When power is applied to the circuit one might expect both transistors to
switch on because both of them have paths to their bases for current to flow
(R2 & R3). If the supply voltage is not applied quickly, i.e. it increases
from zero slowly, this can in fact happen but there is a technique to overcome
this. However, in most cases the supply is established quickly enough
and operation is as follows.
Because of inevitable differences in nominally similar components, and in particular the transistors, one of the transistors will conduct before, or more rapidly than, the other.
Unless stated otherwise any voltage mentioned in the following description implies that it is measured with respect to the negative pole of the supply (0V) and therefore the transistor emitters.
Let’s suppose that TR1 conducts first.
It is held on by the current from the supply through R3 and its base-emitter
junction.
The voltage at its collector will be around +100 mV (Vcesat). Since both capacitors
will not have any charge on them at switch on they will both have zero voltage
across them.
With +100 mV at the left hand plate of C1 and zero volts across it there will
also be +100 mV at its right hand plate and therefore at the base of TR2. This
is not sufficient to cause TR2 to conduct.
Since TR1 is conducting there will be +0.6V at its base. C2 has zero volts across it so TR2 collector will also be at +0.6V.
Two things now start to happen.
Both C1 and C2 start to charge. The right hand plate of C1 starts charging,
through R2, from +100mV towards +9V, and the right hand plate of C2, through
R4, from +0.6V towards +9V.
The next significant thing which happens is when the voltage on the right hand
plate of C1, and therefore TR2’s base, approaches +0.6V.
TR2 begins to conduct.
The collector voltage of TR2, which by now will have increased in value as C2 started to charge through R4, will fall and go down to +100mV.
The voltage across C2 can’t change instantaneously so when the voltage at its right hand plate goes down the voltage at its left hand plate goes down by the same magnitude. If TR2’s collector voltage had risen to, say, +3V, then when it reduces to +100mV (a negative-going change of 2.9V) the left hand plate of C2, and the base of TR1, will also go 2.9V more negative, but starting from +0.6V.
They will end up at –2.3V. This will cut TR1 off.
TR1’s collector voltage will rise towards +9V as C1 initially discharges (left hand plate rising from +0.1V to +0.6V) and then charges (left hand plate rising to +9V while its right hand plate stays at +0.6V).
C2, with its right hand plate at +0.1V and left hand plate initially at –2.3V, will now start to discharge through R3. Its left hand plate rises from –2.3V towards +9V, its right hand plate being held at 0.1V. It will discharge (+0.1V on each plate) and then charge as the voltage at its left hand plate continues to rise.
When this voltage reaches +0.6V TR1 starts to conduct. TR1’s collector voltage then falls from +9V to +100mV. As the voltage across C1 can’t change instantaneously, the same magnitude of voltage change on its left hand plate will also appear on its right hand one. The voltage at TR2 base will therefore go from +0.6V to –8.3V cutting TR2 off.
TR2’s collector voltage rises to +9V as C2 charges through R4. C2’s left hand plate staying at +0.6V.
C1 will now begin to discharge through R2, the voltage at its right hand plate rising from –8.3V towards +9V, reaching +0.1V (C1 discharged), continuing to rise (C1 charging) until it reaches +0.6V. At this point TR2 starts to conduct again.
By this time the operation should begin to sound familiar to you. From now on, however, the negative excursions of voltage at both bases will be to –8.3V.
The above sequence repeats. As each collector voltage alternately changes from +9V to about 0V, the other transistor's base voltage changes from about 0V (+0.6V) to about -9V.
**********************
The above is all very well for the purposes of explaining circuit operation,
but it has one serious flaw.
The max reverse base-emitter voltage of most transistors is around 5 to7. Which
means that operating it with a 9V supply in the circuit as shown isn't a very
clever thing to do.
In order to preserve the life of the transistors if the supply is greater than 5V we need to put a diode, a common or garden signal variety is adequate, in series with the base of each transistor, i.e. between C1/R2 junction and TR2 base and between C2/R3 and TR1 base.
**********************
The voltage at the collectors in the circuit above will fall from 9V to Vcesat rapidly, the waveform at this point being essentially vertical. However, the rise from Vcesat to 9+ will be exponential.
If a vertical edge is required here, a solution is to add two resistors and two diodes as indicated in the diagram below.
A full explanation is in Sec. 11-12 of "Pulse, Digital and Switching Waveforms" by Millman and Taub, published by McGraw-Hill. Library of Congress Catalog Card Number 64-66293.
In applications where the supply voltage rises slowly at switch-on the following modification will ensure correct operation without lock-up.
