Multivibrators?

Do we analyse multivibrators and oscillators in the same way, i.e. beta*A = -1 . For example, if we look at the multivibrator at

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do we need to calculate the gain with a small signal equivalent circuit or not. Is it a voltage-shunt feedback, what approximations do we do to ease the analysis, and how do we get the equations for the frequency of oscillation?

Reply to
kerntoppi
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No. That technique is only applicable to linear circuits e.g. sine wave oscillators.

The base-emitter voltage ramps as the capacitor charges/discharges. As it passes 0.6V, aided by positive feedback, the transistor suddenly switches hard on/off. The frequency can be calculated by working out how long it takes to charge/discharge the capacitor.

Download LTSpice (free) and load the following as an ASC file:

Version 4 SHEET 1 880 680 WIRE 48 64 176 64 WIRE 240 -16 240 16 WIRE 240 112 240 320 WIRE 160 320 240 320 WIRE 240 320 240 384 WIRE 32 320 96 320 WIRE -48 320 -96 320 WIRE -96 320 -96 112 WIRE -16 112 -96 112 WIRE -96 112 -96 0 WIRE -96 -80 -96 -144 WIRE -96 -144 240 -144 WIRE 528 -144 528 -128 WIRE 240 -80 240 -144 WIRE 240 -144 528 -144 FLAG 48 160 0 FLAG 240 464 0 FLAG 528 -48 0 SYMBOL LED 224 -80 R0 SYMATTR InstName D1 SYMBOL res 224 368 R0 SYMATTR InstName R1 SYMATTR Value 51 SYMBOL cap 160 304 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C1 SYMATTR Value .47µ SYMBOL pnp 176 112 M180 SYMATTR InstName Q1 SYMATTR Value 2N3906 SYMBOL res 48 304 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R2 SYMATTR Value 47k SYMBOL npn -16 64 R0 SYMATTR InstName Q2 SYMATTR Value 2N3904 SYMBOL res -112 -96 R0 SYMATTR InstName R3 SYMATTR Value 3.3e6 SYMBOL voltage 528 -144 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 3 0) TEXT -114 506 Left 0 !.tran 10s

Reply to
ajholme

With this sort of oscillator, you have to analyze several completely separate phases.

One phase starts the moment the output transistor begins to turn off, and collector voltage heads toward ground by the pull down action of the 470 ohm resistor. during this phase, D1 first forward biases and clamps the base voltage for Q1 to no more than a diode drop and C1 charges through R2.

Eventually (this is where a bit of RC analysis comes in) the current through D1 goes through zero and C2 changes charge rate because the current through R1 in series with R2 sets the charge rate for C2. During this phase, the base voltage at Q1 swings between negative one diode drop toward positive one diode drop, where Q1 begins to conduct as the current through R1 detours from C2 to the base emitter junction.

Once Q1 conducts enough to start to turn Q2 on and the voltage drop across R3 starts to increase, a positive feedback loop forms that very quickly drives Q1 and Q2 into saturation because the positive voltage change across R3 gets coupled back to the base of Q1 through C1 and R2, making the current from R1 insignificant. This phase lasts as long as C1 R2 can supply enough current to Q1 to keep both transistors well saturated.

Once C1 charges enough that this current is no longer available, Q1 and Q2, while still conducting somewhat, fall out of saturation enough that the voltage drop across R3 starts to sag. At that moment, the current through C1 R2 reverses direction and Q1 shuts down and you are at the starting point of this description.

Each of these phases has to be analyzed to predict both the on and off time.

--
John Popelish
Reply to
John Popelish

I have noticed that if I increase R3 a little the oscillation stops, why is that, and what alternative I have to obtain a longer period / lower frequency?

Reply to
telepele

I have noticed that if I increase R3 a little the oscillation stops, why is that, and what alternative I have to obtain a longer period / lower frequency?

Reply to
telepele

I have noticed that if I increase R3 a little the oscillation stops, why is that, and what alternative I have to obtain a longer period / lower frequency?

Reply to
telepele

The current through R1 multiplied by the current gain of the two transistors must bias the voltage drop across R3 somewhere between cut off and saturated full on so that the positive feedback through C1 R2 can swing the transistors either full on or full off from that bias point.

R2 is the primary timing resistor that can be easily adjusted. Bigger changes require that you change C1.

--
John Popelish
Reply to
John Popelish

Good explanation. The original of that circuit with identical parts is located here:

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And I copied it from an old GE transistor manual around 1968. It was originally a metronome using germanium transistors with a loudspeaker in place of R3.

-Bill

Reply to
wrongaddress

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