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I asked a question here about why I was getting a sawtooth wave from a two transistor oscillator. While that question is answered, I don't understand how the circuit produces oscillation in the first place.

Circuit Diagram:

Oscillator

This is the circuit I've built.

In order to help the reader identify the errors in my reasoning, the following is exactly what I think happens.

1) Random transistor turns on first (Call it T1). Because there is now a pathway to ground for C1, C1 begins to charge with its right plate being positive and its left plate being negative through R2.
2) C1's right plate reaches a threshold where T2 is turned on. Turning on T2 gives a pathway to ground to C2, causing C2 to charge through R3 with a positive voltage on its left plate and a negative voltage on its left.
3) The increase in negative voltage on C2's left causes T1 to turn off. T2 can't stay on because there's no pathway to ground for C1.
4) Everything stops.

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    \$\begingroup\$ Thinking (1) No - The first thing that happens is ONE of the transistors turns on. Which one depends on a number of factors so its pretty random and frankly its not important in most cases. If you need to have a precise start up, use a micro. Issue (1) No. Only one transistor's base is held OFF at any time otherwise the circuit can't oscillate. (2) For charge to flow there must be a voltage difference as well as a pathway. One cap is held at 0.6V (clamped by B-E ) at the base and +V at the collector of the other transistor. The other cap is fixed at about 0V with its neg end charged from -V. \$\endgroup\$ – JIm Dearden May 20 '15 at 19:02
  • \$\begingroup\$ Updating question again. I'm still not quite there, I'm sorry. \$\endgroup\$ – Allenph May 20 '15 at 19:27
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    \$\begingroup\$ I think if you studied simple BJT amplifiers first you might make the small leap to this type of oscillator. \$\endgroup\$ – Andy aka May 20 '15 at 19:44
  • \$\begingroup\$ I don't think you've grasped the really important concept (and its consequence) that you can't instantly change the voltage across a capacitor. Initially (with the circuit unpowered) both capacitors have 0V across them. When you switch ON lets say T1 turns ON. That means that the base voltage of T1 must be 0.6V and that the base voltage of T2 must be LESS than 0.6V (because T2 is off). The collector of T1 falls from +V to about 0V. Now apply the capacitor rule. The other plate of C1 must also drop by exactly the same voltage (from 0 to -V) in order to keep the voltage ACROSS C1 = 0V \$\endgroup\$ – JIm Dearden May 20 '15 at 19:55
  • \$\begingroup\$ @Andyaka here here (+1) \$\endgroup\$ – JIm Dearden May 20 '15 at 21:16
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1) Why are R4 and R1 required at all? Why would we limit current flowing into the transistors and charging the capacitors?

Transistors operate over a range of collector currents - you can look this up[ for different types (see https://www.sparkfun.com/datasheets/Components/2N3904.pdf) . The minimum values used for R1 and R4 will determine the maximum collector current. This value can easily be calculated (using Ohm's law) by assuming that all the supply voltage is dropped by the resistor. In the case of the 2N3904 the maximum current is 200mA. Failure to limit this current would damage/destroy the transistor.

The also act as part of the circuit to produce a switching voltage for one side of the capacitors

2) Why are R2 and R3 included in the circuit? It seems that they have absolutely no function.

R2 and R3 have two functions.

The first is to switch the transistors ON by connecting the bases of the transistors to the positive supply through a suitable resistor that limits the base current to a safe value.

When you first power up the circuit ONE of these transistors turns on first and start the process.

The second function is to charge the other capacitor plates attached to the bases of the transistor by connecting them to the positive supply.

When one transistor turns ON it causes the voltage at the base of the other transistor to go to a NEGATIVE voltage and turn OFF. This negative (base) voltage is also connected to the positive supply through either R2 or R3 (depending on which base we are looking at). The voltage ACROSS this resistor at this time is about TWICE the supply voltage (+V to -V). {this effect is sometimes used to produce a negative supply from a positive supply or to double the voltage of a supply}

The capacitor plate has to charge from about -V to +V and when it gets to 0.6V (just over mid way) it causes the transistor to turn ON. This is about 50% of the final voltage its trying to charge to and takes about 0.7 x time constant (either C1R2 or C2R3)

3) How are C1 and C2 discharged to allow the cycle to repeat? If they are not ever discharged, after one cycle the circuit would stop working because C1 and C2 would have a voltage high enough to limit current flow into the base of either of the transistors, effectively putting halt to the oscillations.

The trick to understanding this circuit is knowing that you can't instantly change the voltage across a capacitor.

When T1 switches ON the base of T2 is taken to a negative voltage and turns T2 OFF. The capacitor (C1) then takes time to charge back up to 0.6V through R2. When it eventually gets there (after about 0.7 C1R2) it turns T2 ON. The voltage at T2 collector suddenly drops and takes the base of T1 negative which turns T1 OFF . The capacitor (C2) attached to the base of T1 takes time to charge up through R3 until it gets to 0.6V at the base of T1 and turns T1 ON. This then results in T2 turning OFF again and the cycle keeps repeating.

The voltage at the bases of the transistor can't get higher than 0.6V because the base-emitter junction acts as a forward biased diode and clamps the voltage at that level. This ensures that the step turn OFF pulse voltage will take the base negative by about the supply voltage - 0.6V every time.

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  • \$\begingroup\$ I think I've got your meaning now. I've edited my question to what I now think happens. \$\endgroup\$ – Allenph May 20 '15 at 18:07
  • \$\begingroup\$ I think it may be worthwhile to note that if R2 and R3 would be small enough relative to R1 and R4 to turn on the transistors fully in the absence of the capacitors the circuit could get into a "locked up" state where both transistors simply sat "on". Having those resistors be too weak to fully turn on the transistors will mean that if something--even a bit of noise--causes one transistor's base current to increase momentarily, that will cause a drop in its collector voltage which will get capacitively coupled to the other transistor's base, causing that transistor to turn off slightly... \$\endgroup\$ – supercat May 20 '15 at 20:47
  • \$\begingroup\$ ...thus further increasing the current available to the first transistor's base. Consequently, while lock-up could occur with some resistor values, properly chosen resistor values will ensure that any equilibrium state will be unstable. \$\endgroup\$ – supercat May 20 '15 at 20:48
  • \$\begingroup\$ @supercat Properly chosen values for resistors are absolutely de rigueur for a fully functioning circuit but the problem here is the understanding the very basics of operation. Its a very robust circuit and quite tolerent to a wide range of values (within reason as you have correctly pointed out). The initial values shown in the OP are fine and would work. \$\endgroup\$ – JIm Dearden May 20 '15 at 21:14
  • \$\begingroup\$ @JImDearden: Indeed, but I remember one of my early reactions to that circuit was wondering why there's no danger of a state where both transistors are on and stay on. Saying that R2/R3 don't have enough "oomph" to turn on the transistors solidly without help from R1/R4, and that R1/R4 can only help turn on one transistor when the opposite transistor is off, would clarify that. \$\endgroup\$ – supercat May 20 '15 at 22:28
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The two capacitors provide a negative coupling between the two transistors: when one goes into conducting, this keeps the other out of conducting.

1) Without R1 and R4 there would not be a negative going voltage on the collectors when a transistor gets into conducting: it would just be the power supply, no change. And a capcitor does no 'conduct' a 'no change'. Also, a conducting transistor would short the power, which is not a good idea.

2) R2 and R3 are there to let the transistors start conducting. without those resistors the two transistors would forever be off (non-conducting).

3) When a transistor starts to conduct, it will cause the other to be non-conducting, hence the collector voltage of that transistor will rise to the power level. This discharges (or charges, depending on your perspective) the capacitor.

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  • \$\begingroup\$ Why would having one transistor on stop the other from conducting? \$\endgroup\$ – Allenph May 19 '15 at 9:18
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    \$\begingroup\$ When a transistor starts conducting, the voltage at its collector drops. The capacitor couples this voltage drop to the base of the other transistor. Untill the base resistor has recharged the capacitor to ~ 0.6V that transistor will be off. \$\endgroup\$ – Wouter van Ooijen May 19 '15 at 9:38
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Let's start with the state where T1 is off, and T2 is on.

C1 has a path to ground through T2.

The left plate of C1 builds up a positive charge, while the right plate of C1 is pulled down to negative by T2.

The base of T1 is slowly building up a positive charge through R3.

Once enough charge builds up on the base of T1, T1 turns on and plunges the left side of C1 to negative.

Due to the coupling effect / displacement current, the right side of C1 goes even more negative. It actually goes below zero. This shuts off T2 due to negative voltage on its base.

Now T1 is on, and T2 is off. The process repeats.

The book Make: Electronics by Charles Platt describes this exact circuit.

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