Why are both leads of the cap at 5v in the beginning?

You always have to look at the voltage across a capacitor (not voltage to ground) and that voltage doesn't change unless there is a current flow (charge transfer) through the capacitor.
Thus at the start the voltage across the capacitor is zero.
At startup the NPN transistor is off, so the right side of the cap will immediately go to 5V through the 22Ω resistor.
This means the left side of the cap will also go to 5V across the 330kΩ resistor (since the voltage across it is still zero as no current has yet flown through it).

 

Some info on battery protection circuits -

Aren't those usually just to prevent over-discharge?
The battery in question shuts off when the output load current is below a certain threshold.

 

You always have to look at the voltage across a capacitor (not voltage to ground) and that voltage doesn't change unless there is a current flow (charge transfer) through the capacitor.
Thus at the start the voltage across the capacitor is zero.
At startup the NPN transistor is off, so the right side of the cap will immediately go to 5V through the 22Ω resistor.
This means the left side of the cap will also go to 5V across the 330kΩ resistor (since the voltage across it is still zero as no current has yet flown through it).

OH WOW! I think I actually understood that bit. So I put a small set of images together because the time thing is what confuses me. I understand what these components do by themselves, but combined and how one interacts with the other tends to confuse me. Opamps were a nightmare and im still not 100% on them, but thats another story. At the start, both ends of the cap are at 0V differential from each other, OK. Moving left to right and top to bottom:

1. It is said that current doesnt flow thru the cap but it reaches the base of the PNP transistor thru the 330kΩ? How does that happen?
2. I see how the PNP turning on, turns on the NPN.
3. How does an emitter-base current turn off the PNP? I thought current at the base of a bjt turned it ON?

 

If you want to take things very slowly, the cap is not charged at the beginning and your first diagram is correct. The capacitor charges via the 22R and 330k. But it only charges a very small amount.
The tiny amount of charging current puts a microscopic voltage across the 22R and 5v across the 330k. Nothing else happens.
The capacitor starts to charge and the left lead drops a little bit as the capacitor starts to charge.
As soon as it drops by about 0.6v, the base of the PNP transistor detects this and it starts to turn ON.
That's how things start.

 

Well as I try to understand, I tried out the circuit and it doesnt work. The LED comes on, then turns ON and turns OFF when I unplug the battery pack.

Why does only a small amount of current flow thru 22R and a larger amount thru 330k of 22R is less resistant?

How does the left side of the cap drop a little? You mean it drops from 5V to about 4.4V before the pnp starts? Why?

 

Current flows out of the base of the PNP and through the 330k ohm resistor when the base voltage drops to about 0.6V below the emitter (battery) voltage.
That turns the PNP on.


There is no e-b current when the base base voltage is higher than about 0.6V below the emitter voltage.
When there is no e-b current, the transistor is off.
So when the NPN turns off and the right side of the capacitor goes to 5V, the left side increase by 5V also.
This reverse biases the PNP e-b junction, keeping the transistor off.

Why does only a small amount of current flow thru 22R and a larger amount thru 330k of 22R is less resistant?

The small current flows when the NPN is off because that puts the 22R and the 330k in series so the high resistor determines the current value.

How does the left side of the cap drop a little? You mean it drops from 5V to about 4.4V before the pnp starts? Why?

It drops from a peak of nearly 5V above the supply voltage to about 4.4V.
That's because the 330k resistor is slowly discharging the capacitor.

You should get the free LTspice download from Analog Devices/Linear Technology and experiment with the circuit.
With that you can simultaneously look at both voltages and current anywhere in the circuit.
That's something you can't readily do in the real circuit.

 

Thanks...I found this which is excellent, but it has no transistors as far as i can tell:

https://phet.colorado.edu/en/simulation/legacy/circuit-construction-kit-ac

Ill look into ltspice.

 

ok ive got it set up on ltspice and i modified the battery, resistor, capacitor values. How do I run the simulation in a way that I can see time slowly?

 

The 3906 is up-side-down.

 

The 3906 is up-side-down.

ok, got it...

 

OK, Ive done a bit of reading up on oscillator circuits. I think Im beginning to understand. I found this oscillator circuit on Stack:

Btw, when you guys get a chance (I know youre busy helping noobs like myself ), please help me with the sim on ltspice. I tried .tran 5m but when I hit Run a window pops up with the x-axis 0-5ms but nothing else happens. Anywhere I click it shows as a straight line thru 0.

Anyway, a couple of things:

1) What's tripping me up is the basic concept of current and electricity. How does electron flow (current) generate positive voltages? So when the circuit is powered on and current flows through R2 towards the base of T2, it turns ON T2 giving C1 a path to ground. So if electrons flowed to the base of T2 it's base should be negative. So all this while somehow T1's base is accumulating positive charge, which results in eventually turning on T1. This super confuses me because if + charge turns on a transistor (T1) then why did negative electron flow (current) turn on T2?

So OK,T1 turns on allowing current flow down R1 which now makes the left of C1 negative. This somehow makes the right of C1 more negative which spreads to T2 base and now this negative flow of electrons turns off T2 instead of ON.


If I can understand these things I believe Ill be better suited to understand my oscillator circuit:

 

OK if I don't use the +/voltage vs (-) current flow concepts and just think of current flow...

1). Current flows from battery to R1 to C1 to base of T2 allowing current down R4 thru an LED to GND. All this time the capacitor charges and ultimately stops conducting. Led goes from being ON to being OFF.

2. Now current flows thru R4 to C2 to T1 and thus turning on the other led.

Right?

 

The Multivibrator. This circuit is also called a flip flop as is turns on the left LED then the right LED in a flip, flop action. This is a very interesting circuit and it employs the features we have covered in the previous sections.

The flip flop circuit is shown above but before we describe how it works, the animation below shows it in operation. The circuit can be used as railway crossing lights or for road-works in a model train layout.

Even though this circuit looks very simple , it has a lot of features that need explaining.
When the circuit is first turned on, both capacitors (the capacitors in the circuit are called electrolytics - see the section on capacitors and electrolytics) are uncharged and they can be taken out of the circuit as shown in the diagram below to simplify the discussion:

The two 10k base resistors will turn BOTH transistors on at the same time and due to the slightly different characteristics of the transistors, (their speed of operation), one transistor will "beat the other" and turn on slightly faster. Suppose Q1 turns on faster than Q2.
Now put the electrolytics back into circuit. The positive end of the 100u electrolytic is connected to the collector of this transistor. At the instant the circuit turns on, the voltage on the collector will be about 0.7v as it cannot rise any higher than the "characteristic" base-emitter voltage of the second transistor and will fall to about 0.3v when the transistor turns on.
This effect is shown in the circuit below with the first transistor turned on fully and the electrolytic keeping the second transistor off.

The voltage on the base of the second transistor falls from 0.6v to 0.2v and the electrolytic charges in the REVERSE DIRECTION until the base of the second transistor rises to about 0.65v. This is shown in the diagram below:

The second transistor turns on and its collector voltage falls, bringing the second electrolytic towards the 0v rail. This action is shown in the diagram below:

The other end of the second electrolytic is connected to the base of the first transistor and this turns off the first transistor.
This "change-over" action happens VERY FAST and since the first transistor is turning off, it is raising the left-hand side of the first electrolytic , thus increasing the voltage on the base of the second transistor, (and thus allowing more current to flow into the base) to turn it on HARDER.
The second electrolytic is now charged in the reverse direction and after a short period of time the first transistor turns on and the cycle is repeated.
You can think of the two transistor "jumping up and down" in the circuit, raising and lowering the voltage on the collector of each. This effect is shown in the animation below and will help you to understand how the transistors "take it in turn" to illuminate the LEDs.

Question 69: Name the component that turns t

 

please help me with the sim on ltspice. I tried .tran 5m but when I hit Run a window pops up with the x-axis 0-5ms but nothing else happens.

Right-click on V1 and set to 5V.
Set the time to 7s, not 5ms.
Click the bottom box in the simulation command:
(Oscillators often don't start in a simulation if the initial operating (DC) bias point is first calculated.)


Run the sim.
Click on the node voltage you want to see.

 

What's tripping me up is the basic concept of current and electricity. How does electron flow (current) generate positive voltages?

You are confused because you are thinking about it backwards.
Voltage or EMF (Electro Motive Force) generates the current flow, not vice-versa.
The value of the current is determined by the impedance in the circuit.

Think of water in a pipe.
It's the water pressure that creates the water flow.
The resistance (such as valve position) then determines the flow rate.

 

Ok so in this circuit, my original circuit:

When the circuit is powered up, the voltage difference between the source causes current to flow from + to - terminal of the battery.
Current can only flow thru 33R or into the collector of the Q1. I assume it cant enter the collector of Q1.
Its path thru 33R is takes it thru C2 and to the base of Q1. This charges C2 and turns on Q1.
[Not sure what R1 and C1 do since they weren't in the original circuit referenced in the post]
This current turns on Q2 and current has a quicker path to GND via Q2.
After a certain time, C2 is charged up and stops conducting, which turns off Q1, turning off Q2.
At this point the C2 begins to discharge thru 330k?

Is that about it?

 

No. It's a lot more complex than that.
There is an action called a REGENERATIVE ACTION that has never been discussed.

 

Awh man!

 

Has or can anyone try this circuit? Its not working for me. The LED doesnt light up when I plug the power in. I checked the LED and it works fine. I switched to other 2n2222 transistors and other 3906 and another electrolytic 22uF cap thinking I might have burnt out my original ones while testing but stil nothing.

Im unsure about the connection of the cap and LED. Wondering if my leads are connected in the right order.