555 for LED Dimming (solved) - Page 1
Forum section: electronics / beginners

author: jGalt
author_rank: Contributor
timestamp: September 08, 2014, 05:01:01 am
content: I watched Dave Jones' video on this but I'm struggling to understand how the circuit actually works. I attached the pinout of the 555 and the dimming circuit that I've been looking at.My thoughts so far: threshold and trigger pin are shorted. Both of these pins are inputs to respective comparators in the 555. I think that current flows from the "IN"rail to the output (pin3) so that when this is first powered up current flows through the diode on the right and through the pot and charges the cap that's placed before ground. Eventually the cap reaches the threshold/trigger voltage? Then things discharge through the Diode on the left? Now I'm confused. I don't quite understand how the 555 pins talk to each other. How can I predict what's coming out of the discharge pin (pin7)? And what's the cap hanging off pin 5 for?

author: westfw
author_rank: Super Contributor
timestamp: September 08, 2014, 05:37:26 am
content: This is the standard 555 "astable multivibrator" configuration, which you can find many explanations for, EXCEPT that the pot and diodes are arranged so that the cap charges through one side of the pot, and discharges through the other, yielding variable duty cycle.

author: jGalt
author_rank: Contributor
timestamp: September 08, 2014, 12:14:14 pm
content: Perfect, thank you for the key words. I'll do some google/youtube searching then; my apologies for posting prematurely.

author: jGalt
author_rank: Contributor
timestamp: September 08, 2014, 08:26:53 pm
content: Well I read thishttp://www.electronics-tutorials.ws/waveforms/555_oscillator.htmland thishttp://www.circuitstoday.com/555-timerand I watched thisand I took out my notes from last year and it still seems to me that the bottom left cap charges until it reaches the threshold/trigger voltage that's determined by the pot. Since the pot controls the amount of current flowing it affects how fast the cap is charging or discharging and that in turn controls the effective duty cycle. All of this works because if the cap is less than the trigger threshold voltage that corresponds to either a high or low output (which one is arbitrary to me atm) and if it's above the voltage it corresponds to the opposite output, so you get a square wave on the output. The transistor on the output is then switched on and off at a rate that's controlled by the pot.I'm still fuzzy on the significance of the diodes. Maybe for the Vd(on) drop?Either way, unless I'm still missing something, I don't understand westfw's sassy remark that "you can find may explanations for , EXCEPT [what I wrote]." It may have been incomplete but my ideas weren't far off and I was just hoping you'd save me half a day's time off researching and write what I just did above. Not trying to be a keyboard warrior here but I was simply trying to show that I had given the circuit some thought and you found it necessary to poke fun at my ideas.

author: Zero999
author_rank: Super Contributor
timestamp: September 08, 2014, 09:00:00 pm
content: The diodes only conduct in one direction so the capacitor charges and discharges via different sides of the potentiometer.The key to understanding this is to look at the potentiometer as two resistors with a total value of 50k and a variable tap at the wiper. Look at the attached schematic, showing what the potentiometer looks like electrically with the wiper at different positions.

author: jGalt
author_rank: Contributor
timestamp: September 08, 2014, 10:26:27 pm
content: Ah that's perfect! Thank you so much Hero. I just recently used a pot for variable voltage dividing, so I understand what you're trying to say.Once again my original idea that the diodes were intended to control the paths that the current takes during charging and discharging was close to being correct westfw.But it makes makes sense now that the threshold and trigger values aren't changing. They're set at 2/3rds and 1/3Vcc from the 3 5k resistors that are internal. What's changing, as Hero pointed out, is the rate that the cap can charge and discharge.

author: jGalt
author_rank: Contributor
timestamp: September 08, 2014, 10:34:50 pm
content: Anyone know what the cap hanging off the control pin is doing?

author: sonic2000gr
author_rank: Newbie
timestamp: September 08, 2014, 11:07:45 pm
content: Nothing much really in this configuration, it will work without it.(unconnected) It just prevents noise entering the chip.Sent from my LG-V500 using Tapatalk

author: mrkev
author_rank: Regular Contributor
timestamp: September 08, 2014, 11:30:15 pm
content: Cap at pin 5 is just decoupling the voltage divider at the input of the 555, look into the datasheet, what it's connected to.It's usually there just because it's recomended, ofc it's gonna work without it too. Leaving that pin open could decrease the frequency stability, but you don't really care for that in circuits like this.The function of 555 was described again and again, I think that you could doo some more googling. Anyway easiest way how to understand this is:Its a simple flip-flop, flipping output pin 3 to gnd (and "connecting" 7 to ground) when the voltage at 6 reaches above 2/3 V+, flipping to V+ (and disconnecting pin 7) when the voltage at 4 is below 1/3 V+.

author: jGalt
author_rank: Contributor
timestamp: September 08, 2014, 11:56:30 pm
content: Thank you for answering my question concerning the capacitor at pin 5 mrkev and sonic.