Multivibrators (Sorry girls, nothing to get excited at here... :) )

Providing the image is able to be seen above..

I've just built that circuit and now i have 2 LED's flashing, I'm still a little confused as to how this circuit works?

  1. On applying voltage, at the start, do both caps start charging..
  2. Resistor 2 and 4 are both connected to the Minus side of the capacitors yet connected to a positive + rail... huh? why?

Don't tell me magic :slight_smile:

Probably doing a google search for 'Flipflop circuit' will save thousands of words :slight_smile:

Edit:
I already did it http://www.talkingelectronics.com/FreeProjects/5-Projects/Page16.html

where do you think i got the picture from?

  1. googling is not going answer my questions that i asked

A transistor doesn't have a "minus" side. It has a base, collector, and emitter. Read up on how a transistor works and you'll understand why the resistors are connected the way they are. And that link that Riva gave will explain in plenty detail how the circuit actually works, how the capacitors work and how they affect the circuit. Google itself might not answer your question directly, however it will give you plenty of references of where you can find the answer.

  1. a capacitor can only charge up if there is a voltage across it. Initially on switch on both sides of the capacitor are connected to +V so the can not charge.
  2. the -ve end of the capacitor can not rise above 0.7V because it is on the base of a transistor, so it will always be more negitave than the other end.

mike ahh ok that's helped me understand it more :slight_smile:

cheers

Don't tell me magic

I was told in the Air Force when we studied this circuit that it indeed was magic. :wink:

That is if your question really is: upon initial power up for this circuit which transistor will switch on first, Q1 or Q2? And will that always be the case upon each powering up event? Lets let the peanut gallery chew on that one for awhile.

Possible answers:

A: Q1 always.
B: Q2 always.
C: It's always random if either Q1 or Q2 starts first.
D: It can be either Q1 or Q2 but it will be the same transistor starting first for each specific circuit built.
E: Some other magic will decide which transistor starts first.

KirAsh4:
A transistor doesn't have a "minus" side. It has a base, collector, and emitter. Read up on how a transistor works and you'll understand why the resistors are connected the way they are. And that link that Riva gave will explain in plenty detail how the circuit actually works, how the capacitors work and how they affect the circuit. Google itself might not answer your question directly, however it will give you plenty of references of where you can find the answer.

transistors? i know how they work...

re read my question...

The key to understanding a multi vibrator is to know that when one end of a capacitor changes the voltage it is at, the other end also changes by the same ammount.
So if you have 5V across a capacitor and you change the positave end to zero the the other end of the capacitor instantly goes to minus 5V.

This is the basis of a lot of circuits like negitave rail generators and voltage multipliers.

  1. On applying voltage, at the start, do both caps start charging..

Both charge up, but at different rate. The one that gets charged up first flips the other transistor first.

  1. Resistor 2 and 4 are both connected to the Minus side of the capacitors yet connected to a positive + rail... huh? why?

Those capacitors see reverse polarities so the polarity marking is likely to indicate that they are larger electrolytics. They should be non-polarity capacitors.

If the base resistors are too small both transistors will turn on and stay on - make them 20 times larger in value than the collector
resistors and the circuit seems to start reliably (and will even run at down to about 0.8V supply! My circuit seems to start up in the
same state every time, for what its worth - I wouldn't rely on this though unless using non-symmetric component values.

The capacitors see about the supply voltage across them for half the cycle, then a reverse Vbe (0.7V or so) for the other
half cycle - most electrolytics will tolerate this small reverse voltage but its not recommended.

The name for all oscillators of this sort is "relaxation oscillator" - there is a sudden change, then an RC (or RL) circuit discharges
or charges (relaxes...) and then suddenly a new flip in state happens. Positive feedback causes the sudden change in state, which
saturates, then the relaxation phase before the next positive-feedback step can happen. The classic 555 timer circuit is another
good example of a relaxation oscillator.

[edit: just for completeness, I'm using 2N3904's, 4k7 on collectors, 100k on bases, 10uF non-polarized electrolytics.
Runs with a period of about 2.0s at 0.8V, 1.6s at 5V and the rapid switches have a 100ns rise time or so.]

retrolefty:

Don't tell me magic

I was told in the Air Force when we studied this circuit that it indeed was magic. :wink:

That is if your question really is: upon initial power up for this circuit which transistor will switch on first, Q1 or Q2? And will that always be the case upon each powering up event? Lets let the peanut gallery chew on that one for awhile.

Possible answers:

A: Q1 always.
B: Q2 always.
C: It's always random if either Q1 or Q2 starts first.
D: It can be either Q1 or Q2 but it will be the same transistor starting first for each specific circuit built.
E: Some other magic will decide which transistor starts first.

I prefer the first answer given to you were in the air force :slight_smile: - magic :slight_smile:

yup, I'm merely trying to get my head around how it works... making them is simple, but how they work..

but this -5volt swing has got me curious, i need an oscilloscope :slight_smile: - thanks all, tomorrow i'll probe this circuit with my multimeter and give this relaxation oscillator / multivibrator / flip flop /astable osccilator / a good going over lol, i'll increase resistor and cap sizes and try and watch the voltages across the caps and bases of the transistor to see this in action, i built it, it worked i just have to slow it down..

p.s

if i see a yellow canary, it's a canary, if i see an orange canary, i don't start suddenly start calling the canary (even though it's a different colour) something else! (electronic engineers seem to do, 1 name is never enough)

if i see a yellow canary, it's a canary, if i see an orange canary, i don't start suddenly start calling the canary (even though it's a different colour) something else! (electronic engineers seem to do, 1 name is never enough)

Then how would anyone else who never saw the canary know what colour it was XD

The opposite is true with men/women and colours. To me green is green, red is red & blue is blue in almost all there shades but to the wife there are several different names :~

To me green is green, red is red & blue is blue

To me, green might be green, or red, or brown, and so might red. Blue might be blue, or purple, or green. Yellow might be yellow, or green, or orange. I hate resistor colour codes.

I wouldn't rely on this though unless using non-symmetric component values.

It is impossible to have anything other than non-symmetrical component values. All components even if marked with the same value have a tolerance. Even when you try and match components you only acheave a match to a certain number of places. Even if you acheave the impossibility of exactly the same component values, transistors with the same gain, and so on. They are physically diffrent things in physically different places. There willbe differences.

Grumpy_Mike:

I wouldn't rely on this though unless using non-symmetric component values.

It is impossible to have anything other than non-symmetrical component values. All components even if marked with the same value have a tolerance. Even when you try and match components you only acheave a match to a certain number of places. Even if you acheave the impossibility of exactly the same component values, transistors with the same gain, and so on. They are physically diffrent things in physically different places. There willbe differences.

That's pretty much what they said in the Air Force, (plus they mentioned random circuit noise) would cause a preferential treatment such that the starting state on initial power up would always be the same, set or reset. But they also said don't worry about it too much, the circuit will oscillate, that's why it's called a astable multivibrator circuit. :wink:

From wikipedia:

Initial power-up
When the circuit is first powered up, neither transistor will be switched on. However, this means that at this stage they will both have high base voltages and therefore a tendency to switch on, and inevitable slight asymmetries will mean that one of the transistors is first to switch on. This will quickly put the circuit into one of the above states, and oscillation will ensue. In practice, oscillation always occurs for practical values of R and C.

Lefty

plus they mentioned random circuit noise

Yes forgot that bit.

MarkT:
The name for all oscillators of this sort is "relaxation oscillator" -

No a relaxation oscillator only has one active component that is in two states, this is a cross coupled bistable oscillator.

The multivibrator was the first circuit I worked out how it worked. I built a very large one on a wooden board with switches and knobs so I could alter the on and off times independently. I also had relays on the two outputs and used them to control disco lights back in 1964. A few years later I hooked it up to a telephone uniselector and was able to produce a long complex sequence for five lights of 3 amps each. The sparks flew.

this is a cross coupled bistable oscillator.

I don't think so, It's called a astable mutivibrator circuit, and oscillates continously. A bistable multivibrator circuit does not have cross coupled capacitors and requires external signal pulses to place the circuit into it's set or reset state. A bistable can latch last position, a astable is freerunning.

Different side of the pond definitions here I think.