How flip - flop works??? Image provided - Page 1
Forum section: electronics / beginners

author: RivaultUser
author_rank: Contributor
timestamp: January 01, 2013, 09:10:13 am
content: Can you explain to me how flip - flop works in this circuit ?http://www.talkingelectronics.com/FreeProjects/5-Projects/The-flip-flop-in-action-complete.gifI know that the web has already explained about how this circuit works, but I JUST DON'T GET IT!!!I'm a newbie to electronics ( like 1 year I tried to learn it by myself ), and I cannot imagine how this circuit works.....I need step by step explanation how this circuit works, so that I know what this component does and why the LED can flip - flopping ^^Please give me step by step explanation ! (*Bow down*)

author: vk6zgo
author_rank: Super Contributor
timestamp: January 01, 2013, 01:03:07 pm
content: Well,for a start,that circuit is not a a flip flop,it is anastableMultivibratorYou will notice that it is continually switching from one state to another---In other words,it has no stable state,which is what "astable" means!A "flip-flop" is abistablemultivibrator----"bi-stable" implies that it has two stable states,which in a practical circuit can be switched between by applying external control signals.For some strange reason,"Talking Electronics" seem to have taken it upon itself to disagree with the rest of the world when it comes to the naming of this circuit.

author: RivaultUser
author_rank: Contributor
timestamp: January 01, 2013, 01:25:01 pm
content: Quote from: vk6zgo on January 01, 2013, 01:03:07 pmI see...... Thx for your reply and correction man^^But could you explain how that astable multivibrator works?

author: lewis
author_rank: Frequent Contributor
timestamp: January 01, 2013, 01:58:42 pm
content: This seems like a good explanation -http://rayshobby.net/?p=1079

author: RivaultUser
author_rank: Contributor
timestamp: January 01, 2013, 04:57:03 pm
content: Quote from: lewis on January 01, 2013, 01:58:42 pmThank you for your reply man, but there are things that still I couldn't understand.......Like this sentence "Therefore C1?s left lead is in floating status, and its right lead is connected to T2?s base, which is about 0.7V due to the forward voltage drop of transistors."What does it mean by floating status??? And is the sentence above tells me that the capacitor is charge about 0.7 V due to the voltage drop of the transistorand this sentence "T2?s collector (output O2) remains low"Does this mean that T2's collector has 0 V"voltage on C2?s left lead will rise"Why only on the left lead of the C2'sWhat about the right lead ?"drop to ground voltage"What does it mean by ground voltage"As C1 is fully charged, it’s right lead will suddenly drop to a negative voltage (-Vcc). This will shuts off T2 firmly."Why C1 right lead will suddenly drop to a negative voltage??? ( well I never heard of negative voltage (-Vcc) )And why when that happened T2 will shut offPlease help me with this..... I'm struggling to learn electronics because I love it, but sadly I don't have ability to learn fast like my friends, that's why I'm keep asking stupid question. There's no one I can ask except everyone on this forum.Thx for your concern!

author: IanB
author_rank: Super Contributor
timestamp: January 01, 2013, 07:10:24 pm
content: OK, let's try to keep it simple.Two facts:1. A transistor (NPN type) will not turn on unless the base voltage is greater than about 0.6 V2. The voltage difference across a capacitor can only change when current flows into it.So, suppose transistor Q1 is turned off. Since only one transistor can be on at a time (you figure out why), then Q2 is turned on.Now, current is flowing through the left 10k resistor and trying to turn Q1 on, but it can't because the current is being absorbed by the capacitor also attached to Q1 base. The current flowing into this capacitor is raising the voltage difference, but only slowly, which is why the circuit can only flip-flop slowly from one state to the other.Eventually, the voltage at Q1 base will increase above 0.6 V and will cause Q1 to turn on. As soon as Q1 turns on the voltage on Q1 collector will drop close to zero. Remembering fact 2, the voltage on Q2 base will also drop rapidly, turning Q2 off (remember it was on before).When Q2 turns off its collector voltage will rise, causing the voltage on Q1 base to rise (fact 2 again). Therefore Q1 will rapidly turn on, causing Q2 to be firmly turned off.The circuit is now flipped, and current is flowing through the right 10k resistor, trying to turn Q2 on. Eventually the voltage at Q2 base will rise above 0.6 V and the circuit will flop back again.Everything is now back where we started and the cycle repeats.