The circuit is much more complex than one transistor turning ON and the other turning OFF.
It has hidden secrets that no-one has explained.

I am going into a lot of detail with this FLIP FLOP circuit to show there is a lot to learn about how a circuit works and its main feature is the secret to it being so reliable and capable.
The most difficult electronic component to understand is the transistor. That's because it has so many modes of operation and it reacts differently when it is passing a small current and then a high current. And when a capacitor is placed in a circuit and not being connected to either power rail, it can create a mystifying effect.
We are going to cover one of the cleverest circuits. It has features that guarantee its operation and it has NEVER been described fully in any text book or lecture. It is called the FLIP FLOP or MULTIVIBRATOR or SQUARE-WAVE OSCILLATOR or ASTABLE MULTIVIBRATOR or ASTABLE OSCILLATOR - or constantly operating oscillator.
Although the circuit is very simple, its operation is very complex and two terms have to be introduced to describe its operation. These two terms have never been mentioned before.
That's because no-one has dealt into the depths of its operation.
The two transistors are just amplifiers. But because they are cross-coupled, one amplifier is controlled by the other and then the second amplifier is controlled by the first.
But when one stage turns ON, the other stage does not just turn it ON FULLY, but turns it on EVEN HARDER than it needs to and this allows it to pass a very high current and makes the transistor a very good conductor. This extra "turn-on" is called SUPER-SATURATION. and allows loads such as globes and motors to be connected as loads, as they require up to 6 times more current to start to turn ON.
The other feature of the circuit is called REGENERATION.
One transistor starts to turn ON the other, and this effect is passed back to the first transistor and allows it to turn on the second EVEN MORE. This "runs around the circuit" very quickly 1,000 times or more and in the end, one transistor is turned on VERY HARD with more base current than it needs and it becomes SUPER SATURATED. The other transistor is fully turned off.
Because the transistors do not work within the parameters of any data sheet, we do not need any mathematics to explain how the circuit works. Just an understanding of the brilliance of the inventor of the circuit.
This was not the first flip-flop circuit to be invented. The others were very complex and were a complete failure. As the voltage of the supply dropped, they stopped working and they did not supersaturate the transistors.  

The two wires in the middle of the circuit do not touch.  That's why we say the circuit is "cross coupled." One side is connected to the other and the second stage is connected to the first.
All the components and the transistors will have slightly different values and one transistor will start to turn ON when the switch is closed.
Suppose the second transistor starts to turn ON via the 10k base resistor and the voltage on the collector drops a small amount. This drop will be transferred through the 100u electrolytic so that when the positive lead of the electro drops a very small amount, the negative lead will fall the same amount.
The negative lead is connected to the base of the first transistor and it will work against the effect of the 10k resistor connected to its base. The 10k was having a slight effect on trying to turn the transistor ON but now the 100u is reducing this effect.
This means the first transistor will be starting to be turned OFF and the voltage on the collector of the transistor will start to rise.
The 470R resistor and the LED will pull the positive lead of the first 100u towards the positive rail and the negative lead will follow. But the negative lead cannot rise more than 0.6v and so, when the positive lead is pulled towards the positive rail, the effect is to charge the 100u.
This charging current flows through the base-emitter junction of transistor 2 to turn it ON more.
Transistor 2 turns ON more and the voltage on the base of transistor 1 becomes so small that transistor 1 is not turned ON at all and now the first 100u electro gets charged by the 470u and LED 1. This current is much more than the second transistor needs to illuminate LED2 but if we had put a high current device in place of  LED2, it would also have operated.
The transistor is being turned ON so hard that anything you put in the collector circuit will work. That's why we say the transistor is SUPERSATURATED.
To get to this super-saturated condition the two transistors "feed" each other with the "turning-ON" effect, with hundreds and hundreds of cycles we call REGENERATION and eventually the effect produces SUPERSATURATION. 
Normally regeneration does not have this enormous effect but in this circuit it makes the circuit very reliable and almost any transistor will drive almost any load.
The circuit is not just one transistor turning on the other. The TWO transistors operate in a cyclic or feedback way with POSITIVE FEEDBACK, to create a very fast switching action that is a SQUARE WAVE with very fast rise and fall time.
One transistor feeds back to the other to turn it OFF more and more and now the components in the circuit take over to supply current to turn one transistor ON FULLY. It is now the components that turn the transistor ON fully and the other transistor does NOTHING. This is quite a surprise. Show me anywhere, where a text book or instructor has even hinted at this during part of the cycle!!!
We have the state where the first transistor is OFF and the second transistor is fully ON.
You can see this action in the following animation:


The animation only shows part of the action. But it does show the base of the transistor seeing a negative voltage from the electrolytic that can be about 6v below the 0v rail.

What makes the circuit change state?
The first transistor has no voltage on the base because the small voltage across the second 100u has made the negative lead put a negative voltage on the base of the first transistor.
The circuit would stay in this state but two things are happening that make the circuit change states.
The 10k resistor on the base of the first transistor is beginning to charge the second 100u and the voltage on the base of the first transistor is starting to rise. When it gets to 0.55v, the transistor starts to turn ON.
When it turns ON a small amount, the voltage on the collector drops a very small amount. This means the voltage on the positive lead of the first 100u reduces slightly and this effect is passed through the 100u to make the negative lead drop by the same amount. Another hidden secret has also occurred in the circuit. The first 100u has charged considerably during this time and the current being passed to the base of the second transistor has been reduced considerably but the transistor is still fully turned ON. But not turned on "Extra Hard."
This means that when the current into the second transistor is reduced, it will start to turn OFF. And that is what happens. The first transistor starts to turn ON and the current through the first 100u has reduced considerably because the capacitor is in a charged state and when the voltage on the positive lead is reduced, this effect is passed to the base of the second transistor to start and turn it off.
The voltage on the collector of the second transistor rises and the second 100u starts to charge via the base-emitter junction of the first transistor and the two transistor form a loop of POSITIVE FEEDBACK called REGENERATION and the end result is a change of state where the other transistor becomes FULLY SATURATED.
It is no wonder this complexity has never been described before but when you build the circuit and add car globes and find it does not work, you need the knowledge of SUPERSATURATION to use transistors that allow a high current to be delivered to the load AND the power supply must be able to deliver the high current. Car globes take 6 times more current when they are starting to illuminate and if the transistors and power supply cannot provide the current, the circuit FREEZES.
It is the hidden secrets of how a circuit works, that you need to know. But these have never been covered in any text book. 
For instance, just before the circuit changes state, the transistor that has been turned ON "with great force," has now had this "force" reduced considerably because the capacitor creating the "great force" is in a charged state and the current flowing through it is considerably reduced,  to a point were the second transistor is turned ON but not excessively. Any slight movement in the down direction of the first 100u will reduce the voltage on the base of the second transistor and remove the "turn ON" effect.  If the first 100u was still  turning on the second transistor "with great force," a slight reduction in its ability to deliver a reduced voltage would simply change the state of the second transistor from SUPERSATURATED to FULLY TURNED ON and the circuit would not start to change state.
One thing you have to understand is the interspersal of voltage and current. The transistor actually works on current but when you deliver a higher voltage via parts of the circuit, this translates to a higher current and the transistor reacts.
You have to forget what you have read, seen and been told as none of it related to understanding how the circuit actually works. It's a mystery that no-one has exposed. You only find out these things when you experiment and find that everything you have learnt has not solved the problem.


We will cover the concept of REGENERATION again.
Regeneration is basically POSITIVE FEEDBACK and when the feedback is provided by another transistor or other components, this feedback can be 100 times stronger than needed.
In general, feedback occurs until the transistor is fully turned ON by another part of the circuit and at this stage, this part of the circuit cannot do anything more. This is when another part of the circuit "reduces its feedback power" and during this time, the cycle generates either the High or LOW part of the cycle. In other words, half the duration of the cycle is generated.
There are two sections of the circuit and they take turns in being turned ON or OFF and when the circuit decides to change from one state to the other, the change takes place very quickly.
This is why the output is a square-wave with steep sides.
But more important, this action is very reliable and has a very strong ability to change states and is impossible to stop the action.
Going to the circuit above, we start with the second transistor turning ON. It turns ON via the low resistance of the LOAD of the first transistor and the uncharged 100u electrolytic. This supplies a high current to the second transistor and it turns ON and the current can be 10 times more than is necessary. This pulls the second 100u down and it puts a zero voltage and a slightly negative voltage on the base of the first transistor to turn it OFF.
The first 100u keeps charging and as it charges, the current into the base of the second transistor reduces. It reduces to a very small amount and the second transistor cannot keep fully saturated. The collector voltage rises a small amount and the second 100u rises. At the same time the second 100u is being discharged via the 10k resistor and the voltage on the base of the first
transistor is rising. When it reaches 0.65v, the first transistor starts to turn ON and pulls the first 100u down.
This turns OFF the second transistor and makes the second 100u rise even more and it rises very quickly.
This produces the effect of both transistors changing state very quickly to repeat the cycle. 
No external pulse is needed for the circuit to "cycle" and produce a square-wave. 


1. Which transistor turns ON first?

Answer:  It mainly depends on the gain of each transistor and the state of charge of the two 100u electrolytics.

2. Does the transistor turn ON more when the capacitor (electrolytic) is charged or uncharged?

Answer: The transistor is turned ON the most when the capacitor (electrolytic) is starting to charge as the voltage across it is the highest.
When it is nearly charged, the current into the transistor is so small that the transistor does not turn on as much as this is when the circuit starts to change states.

3. For a 9v supply, what will be the approximate initial base current?

Answer: We take the characteristic voltage of 1.8v for a red LED from the supply voltage and the base-emitter voltage of 0.7v to get an effective charging voltage of 6.5v  This voltage appears across the 470R resistor to get a current of  about 13mA  The transistor will have a gain of 100 or more and will theoretically be able to deliver a current 1,300mA and that's why a high current device can be used in the LOAD. Obviously the LED only requires 10mA and the circuit will not deliver any more than about 13mA, so nothing will be damaged.

4. If a transistor (in the circuit above) is capable of passing (delivering) more than 1 amp, why doesn't the LED receive the 1 amp and get damaged?

Answer: The resistor (in series with the LED) is called a CURRENT LIMITING RESISTOR and its job is to limit the current to about 10mA to 15mA so the LED is not blown up.  The resistor uses Ohm's Law to limit this current.

5. What are the two new terms you have learnt in this discussion?

Answer: Super Saturation and Regeneration.

6. What does Super Saturation mean?

Answer:  Super Saturation means the transistor is turned ON with a higher base current than needed to make sure the transistor will turn ON fully.

7. What does Regeneration mean?

Answer:  Regeneration means a signal is passed from the output of a stage to the input to increase the amplitude of the signal.

8. Is Regeneration a Negative or Positive feedback signal?

Answer: Regeneration is a Positive feedback signal. A positive signal makes things: "worse and worse."  In other words it turns on the circuit: "more and more."

9. Why is this circuit so reliable?

Answer:  It revolves around the fact that each transistor is turned ON fully and this is classified as a DIGITAL condition or a POSITIVE or DEFINED STATE or ACCURATE STATE or GUARANTEED STATE.

10. How is this Digital State produced?

Answer:  The Digital State is produced by Regeneration producing many feedback cycles that result in the transistor being more than fully turned ON. And this process occurs very quickly.

11. Will this Digital State damage the transistor?

Answer:  No. The surrounding components will limit the maximum base current to an acceptable level.  . . .  even though the level is 10 times higher than needed.

12. What determines the duration of each cycle?

Answer: The 100u electrolytics and 10k base resistors. 

13. If a 100u is replaced with 47u, what happens?

Answer: One half of the cycle will be much shorter than the other (in duration - in timing) In other words one LED will flash for a shorter time than the other.  This is called an "imbalance." You can also connect a single LED and get a flashing LED "road sign."

14. Will this circuit work?

Answer: Yes.  The circuit above will work. The circuit is so tolerant and so reliable that it will work with almost any value components and the LEDs can be placed in the emitter circuit.
This will change the timing when the transistor turns ON,  but this will not be noticed in the flashing of the LEDs. The LOAD is normally placed in the collector-circuit.

15. Will this circuit work?

Answer: The circuit will not work because a transistor cannot be turned ON during part of the cycle.

Firstly, the circuit has to be redrawn so we can see the fault.
In our circuit, and from the animation above, you can see the left transistor has the negative voltage on the base removed by the 10k resistor and when the base is zero, the 10k starts to charge the 100u in the opposite direction, so that when the base voltage reaches 0.65v, the left transistor starts to turn ON.

In the circuit above, the 10k resistor discharges the 100u, but it cannot increase the base voltage above 0.3v as the right-hand transistor is fully turned ON and it has 0.3v on the collector..
This means the left transistor will not be turned ON and cycle will stop.

15a. Do you understand the concept that the 100u electrolytics are rising and falling in the circuit by an amount equal to approximately the value of the supply voltage.
This is the basis of seeing how the circuit is operating.

16. Name a benefit of the circuit:

Answer: It will illuminate globes and motors as they need a very high current (up to 6 times more) to start the illumination or revolving of the armature.

17. How is a transistor turned on more than "normal" in this circuit?

Answer: The current delivered to the base is very high when the 100u is uncharged and supplied via a 470R resistor and the LED.

18. Does this circuit start every time?

Answer: Yes.  It may start with both transistors turning ON at the same time but very soon one transistor will turn ON more and prevents the other from turning ON any more and the circuit will generate the FLIP FLOP action.

The Flip Flop circuit shows the effect of a capacitor (electrolytic) when placed in a circuit and is not connected to either power rail. In this arrangement it connects two sections of a circuit when charging, but when fully charged it provides SEPARATION.
The circuit is able to provide a high current to a LOAD component, but a LED does not require this feature. It only needs a few milliamp.

You can connect a single LED to the circuit, but that is a waste of energy as half the energy is lost in the dummy LOAD.
A single output version of the FLIP FLOP uses a different arrangement of parts and is called a FEEDBACK OSCILLATOR.
Here is the circuit:

You can add a LED in series with a 470R current limiting resistor to produce a very bright flash for a "road-works" sign.

We are going to study more on the effect of a capacitor in a circuit.
This is a very important feature that has never been covered in any text books.
The circuit we will be studying is called a FEEDBACK OSCILLATOR and is basically a HIGH GAIN AMPLIFIER that is turned ON and OFF by the effect of a capacitor charging and when it is charged, it does not have any effect of keeping the circuit fully conducting and so the circuit starts to turn OFF and it makes the capacitor turn the circuit OFF more and more. It does this by making the charge in the capacitor "turn the circuit OFF". Eventually the charge in the capacitor becomes zero and a resistor in the circuit starts to turn the amplifier ON for the next cycle.

Here is the circuit we will study:

The circuit starts in the OFF state and the capacitor charges via the 10k and 330k.
This takes a long time and produces the long OFF time.  When the base of the PNP transistor sees a voltage about 0.6v below rail voltage, it turns ON and this action turns on the NPN transistor. This pulls the positive lead of the 1u electrolytic down and the small voltage it contains is delivered to the PNP transistor to turn it ON harder. The right lead of the capacitor is now pulled down to almost 0v via the NPN transistor and now the capacitor starts to charge via the 10k resistor and the base-emitter junction of the PNP transistor. Eventually the capacitor is fully charged and the current through the PNP transistor base reduces to a point where the transistor is not turned ON as much and this starts to turn OFF the NPN transistor. The voltage on the collector of NPN rises and raises the 1u and this completely turns OFF the PNP transistor. The voltage on the base of the PNP is actually higher than rail voltage and the 1u is gradually discharged via the 10k and 330k to repeat the cycle. Because this circuit has such a fast change from one state to the other it is classified as a flyback arrangement or flyback circuit.

But we mainly want to explain how the capacitor works . . the electrolytic
At the start, the 1u is charged via the 1k, 10k and 330k. As soon as the PNP transistor sees a voltage on the base that is 0.6v lower than the supply of 6v, it starts to turn ON. This action will turn on the NPN transistor.
The 1u electrolytic will be pulled down by the NPN transistor, and cause it to be charged at a higher current via the collector-base junction of the PNP transistor and 10k resistor. This higher current makes the PNP turn ON more. Eventually the 1u will be fully charged and the PNP will turn off a small amount and this action will turn off the NPN transistor.
The fully charged capacitor will rise high in the circuit and produce a voltage on the base of the PNP transistor that is higher than rail voltage.
At this point the two transistors are "out of circuit" and the components will discharge the capacitor to produce the "OFF time."
When the cap fully discharges it begins to charge in the reverse direction to start the cycle again.

Feedback Oscillator Animation

You need to view the animation a number of time to get the full concept of what is happening. The voltage on the base of the PNP transistor will be higher than rail voltage for part of the cycle. This is only obvious when you start to measure the voltages or see the circuit working with a Digital Oscilloscope.
One way to "see" what is going on in a circuit is to add a speaker.
Here is a circuit. Each half of the circuit has different timing components and you can change these to see the effect it has on the sound.


These circuits show how the driver transistor turns ON with a high "current capability" that it is able to illuminate a globe.

When the two transistor are directly connected together, as shown in the circuit below, they form a single transistor (called a Uni-Junction Transistor) and this circuit simulates a UJT. The output is a very short duration spike and has a very low current drain.
We are not going into UJT designs as this is a different area to study and the capacitor in this circuit is simply a storage device and not one that "jumps up and down.".

There are many ways to produce an oscillator. You will be able to recognise some of the ways and one of the ways is to understand the action of a capacitor that is not connected to a power rail. 
More circuits using this feature are covered here:
The Transistor is a POTENTIOMETER.
and this link

The "how and why" of a capacitor works in a circuit has never been discussed in any text book or University demonstration and that's why we have covered it in detail.
You have to understand a capacitor "jumps up and down in a circuit" and sometimes connects two parts of a circuit and then separates two parts when it is charged. And it can produce a negative voltage or a voltage higher than the supply.
Unless you can "see" a circuit working in your mind, you cannot design or repair ANYTHING. It will just be a mystery.

A capacitor is simply a rechargeable battery.

Understanding the operation of a capacitor is very basic but very complex. If you don't understand the principles of electronics, it is pointless proceeding to higher levels.
It's like learning to repair a car and not learning how to unscrew a nut.

Page-1 Common-Emitter stage
Page1A. more on the stages
Page-2 emitter-follower - common collector stage
Page-2A common base stage
Page 3  Coupling stages
Page 4  More on Coupling Stages
Page-6 All the "YouTube" mistakes

The REAL Transistor Amplifier

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