Still assuming Q1 gets turned on first, R1's bottom side is electrically common to ground. HOWEVER there is still current flowing. Meaning C1's left plate will be positive. If C1's left plate is positive, its right one must be negative.

Ok, in reviewing our discussion, I gave this statement a little more thought. When Q1 turns on, the right side of C1 indeed becomes negative, but for a reason I was going to get to after I felt you had a better grip on capacitor operation. One thing to think about is that voltage across a capacitor cannot change instantaneously. so when the left plate goes from Vcc -> ground, the condition stated above requires the right side to go from V(whatever it was) -> V(water it was) - (Vcc - 0) So, if the right side was at .7v, it would go to .7v - vcc, which would most likely be negative.

 

I'll review the discussion, and see where I am after work tomorrow. Thanks!

 

Going by the old saying that a picture is worth a thousand words, below is another simulation of the Astable using LTspice.
Looking at the waveforms of the two transistors' bases and collectors, along with the previous discussions, should help you better understand what's happening, or at least provide the basis for some further questions.

 

Here you have the brief description of how astable generator work.
http://forum.allaboutcircuits.com/sh...183#post200183

Any further questions?

 

@crutschow, thanks but I already built one and played around with it on the oscilloscope. I was wondering why I was getting a sawtooth-ish wave instead of a square wave, that diagram helps a bit. The probe was in the wrong place. I'm still at a loss as to why the circuit works. Also, it did bring up a question. If the base of each transistor goes through a gradual rise in current instead of an instantaneous one, won't we be letting a rising amount of current through each resistor?

I also want to make sure I understand something. The base current of the transistors is rising slowly when the capacitor is charging because the charge time is limited by the RC constant, but the discharge time is instant since the path through the transistor and into ground has no resistance on either branch of its path. That would be why the right side of the pulse is flat, and the left is not, correct?

 

@Jony130, and @Brownout I think I got it from a combination of your explanations. I don't know why I didn't realize this before, but Jony, calling an uncharged capacitor a "short" for some reason connected something in my brain. (I think.)

1) For the purposes of the discussion, Q1 turns on first through R3.
2) C2 is acting as a short at this point. Current through R3 keeps Q1 on. C2 charges with its positive plate on the right and its negative plate on the left. When C2 becomes charged enough, the negative voltage on its left plate "cancels out" the current coming through R3. Q1 turns off.
2) During the events of step 2, C1 was charging with its left plate being negative and its right plate being positive. It charges to a point where it can saturate the base of Q2. When that happens C2's right plate suddenly becomes electrically common to ground. Its right plate goes from Vcc+ to Vcc- and thus the left plate builds up charge from Vcc- to Vcc+ until Q1 turns on again.

Some Questions:

1) If this is correct, it seems that there would be moments when both transistors would be on. While I think I have acquired the theory at this point (hopefully) perhaps I haven't figured out the timing yet.
2) I didn't account for capacitors resisting changes in voltage anywhere in my explanation. I'm guessing that's where my timing issue comes in.
3) How do we have positive voltages to play with in the first place? Everything active in this circuit is connected BELOW resistors. While we may have current, we don't have any positive voltage anywhere, do we?

 

2) When C2 becomes charged enough, the negative voltage on its left plate "cancels out" the current coming through R3. Q1 turns off.

Wrong. Q1 turns-off only when Q2 starts to conduct current. And previously C2 charged to +V "pulls down" the Q1 base voltage bellow ground.
http://forum.allaboutcircuits.com/threads/capacitors-how-do-they-work.80390/#post-570788

2) During the events of step 2, C1 was charging with its left plate being negative and its right plate being positive. It charges to a point where it can saturate the base of Q2.

Saturate ?? Wrong world. Q2 will start turning on phase only when his base voltage reach 0.6V.

1) If this is correct, it seems that there would be moments when both transistors would be on. While I think I have acquired the theory at this point (hopefully) perhaps I haven't figured out the timing yet.

Yes, for the short period of time both transistor are ON. But capacitors almost immediately turns off one of a transistors thanks to a "positive feedback". Also you need to learn a lot more, before you understand how this circuit work.

3) How do we have positive voltages to play with in the first place? Everything active in this circuit is connected BELOW resistors. While we may have current, we don't have any positive voltage anywhere, do we?

I do not understand this question.

 

http://www.homofaciens.de/technics-base-circuits-multivibrator_en_navion.htm

 

...........................
2) I didn't account for capacitors resisting changes in voltage anywhere in my explanation. I'm guessing that's where my timing issue comes in.
3) How do we have positive voltages to play with in the first place? Everything active in this circuit is connected BELOW resistors. While we may have current, we don't have any positive voltage anywhere, do we?

2)Capacitors don't resist change, they couple any voltage change from one side to the other.
Inductors resist change.

3)Being BELOW the resistors does not change the polarity of the voltage.
That's a fundamental error in your thinking.
A resistor will drop voltage across it so the voltage on one side will be less positive then the positive supply but it will still be positive.
The negative voltage on the base is generated by capacitor action from the collector to the opposite base.

 

Being BELOW the resistors does not change the polarity of the voltage.
That's a fundamental error in your thinking.
A resistor will drop voltage across it so the voltage on one side will be less positive then the positive supply but it will still be positive.
The negative voltage on the base is generated by capacitor action from the collector to the opposite base.

Wait...wait...wait. I thought just one resistor would have a voltage drop of 100%?

Example: 9V Voltage Source, 1K Ohm Resistor.

I = V/R
I = .009A
I = 9mA

V Drop = .009(1k)
V Drop = 9V

So...the current at the bottom end of the resistor is 9mA...but at 0V...

Current is being "pulled" through the resistor towards ground...The potential difference (Voltage) between the top of the resistor and the bottom is 9V so current flows through the resistor...but the potential difference between the bottom of the resistor and ground is 0V...so why does the current continue to flow? How can there be a positive voltage on anything below the resistor?

 

..................
How can there be a positive voltage on anything below the resistor?

If the bottom of the resistor is not connected to ground.

 

If the bottom of the resistor is not connected to ground.

But if one end of the resistor is not connected to ground there's no potential difference, thus no voltage, thus no driving force behind the current...

 

But if one end of the resistor is not connected to ground there's no potential difference, thus no voltage, thus no driving force behind the current...

I didn't say to leave it open...
And there is still a driving force (voltage) to ground, just no current.
If one end is open then the voltage on both ends of the resistor would have the same (positive) value.

But the resistor can be connected to another component such as a resistor , transistor, capacitor, inductor, etc.
In that case the voltage on the side of the resistor opposite the positive supply will have some value (often of positive polarity) depending upon the impedance of the device connected to it and where that device is connected.

 

I didn't say to leave it open...
And there is still a driving force (voltage) to ground, just no current.
If one end is open then the voltage on both ends of the resistor would have the same (positive) value.

But the resistor can be connected to another component such as a resistor , transistor, capacitor, inductor, etc.
In that case the voltage on the side of the resistor opposite the positive supply will have some value (often of positive polarity) depending upon the impedance of the device connected to it and where that device is connected.

I.E., if the device has a pathway to ground then what I said would apply. Otherwise, it's rail voltage at a current specified by the resistance?

If this is true I think things are going to start making more sense.

 

The voltage drop across a resistor is proportional to the current, so if it's connected to something that restricts the current, then the drop will be lower.
For example, it should be apparent that if you connect two identical resistors in series from V+ to ground, then the voltage (to ground) at the junction of the two resistors will be 1/2 V+ since the total resistance has doubled, dropping the current in half.

 

The voltage drop across a resistor is proportional to the current, so if it's connected to something that restricts the current, then the drop will be lower.
For example, it should be apparent that if you connect two identical resistors in series from V+ to ground, then the voltage (to ground) at the junction of the two resistors will be 1/2 V+ since the total resistance has doubled, dropping the current in half.

Say you have a 9V battery. Then say you have 1 1k resistor connected to ground and V+.

I = 9/1k
I = 9mA

VDrop = 9V

If the resistor is NOT connected to ground then you still have 9V at the end not connected to V+, but the current you can take out of that end is still 9mA? If that's true then you could use that to your advantage with a capacitor not connected to ground. One plate connected to the end of the resistor NOT at V+ would be at 9V and the other end would be at -9V. The only problem with that theory is where do the electrons in the other plate go in order for that plate to have -9V?

 

If the resistor is NOT connected to ground then you still have 9V at the end not connected to V+, but the current you can take out of that end is still 9mA?

That's not what crutschow is saying. The resistor is not connected to ground, meaning the low end is not at ground potential. But something else is connected between that end and ground, further restricting current. So you have in effect, a voltage divider. The voltage at the low end of the resistor is between V+ and ground. Please google "voltage divider" for more information.

 

................
If the resistor is NOT connected to ground then you still have 9V at the end not connected to V+, but the current you can take out of that end is still 9mA?

Absolutely not. Resistor current is proportional to the voltage drop across a resistor. If one end of the resistor is open then both ends of the resistor are at 9V (to ground) and no current flows.
If you connect the other end of the resistor to, say 5V, then (9V-5V) / 1k = 4mA will flow.
It's simple Ohm's law I = V/R, where V is the voltage drop across the resistor.
Don't understand why you are having so much trouble with that concept.

 

Absolutely not. Resistor current is proportional to the voltage drop across a resistor. If one end of the resistor is open then both ends of the resistor are at 9V (to ground) and no current flows.
If you connect the other end of the resistor to, say 5V, then (9V-5V) / 1k = 4mA will flow.
It's simple Ohm's law I = V/R, where V is the voltage drop across the resistor.
Don't understand why you are having so much trouble with that concept.

I'm sorry. I don't know how I am confusing myself. In the case of this circuit there are two posibilites for the resistors...

1) Connected to ground through the base or collector of a transitor that is ON.
2) Not connected to ground because the transistor is off.

In the second case the capacitor simply can't charge because it's connected to V+ ok both sides, correct?

 

I'm sorry. I don't know how I am confusing myself. In the case of this circuit there are two posibilites for the resistors...

1) Connected to ground through the base or collector of a transitor that is ON.
2) Not connected to ground because the transistor is off.

In the second case the capacitor simply can't charge because it's connected to V+ ok both sides, correct?

That's true for the resistor connected to the transistor collector.
The resistor connected to the base is always connected.

When the circuit is operating, the capacitor is being peridically charged and discharged.
If the voltage across a capacitor is constant than the capacitor won't charge or discharge. But any change in the voltage across it will cause a charge flow.
---------------------------------------------
Let's take another shot at the simulation with different parameters displayed:

Below shows the voltage at the collector of Q2 [V(c2)], the voltage across capacitor C2 [V(C2)-V(B1)], the voltage at Q1's base [V(b1)] and the current through C2 [I(C2)].

Note the positive spike of current (positive is current from left to right) as the capacitor is charged when V(c2) goes high (the current on the right of the capacitor goes through Q1's forward biased base).

During the time V(C2) is high there no voltage change across the capacitor and thus no further capacitor current.

When V(C2) goes low there is a short negative spike due to stray circuit capacitances (primarily the charge from the base of Q1). This is small so has a negligible effect on the capacitor charge or voltage.
Note that the positive to negative transition at V(C2) is coupled directly through C2 to the base of Q1 (V(b1) but there is no significant change in voltage across the capacitor (since little charge flows during the transition).

During the time V(C2) is low the capacitor slowly discharges through R3 [voltage across R3 being the V1 supply minus V(b1)] which is the small I(C2) negative capacitor current.
No current goes through Q1's base since it is now reverse biased.

When the voltage across C2 becomes low enough, V(B1) becomes sufficiently high to forward bias Q1's base, turning it on and causing the circuit to flip to its other state.

Hope that helps your understanding a little.
Do you have any questions about that?