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I am playing around with the following circuit: Falstad's simulator

enter image description here

What I understand is:

  1. First, the +5V terminal at the top starts "charging up" the 100F capacitor. While charging, the capacitor "lets electricity flow through", that's why the upper NPN transistor is open.
  2. The mentioned NPN which is open makes the LED light up.
  3. In the meanwhile, the 100F gets fully "charged up" and this acts like a gap in the circuit. The upper NPN reaches cut-off state and the LED goes out.

What I am not sure about:

  1. The 10F (lower) capacitor which should be full after this all opens the lower NPN.
  2. By opening the lower NPN, the 100F (upper) capacitor should empty its charge.
  3. After the 10F (lower) capacitor is empty, the lower NPN should be closed (which is not the case) and the whole process should "restart" with 1.

In the simulator, I can see that V(b) of the lower NPN transistor is between V(c) and V(e) (I do not understand why) - and maybe this is what causes 6. to NOT happen and the whole process to "fail".

Do I understand the situation well? Could you help me with some more explanation? Why isn't the process happening as I described in my points?

A redrawn, possibly cleaner variant here

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    \$\begingroup\$ This is a case of garbage in garbage out. Look at the currents. Do you think giga-amps are flowing through these transistors? \$\endgroup\$ – τεκ Mar 15 '18 at 23:09
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    \$\begingroup\$ your circuit is badly drawn ... here is a redo .. falstad.com/circuit/… \$\endgroup\$ – jsotola Mar 15 '18 at 23:17
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    \$\begingroup\$ Why should the 10F capacitor be full? \$\endgroup\$ – user253751 Mar 16 '18 at 2:29
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    \$\begingroup\$ Please see this (especially item #3) and redraw your schematic accordingly. \$\endgroup\$ – Dave Tweed Mar 16 '18 at 18:47
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    \$\begingroup\$ @jsotola It was quite hard for me to understand the equivalence between the two circuits but thanks for the redraw. \$\endgroup\$ – user1724641 Mar 17 '18 at 7:50
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Welcome to the site and congratulations on starting the journey into understanding electronics.

I'm just going to touch on the biggest issues you appear to not be understanding.

  • First is why your 10F cap never charged. Current tends towards the path of least resistance, and any two connected points without a resistance between them will have the same voltage. These are fundamental rules directly tied to, and as important to remember as, Ohm's law. Whatever current might flow through your upper transistor is going straight to ground and completely ignoring your cap. Ground is 0V. The emitter of your NPN and the top of the 10F cap are tied directly to ground, so both will be seeing 0V. This is why when you redraw your schematic in a more conventional way it appears as if the 10F cap has no power source.

  • Second is how caps respond to DC voltage. It's true that current will briefly flow to some extent while the cap is charging. What you're missing is that once it is charged, it will not further charge or discharge until its INPUT voltage is changed. In simple terms, you can think of it like filling a bucket. Once the bucket is full the water must come back out the top, and you will have to tip it towards ground to do it. For practical purposes, a cap in series with DC is treated as an open circuit after the initial inrush current is resolved.

  • You'll almost never find a circuit design without resistors to manage current and develop voltages. Those cap values are also pretty absurd. You're welcome to use whatever size cap and resistor values you want for exploring theory of operation. In practice, a 1F cap weighs about 3lbs and even if you could build a circuit that works in theory without resistors you'll almost certainly burn things out due to unmanageable current and power.

The closest working circuit I can think of that nearly matches what you're trying to build is a type of oscillator called an "astable multivibrator." Here's a simulated version with some explanation if you want to play around with adding LED's and changing values.

Hope this helps and was worth both of our time.

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100 and 10 Farad capacitors are huge, I can only assume you are dealing with 100 and 10 mFd capacitors. If they are 100 & 10 Farad capacitors this complementary oscillator would have a period of many days. I see no purpose for such a slow oscillator. (long time digitally for accuracy) One redrawing has the 100 F connected to ground instead of +5 Volts, the other I couldn't locate. ???

"What I understand is:

The mentioned NPN which is open makes the LED light up."

IS wrong! Series transistors must be conducting well (on) to allow enough current to light a LED.

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  • \$\begingroup\$ Many beginners seem to use "Open" to mean "conducting" when discussing transistors. This confuses more experieced readers who use "open" to mean "not conducting", particularly with regard to switch and relay contacts. \$\endgroup\$ – Peter Bennett Mar 31 '18 at 15:58

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