: My redraw has the identical topology to the original. I just moved the base resistors so they all pointed in the same direction.As for the design - it may not be elegant, but if it gets the job done....

author: tecman
author_rank: Frequent Contributor
timestamp: April 27, 2017, 12:56:09 pm
content: In the original circuit, the current in the first LED must reach a threshold where the voltage across the emitter resistor reached 0.6 volts to trigger the next LED. The cascade continues as each LED reaches the trigger current. That should allow each LED to reach a certain brightness in each step.paul

author: james_s
author_rank: Super Contributor
timestamp: April 27, 2017, 07:25:19 pm
content: That is a strange circuit, it's so unusual that I'm actually curious to try it. If I were asked to build something like this, I would have used a bargraph IC or counter, either that or a microcontroller. This reminds me of some of the clever transistor circuits that were floating around in the 1970s.

author: yada
author_rank: Frequent Contributor
timestamp: April 27, 2017, 07:44:58 pm
content: So whats the purpose of the two diodes? Also does the cap on the left control the 555 timing and the one on the right just act as the rc circuit?Does one cycle of the 555 last the whole time the 1-5 LED are on or is it turning on and off as each LED comes on? So if it takes 5 seconds to turn on all LED is it turning on for five seconds then off for less? Or is it turning on and off faster then that? I dont have a scope yet.When you say Vbc could you measure that by putting a meter on the B and C leads and measuring the voltage across or is that with one lead attached to the B and ground and also second 0.6v measurement from the C and ground?Another thing is when he first LED comes on it shuts off then comes on again a second time before turning on LED #2. That's the only LED that does that; other LED come on smoothly. It looks like its stuttering. How would you fix that?

author: BrianHG
author_rank: Super Contributor
timestamp: April 27, 2017, 08:04:05 pm
content: The purpose of the diode in the original schematic is to immediately discharge the 100uf cap when it is time to turn off the LEDs, ie the 555 output goes low, making it look like they will all go off instantly at once. When the 555 output goes high, the diode doesn't conduct. This means that the 100uf is slowly being charged through the 4.7k resistor, meaning, you get the one LED row after the next as the charge builds turning on each row one after the other. This is purely intended to give a visual effect...

author: yada
author_rank: Frequent Contributor
timestamp: April 28, 2017, 02:57:16 am
content: So I built this circuit with 8 LED and it doesn't work quite right. It still has that stutter and when it does that it lights the 1st (rightmost) LED bright and the last two LED very dimly as well as making a windscreen wiper noise on my FM radio. Then It starts to light up the LED normally stopping at the red coloured on and the little green next to it wont light up at all, neither will the last two blue LED. I have tried swapping LED around and it always does the same thing. I also put in 330resis. on the first three LED to keep the rightness down then 220resis for the rest. Is this just a bad design? Caps are 220 and 100uf. If you yank out the black 100uf cap out when its stuttering it gets stuck in that state and the RF noise stays constant. It pulls 35ma in this state and up to 155~ma when its (supposed to be)powering all LED. Still at 12.00V.

author: james_s
author_rank: Super Contributor
timestamp: April 28, 2017, 03:04:18 am
content: What happens if you disconnect the wire from pin 3 of the 555 and connect it to the wiper of a 1k pot between Vcc and Gnd?

author: ebclr
author_rank: Super Contributor
timestamp: April 28, 2017, 03:26:54 am
content: You have several problems do be solved1) You are using different color Led, Led from different colors have different requirements2) each resistor on each LED must be different value,Let's assume a 15 ma current to have a bright LEDI = V /R0,015 = V / R => R = 0,015 / VFor Led 1V = 12 X - 2 Vled( depending om color assuming 2 leds in series as schematic) -0,6 ( 1n4148) - 0.2 ( VCE sat)For Led 2V = 12 X - 2 Vled( depending om color) -0,6 ( 1n4148) - 0.2 ( VCE sat) -0,6 (bc547 vbe )For Led 3V = 12 X - 2 Vled( depending om color) -0,6 ( 1n4148) - 0.2 ( VCE sat) -0,6 * 2 (bc547 1 vbe )For Led 4V = 12 X - 2 Vled( depending om color) -0,6 ( 1n4148) - 0.2 ( VCE sat) -0,6 * 3 (bc547 1 vbe )For Led 5V = 12 X - 2 Vled( depending om color) -0,6 ( 1n4148) - 0.2 ( VCE sat) -0,6 * 4 (bc547 1 vbe )

author: yada
author_rank: Frequent Contributor
timestamp: April 28, 2017, 09:50:04 am
content: Quote from: ebclr on April 28, 2017, 03:26:54 amAre you missing the final ohms out of your formulas? Or am I missing something?That seemed like it was the problem but I'm not sure. I fixed the stuttering problem completely by adding a 0.01uf mylar cap to pin five and grnd completely by accident.I disconnected the base resistor of transistor (#1 on right) four and the circuit seemed to still work lighting up LED 1-4 (#1 in on right). I tried higher and lower value (instead of 330 ohm) resistors to the LED and not much difference.

author: yada
author_rank: Frequent Contributor
timestamp: April 28, 2017, 11:42:19 am
content: Whats also interesting is if I remove the diode that parallels the resistor the LED go off in the order they came on, which is a cool effect. Still haven't figured out why that 0.01uf cap fixed the stuttering effect.

author: Brumby
author_rank: Unknown
timestamp: April 28, 2017, 12:28:21 pm
content: Quote from: yada on April 28, 2017, 11:42:19 amThat's why the diode is there - to prevent this (better for the original purpose).