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what makes this oscilator oscilate?

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julian403

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what makes this oscilator oscilate

Hello All.

I've a question regarding this circuit (how its works)

Capture.JPG

What I could see by the simulator and implementing this circuit (measuring the voltages and currents) its:

At first both transistor get the saturation condition: Vbe = 0.7 V and Vce < 0.7V but Vce starts to decrease by time (for both transistor) until Vce reach a low value (I measured 0.150mV) after that one of the transistors get cutoff state (increasing the Vce). But Why this?

I means if both are in saturation. What makes that Vce increase after it got a low Vce?

I think that it has to be because Vcb (for this pnp transistor) is equal to 0.6 V (direct polarization) and there are both junction in direct polarization and it makes Ib goes to 0A. Following ebers moll model
 

Re: what makes this oscilator oscilate

It is difficult to know what parameter variation made the simulator in order to start the oscillation. If you check it with different simulators, other transistor will start up first. In real life due to parameter variations one cap will charge first and pull down the other transistor and so on. That is the startup.
 
Re: what makes this oscilator oscilate

Thanks but I dont know what is the charge path.

When we power on the circuit, it stablish the Vbe potential. For Q1 it will be:

Vbe1 =Vcc - I R3 and for Q2 Vbe2 = Vcc - I R2

So, if we take a look at C1

one potential will be Vbe2 and the other Vce1

So, the voltage across C1 will be

Vc1 = Vbe2 - Vce1

And for almost all transistor Vce(sat) is less than Vbe

vc1(t) = 1/C * ∫ ic1(t) dt

(Vcc - 0.7= = i(t) R2+ 1/C * ∫ ic1(t) dt

ic1(t)= (Vcc-0.7) * exp (-t / R2 C1)

and

ic2(t)= (Vcc-0.7) * exp (-t / R3 C2)

When the capacitor is completed charge, ib for will increase (lets say that the C1 charge first)

ib2= (Vcc - 0.7)/R2

where the diference was

Δ ib = (Vcc - 0.7)/R2 - ic1(t)

and as ib2 increase, so Ic2 will increase

Ic2 = β ib2

and Vce2 will decrease because on R4 the voltage will be bigger. and then?

So. What is the charge path?
 
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Re: what makes this oscilator oscilate

The capacitors charge and discharge alternately. If you watch the voltage on the bias wires, you'll see them go briefly into negative polarity, shutting off a transistor (while the other transistor turns on).

You can watch an animated video of the astable multivibrator on my Youtube channel (patientbrad). It's based on my homebrew animated simulator program. Current flow is portrayed moving through wires. Capacitors visibly charge and discharge.

'Astable Multivibrator simulation (Animated)'

www.youtube.com/watch?v=HjSgdcgYnFk
 
Re: what makes this oscilator oscilate

The capacitors charge and discharge alternately. If you watch the voltage on the bias wires, you'll see them go briefly into negative polarity, shutting off a transistor (while the other transistor turns on).
a negative voltage?? why there is negative voltage if its a simple voltage source?

I think that the simulation is wrong. Why there isnt current through the first base transistor at min 0:54? that only can happen if Vb = 0[V].

- - - Updated - - -

ohh I think I got it.

Let me know if this is right. OK?

At first when both transistor are off. The voltage are:

capture1.gif

One of the transistor turn on, lets say it the right hand

So, the Vce goes to saturation and we can say that there isnt voltage drop from colector to emitter, so:

capture2.gif

But the question here is this:

if the collector goes to 0[V]. Why the positive capacitor plate doesnt send all charge to ground? or it has to send the charge to the other plate?

I mean, I think that the charge which is on negative capacitor plane has to go to positive voltage source potential and the positive capacitor charge has to go to ground. How it really Is this?
 
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Re: what makes this oscilator oscilate

Hi,

This is oscillator circuit. We are using two symmetrical circuit, but the there is some change in parameter(characteristics) of this symmetrical circuit. So one of them will start first ans one will stays off.

But the question here is this:

if the collector goes to 0[V]. Why the positive capacitor plate doesn't send all charge to ground? or it has to send the charge to the other plate?

Yes, it will send it's charge to ground. This is the reason why it oscillates.
Want to understand whole process? or you have difficulty in this mode of operation only?


Thanks.
 

Re: what makes this oscilator oscilate

So. What is the charge path?

The astable multivibrator is based on cross-coupled transistors. This cross-coupled arrangement is also the basis for the RS flip-flop (bistable multivibrator).
By adding two capacitors in strategic locations, it cleverly morphs the RS flip-flop into oscillating charge/discharge action (astable multivibrator).

As for current paths through C1, it is R1-Q2_bias during one half of the cycle.
Then R2-Q1 during the other half of the cycle.
 

Re: what makes this oscilator oscilate

Assume one transistor turns ON faster than the other.
Voltage across C1 "Vc1" has the positive ref voltage in the collector of Q1. Voltage across C2 has the positive ref voltage at the base of Q1.

At startup, Vc1=Vc2=0.
You know in a RC circuit this: v(t)=Vfinal+(Vinitial-Vfinal)*e^(-t/tau)

Assume Q2 ON and Q1 OFF (because happened for Q2 to turn ON faster):
Transistors will switch when Vc2=0.7, so we need the time it take for Vc2 to reach 0.7 V starting at 0 volts => 0.7=5+(0-5)*e^(-t/(C2*R3)) <=> t=0.156 s
Vc1(t=0.156)=4.3+(0-4.3)*e^(-t/C1*R1)=3.35V which is the IC for the next state.

Now Q2 OFF and Q1 ON (because it has switched from the last state):
Transistors will switch when Vc1= -0.7V, so we need the time it takes for Vc1 to reach -0.7V starting at 3.35 V => -0.7= -5+(3.35+5)*e^(-t/C1*R2) <=> t=0.686 s
Vc2(t=0.686)= -4.3 +(0.7+4.3)*e^(-t/C2*R4)= -4.29 V

Now Q2 ON and Q1 OFF again:
Transistors will switch to the next state when Vc2=0.7 V, so we need the time it takes to reach that => 0.7=5+(-4.29-5)*e^(-t/C2*R3) <=> t=0.796 s
Vc1(t=0.796)= 4.3+(-0.7-4.3)*e^(-t/C1*R1)=4.297 V

And so on..
 
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Re: what makes this oscilator oscilate

You cannot use 9V to power that multivibrator oscillator circuit because the maximum allowed reverse E-B voltage of most transistors is 5V, above which they have avalanche breakdown. Additional diodes are needed to prevent each E-B from becoming higher than 5V.
 
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