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 http://www.uoguelph.ca/~antoon/circ/2qflash.htm
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?

 

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?

 

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?

 

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?

 

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

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

http://ourworld.compuserve.com/homepages/Bill_Bowden/page5.htm#flash2.gif

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

 

kerntoppi@btinternet.com wrote:

Do we analyse multivibrators and oscillators in the same way,
i.e. beta*A = -1 . For example, if we look at the multivibrator
at http://www.uoguelph.ca/~antoon/circ/2qflash.htm

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

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?

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 .47ľ
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

 

[John Popelish]

kerntoppi@btinternet.com wrote:

Do we analyse multivibrators and oscillators in the same way,
i.e. beta*A = -1 . For example, if we look at the multivibrator
at http://www.uoguelph.ca/~antoon/circ/2qflash.htm
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?

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

 

[John Popelish]

telepele@btinternet.com wrote:

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?

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