Talk:Multivibrator

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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)

Contents 1 astable 2 Merge 3 Definition 4 3-way

[edit] astable

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

Well this could be prettier, but here is what the waveforms look like: [1] Green or white = voltage, yellow = current. Pfalstad 02:18, 12 January 2006 (UTC)

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)

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)

Thank you for the above explanations. My apologies to Graham if I actually introduced more problems when I thought I was "correcting" the text. I can now see where I was going wrong. I was thinking of the base of the transistor as being high impedance. Of course this is wrong, you are right the BE junction will look like a diode so will conduct if the base voltage exceeds 0.6V. That error was what led me to think the capacitor will be discharging, when it will in fact be reverse-charging as you say, and I think the other problems in what I wrote are all follow-on consequences of assuming zero base current. Sorry, it's too long since I've done any electronics, and maybe it was foolish of me to make such a major edit when I'm so rusty... Anyway, I wholeheartedly agree with you that the best thing would be some annotated plots of waveforms. If you are willing to do that, then that would be fantastic. TerraGreen 22:30, 11 January 2006 (UTC)

Another thing which will almost certainly still need changing is what I wrote about the output voltage during state 1, as this too was based on the assumption of zero Q1 base current. But I'll leave it to others to make the change, because I'm now getting muddled when trying to think about it. TerraGreen 23:01, 11 January 2006 (UTC)

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)

[edit] Merge

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

[edit] Definition

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)

[edit] 3-way

There's an interesting circuit here: [[2]] (about 2/3 down the page), which uses 3 transistors to have 3 states. Would be worth addin