Understanding the following flip-flop

I'm trying to understand the following flip-flop circuit using 2 NPN transistors as explained in this video : My apologies for the lengthy post. I noted down all the things that weren't clear to me as questions in this post (Q1,Q2,...)

http://www.youtube.com/watch?v=IykOrxVcdyg

The full circuit diagram:

After applying power, 1 LED (the right one) is lit. Reasoning here is that this is the case because only 1 transistor is switched on. In this case the right transistor is turned on causing the right LED to turn on.

Q1 : Am I correct in assuming the LED is lit up because of the following current flowing:

Vcc -> right LED -> right 470ohm -> right Collector -> right Emitter - 100Ohm - GND
  • The transistor allows the current to flow through the right LED to ground because it is turned on.
  • Had the transistor been turned off, there would have been no path to ground for current wanting to flow trough the right LED).

Q2 : Current that's passing through the right 470ohm resistor can move in 2 ways :

Vcc -> right LED -> right 470ohm -> right transistor Collector -> right transistor Emitter - 100Ohm - GND   
Vcc -> right LED -> right 470ohm -> left 10K resistor ->left base transistor.

Wouldn't that second current flow also turn on the left transistor, causing the left LED to turn on ?

This is the flow when the right transistor is turned off, but even when it is turned on, won't you still have current flowing through that path ?

  • When the right transistor is turned off, current can only follow the path to the left 10K resistor. (as the transistor is wide open).
  • However, when the right transistor is turned on , current can flow through the transistor OR through the left 10k resistor

Q3: I know current is lazy and that it picks the path of least resistance, but in parallel circuits, you always have current going through all branches right ?

  • More current would flow into the branch that has less resistance (the transistor being turned on).
  • Less current would flow into the branch that has higher resistor (the 10K path from point nr 2).

At 2:50 into the video, it is stated that closing the trigger doesn't change the situation (same LED remains lit) but ....

the 22 microFarad cap is being charged "a little bit more" because the 100ohm resistor is out of the picture.

Q4. Can somebody explain this ? Why is it charging a little bit more, and what exactly is that 100ohm resistor doing.

What happens to the circuit when the 100ohm resistor is out of the picture (by closing the switch).

The capacitor, due to the 100ohm resistor generates a negative voltage pulse on the base of the right transistor, turning it off. I can understand that a negative voltage on the base would turn off the transistor.

Now that the right transistor is turned off, current can flow like this

Vcc -> right LED -> right 470ohm -> left 10K resistor ->left base transistor.

This results in positive voltage that is applied to base of left transistor , turning it on

Q5: Am I correct in saying that this causes the left LED to light up because we have current flowing through

Vcc -> left LED -> left 470ohm -> left Collector -> left Emitter - 100Ohm - GND

Q6: What happened to the flow of current that caused this transistor to turn on ?

Vcc -> right LED -> right 470ohm -> left 10K resistor -> left base transistor

Is there no current flowing through that path anymore and why ?

Q7: Why don't we see any current branching off using this path

Vcc -> left LED -> left 470ohm -> right 10K resistor -> right base transistor.

Current can go to either the left Collector or the right 10K resistor right ? Wouldn't that also turn on the right transistor (because a positive voltage would be applied to the right base)

Q8 : Breadboard setup

When I did the setup on a breadboard (same values, but using a 5V supply or a 9V supply), I could simulate the flip-flop as sown in the video , however I did notice 2 other things :

  • When one particular LED was turned on , I did notice the other one was also on, albeit very dim
  • When I did the "push down" on the switch, I noticed a flicker in the dim LED

Q9 : Practical use

I imagined that a practical use for this would be to have a latching switch with a momentary push button. The ability to turn something on / off (applying either 5V or 0V to another part of the circuit ) ?

In this case it's obviously a simple LED that is turned on / off, but could you replace that LED + resistor with a complete circuit that you would want to turn on/off using that temporary.

Thanks a lot if you made it this far ! :)

You need to realise that current flows into the capacitors too, until they charge up enough for the parallel resistor to fully take over. When the current changes suddenly the capacitor will take all the current change initially because it initially holds the voltage constant, and by ohms law that means the resistor doesn't see the step change in current.

Rather than thinking in currents everywhere, its usually more intuitive to think about the voltages in a circuit.

Put simply when the transistors switch over there is positive feedback. if the left transistor starts to switch on its collector voltage will drop, which then brings the base of the other negative by the same amount (capacitor), so the right transistor switches off, which causes its collector to rise, bringing the left transistor base high (turning it on more).

This feedback is very rapid, the circuit snaps over to the over sense. The capacitors start to discharge through their respective capacitors till the opposite switch over is initiated as the negative drive peters out on the right transistor, causing it to start conducting, pulling its collector down (which pulls the base of the left transistor lower and 'snap' again.

Vcc -> right LED -> right 470ohm -> left 10K resistor ->left base transistor.

Current will only flow through the left transistor base if the Base voltage is at least 0.6V higher than the emitter voltage. Since the Right transistor is ON, it's Collector voltage (which is the same as the left transistor's base voltage) is only about 0.2V, so it does NOT turn on, and no current flows. (Note that these voltages are relative to the emitters of the transistors. Since there's an extra 100ohm resistor to the actual circuit ground, the voltage compared to "ground" will be higher.)

This circuit is kinda confusing due to it’s design…

Basically the 2 capacitors slowly fill when the voltage on the base is high enough (negative side of the cap) 0.6v the transistor switches on causing the led to glow and the other cap to discharge. . The cycle begins again the other cap starts charging.

Get a circuit emulator or build it…

westfw:
Current will only flow through the left transistor base if the Base voltage is at least 0.6V higher than the emitter voltage.
Since the Right transistor is ON, it’s Collector voltage (which is the same as the left transistor’s base voltage) is only about 0.2V, so it does NOT turn on, and no current flows. (Note that these voltages are relative to the emitters of the transistors. Since there’s an extra 100ohm resistor to the actual circuit ground, the voltage compared to “ground” will be higher.)

Am I correct in saying that a transistor that is turned off ( < 0,6V on the base pin) :

  • doesn’t do anything, and for as long as it is turned off it could just as well be pulled out of the circuit ?
  • It doesn’t connect anything through its pins and no current is passing through it.
  • doesn’t let anything pass from Collector to Emitter (effectively creating an open circuit)
  • blocks any current flowing into the base (again creating an open circuit)

Am I correct in saying that a transistor that is turned off ( < 0,6V on the base pin) :
:
blocks any current flowing into the base (again creating an open circuit)

It’s not BLOCKING current from flowing into the base; it’s just not happening. For most practical purposes, your other comments are true, and an “off” transistor behaves like a diode the Base terminal to Emitter terminal.

Not that the bipolar transistor is a CURRENT-DRIVEN device. No current flows through the BE junction unless the voltage is sufficiently high, but it is NOT the voltage that is the controlling parameter.

But that's a meaningless statement, the current and voltage for the base-emitter circuit cannot be controlled independently - the voltage and current are completely interdependent.

The BJT is a current-controlling device, the collector current is set by the base current (approximately, when not saturated), but it is no more current controlled than it is voltage controlled, since they are related by the base input impedance (static or dynamic for DC and small-signal AC).

You could argue that in common-base configuration it is current-controlled since the input impedance is very low and the collector current is basically the same as the emitter current, that would be closer to the mark, but if you take the common long-tailed pair differential amplifier (as commonplace as op-amp input stages) you see that you have a nice well behaved voltage-controlled system, and input current is irrelevant.