With a clear lens, a red LED with 4 uA flowing can be seen internally as a small red dot.

At night in the dark?

 

Thanks for all the replies, but one thing still confuses me is, if all capacitors are charging at the same time, and FET, capacitor rating, resistance they are all same, why all FETs are not triggering at the same time? If that happens all leds will dim and light up at the same time, right? Or is there any other tricky part to that?

In theory - yes, circuit can not start oscillate.
But in reality - not exist two absolutely identically things.

And which software is this?

Software is LTspice.
Free download it from http://ltspice.analog.com/software/LTspiceXVII.exe
This software will answer all your questions.
ADDED:
LTspice demo circuits collection: https://www.analog.com/en/design-ce...ltspice-simulator/lt-spice-demo-circuits.html

 

In theory - yes, circuit can not start oscillate.
But in reality - not exist two absolutely identically things.

Software is LTspice.
Free download it from http://ltspice.analog.com/software/LTspiceXVII.exe
This software will answer all your questions.
ADDED:
LTspice demo circuits collection: https://www.analog.com/en/design-ce...ltspice-simulator/lt-spice-demo-circuits.html

That's what I thought first, but your simulator also showing as led as turing on sequentially. In simulaor all similar components will be identical, so how they are oscillating there?

 

You can assign correct Vfd to the LED's.

 

You can assign correct Vfd to the LED's.

Can you elaborate?
Are you taking about the simulator ?

 

in LTSpice and other spice based simulation you can change the values for your component.

more here
https://www.youspice.com/spice-modeling-of-a-diode-from-datasheet/

 

... when the schematic is drawn poorly in the first place!

There is nothing wrong in the way the circuit is drawn.
There is no need to put dots at a T-junction, i.e. where capacitors C1 and C3 are connected.

If the crossed lines were meant to be connected, the gate of the MOSFET would be directly connected to +6VDC and the MOSFET would be permanently ON. Resistor R6 at 1MΩ would serve no purpose.

It should be very obvious that Q1-Q2-Q3 is a three-stage ring oscillator. Each stage is identical and no one is the "first" stage that triggers first.

 

It's just not consistent but we can see how it's connected. I believe the forward voltage drop of the diodes will cause different dv/dt

 

All three stages are identical.

From power ON, all capacitors C1, C2, and C3 start from a discharged state at 0V.
All MOSFETS Q1, Q2, and Q3 are turned off.
All three capacitors begin charging at the same rate through resistors R4, R5, and R6.

But wait a minute!
Are all three capacitors starting at the same voltage?
Are resistors R1, R2, and R3 identical?
Are resistors R4, R5, and R6 identical?
Are capacitors C1, C2, and C3 identical?
Are the diode currents through LED1, LED2, and LED3 identical?
Are the turn-on characteristics of Q1, Q2, and Q3 identical?

There are so many different ways to show that the RC time constants and turn-on time of the MOSFETS are not the same.
One MOSFET is going to conduct before the other two conduct. This starts the oscillation cycle.

 

Even with perfect components, the charging rate will be different because the circuit uses a red, green and blue LED.

 

Even with perfect components, the charging rate will be different because the circuit uses a red, green and blue LED.

+1
I did not notice that one. Thanks.

 

Hello,

I posted a "dotted" schematic in post # 14

Bertus

 

Exactly - fwd voltage is different for the different colored LED's that's the biggest difference. In actual circuit those caps have been known to vary quite a bit.

 

There is nothing wrong in the way the circuit is drawn.

It's not often I'll argue with people who are better qualified and informed than I am (which you definitely are,) but I disagree with you in this case.

I immediately figured out what was going on in this circuit because the pattern of the identical sub-circuits was pretty clear, and because it would make no sense for there to be a connection there (like you said.)

Nevertheless, I think drawing conventions serve a useful purpose. I don't think you should start ignoring them based on what "should be very obvious." The whole point of sharing schematics in this forum is for education, which presumes that the people being educated don't already have it all figured out. I personally don't have any familiarity with a three stage ring oscillator. It's brand new to me There's nothing obvious about that, unless you already know a lot about circuits. I've been reading schematics for many, many years, and I've had plenty of success in basic electronics, but I never learned about ring oscillators, so I don't recognize one when I see it.

In my opinion, the schematic should provide clear, precise information about what connections do and don't exist. It shouldn't require interpretation, nor should you have to already know the circuit in order to determine what the connections are. To me, a good schematic should be clear and unambiguous.

The existing drawing conventions, using dots at all intersections, and omitting them at all non-intersecting apparent joints, provides this clarity. Why step away from such a useful convention?

Beyond all of that, no matter what drawing convention you might prefer, this particular drawing is inconsistent. It uses dots for two of the capacitor T intersections, but no dots anywhere else. That's just silly, even if you disagree with me about the need to dot T intersections in general!

 

All three stages are identical.

From power ON, all capacitors C1, C2, and C3 start from a discharged state at 0V.
All MOSFETS Q1, Q2, and Q3 are turned off.
All three capacitors begin charging at the same rate through resistors R4, R5, and R6.

But wait a minute!
Are all three capacitors starting at the same voltage?
Are resistors R1, R2, and R3 identical?
Are resistors R4, R5, and R6 identical?
Are capacitors C1, C2, and C3 identical?
Are the diode currents through LED1, LED2, and LED3 identical?
Are the turn-on characteristics of Q1, Q2, and Q3 identical?

There are so many different ways to show that the RC time constants and turn-on time of the MOSFETS are not the same.
One MOSFET is going to conduct before the other two conduct. This starts the oscillation cycle.

Aside from my rant about schematics, I agree with the rest of your comments on the circuit itself. Cheers!

 

@ebeowulf17

I do agree with you about the aim for consistency.

However, the ambiguity only arises at crossed lines. Are the lines connected or not connected?
It should be clear in the drawing if crossed lines are meant to be connected, otherwise it leaves it open for error.

I witnessed a real life situation where a summer student was given a circuit schematic and was given the task of assembling the circuit on a wire-wrap board. (Not my student and not my circuit). I just happened to be passing through and noticed that after a couple of hours of assembly, all crossed lines were wired together into a common node.

Edit: My reference to the three ring oscillator was in reference to someone asking which transistor fires "first".