Skip to content

Astable – The completion

Astable - Final circuit?

This is the last circuit proposed as working for an astable multi-vibrator.
It's now time to understand why this can or not correctly operate and if some issues are still present.

Due to the presence of R and C for each branch we're sure that the timing feature is somehow guaranteed; but how exactly is a matter we're going to see.

As already said at the switch-on one of bjts surely goes in saturation before the other which is so forced to go to interdiction.
We added the capacitors not only to make a timer as proposed but to alternates the states for the outputs too.

Let's suppose bjt2 wins the lottery and goes to saturation: its collector drops form Vcc to nearly 0.2V.
Due to the transitory nature of this falling signal, the capacitor won't block it at first instance so that it (the drop of  ~Vcc) will be replicated at its other top which connects directly to base1, the new value of which becomes:

newVb1 = Vb1 - ~Vcc

Astable - "What if"

where Vcc is known, newVb1 can be measured on the run and Vb1 is the initial value unknown yet.
Vb1 can be calculated within a certain range because it can be at maximum equal to Vcc so that newVb1 becomes quite zero, enough to set the bjt1 in interdiction.

Next step necessary wants the capacitor2 to charge in order to get enough voltage to drive the bjt1 in saturation from the interdiction.

During the transitory bjt1 is electrically open so its collector doesn't experiment as the same drop as the other branch and the cap1 keeps its high value on both its top... or at least it should.

Indeed there's a little problem as suggested at the end of the previous post.
Can the circuit as shown in the first picture really work?

No, or at least not in the way we want.
The clue lies just in capacitors and how they get charged after the drop in voltage at the start.

If you take a better look their inner tops are connected only to the opposite bjt bases, so they have not a strong voltage reference: they're floating.

This entails mainly two things:

  1. both the bases have unknown values which this way remain so;
  2. the capacitors cannot be charged after the drop.

How to solve it?
We need a stable reference and we have only two possible: Vcc and ground.
Clearly ground cannot be the chosen one because otherwise both the bases would be fixed to it so totally disabling the circuit from working; Vcc remains.
Again we have the same problem: both the bases fixed at Vcc.

So what?
We need a resistor to get a controlled connection: but now which connection between the pull-down and the pull-up?

With the pull-down then the minimum voltage on a base can become:

0 - ~Vcc = - ~Vcc

which could lead to problems if the bjt in use has a Vbe reverse voltage less than ~Vcc.

Astable - Final circuit

Let's try the pull-up.
It's obviously the right solution: they still limit the minimum base voltage to nearly 0V and give the capacitors the way needed to get recharged when required.

This leads to the final circuit as shown here beside.

Hope the posts can be useful to approach the right way of thinking a problem from the start, then how to analyse the weak points and bottle-necks.
Next time, a realisation on breadboard.

Thank you!