Talk:Multivibrator

message: the sentences here are awful. can someone with a good knowledge of the English language pls have a look at it?
--zeno
Completely rewritten 5/11/03 - i hope it meets with approval from former contributors GRAHAMUK 23:05, 4 Nov 2003 (UTC)
thread_title:  Poor wording 

message: I've expanded this section considerably.  Sorry to delete existing text, but I think it had some problems, particularly references to capacitors discharging when they were actually charging.  TerraGreen 20:10, 10 January 2006 (UTC)
replies: text: The changes are fine by me, though as the author of the original text you removed, I am pretty sure it was correct. Perhaps it wasn't clear, but it definitely wasn't wrong. Graham 22:10, 10 January 2006 (UTC)

text: In state 1, C1 is discharging, not charging as the text now says.  The voltage is lower on the right side of C1, and the current is going to the left, reducing the voltage across the cap.  And C2 is charging.  Pfalstad 23:03, 10 January 2006 (UTC)
replies: text: I don't think so - C1 is charging. One problem at present is that the circuit (which I drew, so my apologies for this) shows electrolytic capacitors, which might be misleading - a real circuit shouldn't use electrolytics here because they are charged in both directions. However the reverse-bias charge only ever reaches ~0.6V, so it will probably function... anyway, as far as I can see the description of the operation of the circuit is correct both now, and in my original version. Graham 01:35, 11 January 2006 (UTC)
replies: text: Do you agree that when Q2 turns on, then C2's right side is ~0v and C2's left side is negative?  And when Q2 turns off again, both sides of C2 are ~0v?  Pfalstad 02:33, 11 January 2006 (UTC)
replies: text: Yes, I agree with the first part. Before Q2 turns on, the left side of C2 must be at ~0.6V, and the right side at +V. When Q2 switches, the right side is brought to 0v. The left side has no choice but to follow, so at that instant its voltage will be 0.6 - V, which will be quite negative. This switches off Q1... C1 now goes to +V on the left side. The right side tries to follow but will be clipped at 0.6V by the BE junction of Q2 - previously C1 was charged to 0.6V the other way, now it will be charged to +V via R1, rather rapidly, since R1 >> R2. But when Q2 turns off again, C2 will be 0v on the right, and +0.6V on the left, which is what turns Q1 back on. It has charged to 0.6V via R3. This is the charging that the description is referring to - the important one, since it's what the time constant comes from. I think I can see what you're getting at actually - both capacitors are in fact 'charging', but at different rates and in different directions. That makes it hard to really talk about charging without specifying what direction we are talking about. The charging that sets the time constant is as the text describes it - the charging in the other direction isn't mentioned, which is sort of OK since it is really just getting rid of the previous charge and doesn't change the circuit behaviour. You know what would help? Some annotated waveforms and perhaps current arrows around the circuit. I think this seemingly simple circuit is actually a lot more subtle than it looks, and understanding it isn't always that easy. Graham 06:40, 11 January 2006 (UTC)
replies: text: Well since C2's + side is on the right, it's got +V-.6 across it when Q2 swiches on and -.6V across it when Q2 switches off (going through 0V on the way), so it seems to be discharging when Q2 is on.  Pfalstad 23:43, 11 January 2006 (UTC)
replies: text: I'm having trouble understanding what you mean - please bear with me! When you say "when Q2 switches on" are you talking about it having just switched on, or just about to? This is where I think a steady-state explanation for the circuit falls down - it's a very dynamic circuit with everything changing all the time, so you have to be very precise about exactly which moment in time you mean. So, from what you say about the voltage on C2, I assume you mean the moment just before Q2 switches on. When Q2 switches on, the existing charge (+V - 0.6) on C2 is first dumped - so yes, it is discharged at that moment, quite rapidly. It then goes on to charge, more slowly, in the other direction via R3 to +0.6V, and it is this charging which matters. The fact that it is a charge opposite in sign to the earlier charge doesn't to my mind make it a 'discharge' - the capacitor is still storing charge, just of opposite sign to its previous state. As I said, showing electrolytics here doesn't help, since we are not used to the idea of charging these in reverse... I really must fix that I think, and maybe add some waveforms as well. Graham 00:16, 12 January 2006 (UTC)
replies: text: The voltage across C2 is the same before and after Q2 switches on.  When Q2 switches on, the absolute voltages on either side of C2 change rapidly, but the relative voltage across C2 does not change at all.  At least that's how it works in the simulation I am looking at.  C2 discharges rapidly?  How rapidly..  What's the time constant, what does the current look like, etc.?  Where is the current coming from to make this happen?  Pfalstad 00:33, 12 January 2006 (UTC)
replies: text: On further thought, I think I see why you want to call it a discharge... When Q2 switches on, the LEFT side of C2 goes negative (0.6 - V), but since the base of Q1 is reverse biased in this condition, it does look like a high impedance... so C2 ramps up from ~-V to 0v to +0.6v via R3... so yes, I would agree that you could call this a discharge for the most part, since it is a gradual reduction of negative charge... I just tended to look at this as a charge with opposite sign, but perhaps a discharge is more accurate. Urk, my head is spinning... Again, showing the waveforms will cut through all of this - perhaps a text explanation isn't such a great help. Does your simulation feature the ability to create a chart of the circuit waveforms? Graham 00:38, 12 January 2006 (UTC)
replies: text: Well this could be prettier, but here is what the waveforms look like:   Green or white = voltage, yellow = current.  Pfalstad 02:18, 12 January 2006 (UTC)

text: Actually, there is an error, though it's not the one you mention. It states "Q1 base at slightly below the supply voltage...". That can never be. Q1 or Q2 base can never exceed 0.6V, because the B-E junction of the transistor "looks like" a diode to ground. I think it will be best to simply remove those lines. Graham 01:39, 11 January 2006 (UTC)

text: OK, another little problem (I should have checked the change more carefully!). One of the keys to the astable operation is that when the circuit changes state (Q2 turns on, say), the sudden change at the collector from +Vs to ~0v is transmitted via C2 as a negative pulse to the base of Q1, thus rapidly switching it off. Otherwise there is no obvious reason why Q1 should switch off when Q2 switches on. This probably needs adding to the description of the operation of the circuit - I'll have a go. Graham 01:45, 11 January 2006 (UTC)

text: Ah yes, if you assume the transistor has no base current then the working will seem rather different. In fact, I'm not 100% sure, but if that were the case I don't think it will oscillate - both transistors would just end up switched on. You can visualise this by putting enhancement-mode MOSFETs instead of bipolars for Q1 and Q2 - a stable state would be both Q1 and Q2 on, no oscillation. C1 and C2 would be both fully charged and there is no discharge path available to them, as it is the base current in the bipolar case that provides this. Graham 00:16, 12 January 2006 (UTC)
thread_title:  Astable 

message: Someone suggested merging the stubs into this article.  This is a good idea. --Wtshymanski 02:20, 27 July 2006 (UTC)
thread_title: Merge

message: A definition is mysteriously not there. A multivibrator is not defined as something thats used to implement a variety of simple two-state systems. This needs to be more explicitely defined in the intro. Fresheneesz 22:20, 25 October 2006 (UTC)
thread_title:  Definition 

message: I added details to the monostable multivibrator and redefined it with One-Shot as well.  I think a One-shot link should be created to redirect to this page if they are looking for the electronic definition of a one-shot.  Dante 16:26, 22 June 2007 (UTC)
thread_title:  Monostable 

message: for the multivibrator equation
f=1/(.693*R*C)
I'm wondering, is .693 more accurately represented by ln(2)?
I'm no expert, just a hobbyist, but knowing that the capacitor equation is exponential, it seems to make sense.
??
Thanks in advance, and my apologies if I am mistaken.
--Smiley325 20:23, 30 October 2007 (UTC)
thread_title:  Multivibrator equation 

message: Somewhere, somehow, apparently the component values for the circuit[s] were removed. It would be nice to give practical ranges for them.Nikevich (talk) 09:32, 24 November 2009 (UTC)
thread_title: Missing component values

message: Moved from User talk: Oli Filth:
I saw that you deleted a section calling it "...one particular example circuit".  Isn't this a derivation of the time-constant?
It would be nice to have the derivation in the article somewhere.  Mikiemike (talk) 01:29, 28 December 2010 (UTC)
replies: text: It's the derivation of the time-constant for one particular implementation of one particular type of multivibrator.  IMHO, this derivation bloats the article with tangential information.  Regards, Oli Filth(talk|contribs) 09:42, 28 December 2010 (UTC)

text: I see you've restored the material, and you seem to have misquoted me in your edit summary!  I don't claim that this is "an example", I claim it's the derivation for one particular example circuit.  This derivation offers little insight into multivibrators or oscillators in general, so I don't see much purpose in it being here.  That is why I removed it.  Your new section headings don't help this situation!  Oli Filth(talk|contribs) 16:51, 28 December 2010 (UTC)

text: I have added a link to RC circuit (time domain). Maybe this is a more suitable place for this derivation? You have considered a classical problem - determining the charging time (0,7RC) of a capacitor to half the maximum voltage. Circuit dreamer (talk, contribs, email) 18:17, 28 December 2010 (UTC)
thread_title:  Derivation 

message: The general flavor is WP:NOR.
This article makes several general statements about MVs based upon specific instances. The results are often false. Multivibrators must have two passive networks. Multivibrators must have a capacitor.
Ref 1, the etuttle blog, is not WP:RS.
Glrx (talk) 05:07, 24 January 2011 (UTC)
replies: text: Yes, the article has been edited a lot lately; perhaps winding back a month or so would be a better starting point. Lecture notes don't make the clearest encyclopedia style. --Wtshymanski (talk) 05:35, 24 January 2011 (UTC)
replies: text: Excellent work! It's a good idea to limit the scope of multivibrator only to the classical cross-coupled transistor pair. Circuit dreamer (talk, contribs, email) 14:18, 24 January 2011 (UTC)
replies: text:  Articles are always simpler when they stick to their subject.  "Hysteretic" circuits, Christmas light flashers, and ballroom dancing are not multivibrators. If it's not a cross-coupled pair of amplifiers, it's not called "multivibrator" in the books. Wish I could find a clear description of relaxation oscillator but as usual that article has no references either. --Wtshymanski (talk) 14:54, 24 January 2011 (UTC)
replies: text: It does now. --ChetvornoTALK 22:04, 2 December 2016 (UTC)
thread_title:  Dubious tags 

message: "The name "multivibrator" was initially applied to the free-running oscillator version of the circuit because its output waveform was rich in harmonics"
I always thought it was because earlier oscillators were electromechanical, i.e. they were literally vibrators. Gigs (talk) 17:53, 22 May 2012 (UTC)
thread_title:  Name 

message: This article really needs a diagram showing the waveforms , it's much more obvious when you can see the base potential being negative for a long time, and ramping up to the trip point.  like on this page schoolphysics.co.uk 
"Multivibrator" definition is probably lost in history. You can make one with a pair of RC cross coupled relays, possibly someone did this back in the 1800's but failed to publish a web-page. While you can make a "vibrator" with just one relay Trembler_coil (Model T ford), two make it more accurate as a timer. 
I would definitely define a multivibrator as requiring two cross coupled RC stages and two active inverting elements noting the classic multivibrator has two complementary outputs (this might not be a necessary requirement?) so for example you can alternately flash two light bulbs (as on a railway crossing)(a 555 can't do that). With two RC stages you can set each of the two delays independently of the other (a 555 can't do that). 
The 555 is definitely not a multivibrator, it belongs to the class of integrator + schmitt trigger oscillators, and has only one capacitor there are two basic topologies for the 555 , producing approximately sawtooth (classic relaxation oscillator) or triangle wave timing waveforms, the 555 explicitly sets the threshold voltages. 
The other thing that isn't mentioned is that the output voltages on the collecters have a soft leading edge, and this may be undesirable in some applications (transistors get hot).
Historically, I have located Charles Adlers probably first patent on alternating flashing railway lights US patent 1622018 and this is not a multivibrator, just a mono-vibrator with a heavy weight. So no real evidence for my earlier suggestion of cross coupled relays being the precursor to multivibrators.
Salbayeng (talk) 02:34, 14 February 2016 (UTC)
thread_title:  More comments 

message: Hi Has anyone considered adding some of the content that appears on the polish version of this article https://pl.wikipedia.org/wiki/Multiwibrator ?. It is about twice as long, and has more schematics including the triple-vibrator.
Salbayeng (talk) 04:19, 14 February 2016 (UTC)
thread_title:  Consider updating with information from the Polish version of this article