# Operation of an astable multi-vibrator?

Discussion in 'General Electronics Chat' started by Allenph, May 27, 2015.

1. ### Allenph Thread Starter Member

May 27, 2015
76
2
I'm having an awful time trying to understand how astable multi-vibrator works. I've asked questions on stack exchange, but the help I've gotten there has been limited and largely negative.

This is a circuit diagram of a basic astable multi-vibrator utilizing two BJTs. I'm fairly new to electronics, but have bought all the necessary equipment, and have a basic understanding due to reading of the textbooks on this site, tutorials, etc. I would appreciate it if you take it easy, and dumb down your answers to the absolute minimum, haha.

As I'm having trouble understanding the operation of this circuit, it will be difficult to tell me exactly how it works in a manner I can understand. So, the following is how I PERCIEVE operation.

(Due to minute differences in components one of the transistors will turn on first. For the purpose of this discussion lets assume Q1 turns on first.)

1) Q1 turns on through R3.
2A) Current is flowing from the positive rail, through R1, through Q1 and into ground. Therefore, the current applied to the left plate of C1 is positive. Because there is no path for current to flow through R2 to ground, C1's right plate starts to go negative. This negative voltage suppresses the current flowing through R3 and keeps Q2 off.
2B) Simultaneously to the events occurring in step 2A, R3 begins to charge C2's left plate positively because current is flowing through R3, through Q1 and into ground. This keeps Q1 on.

That's as far as I get.

I also have a question in addition to understanding the operation of this circuit. Why does C1 charge when Q1 is on? It has positive voltage at the left plate, but where is it getting the negative voltage it needs on the right plate to "hold" the positive voltage on the left plate in place?

2. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Please expand comment block to see responses in red.

Last edited: May 27, 2015
3. ### Allenph Thread Starter Member

May 27, 2015
76
2
Unfortunately, I didn't understand your response. Sorry I'm such a brick-wall. It appears that my misunderstanding is cascading from a misinterpretation of what happens at step 2A. In my mind, I don't understand why the right plate of C1 would gravitate to 0V. In my mind, it seems like it should be like this...

2A) Everything on TOP of R1 is electrically common to the positive rail. Everything BELOW R1 is electrically common with ground. R1 is acting as a current shunt. I.E. the entire supply voltage is dropped across R1. R1 limits current based on its resistance value in ohms dictated by I = V/R1. Since C1's left plate is electrically common with ground, it cannot build up a positive voltage on its left plate. Since the left plate is essentially at ground, the right plate of C1 can charge to the positive rail voltage through R2. R2 dictates the RC constant of C1 dictated by T = R2C.

Jan 10, 2012
2,375
998

5. ### Allenph Thread Starter Member

May 27, 2015
76
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If I understand correctly, then current is flowing in through the right plate, and out of the left plate. How is this accomplished? Since the plates are not physically connected, current can only give the illusion of "flowing" and that illusion is only maintained until C1 is filled to capacity. Perhaps it is the following...

If the left plate is at ground, then all the free electrons in the left plate are rendered "irrelevant." I.E. the left plate being electrically common to ground produces the effect of the left plate having an unlimited capacity to "hold" charge that is on the right plate. Therefore, all the current flowing through R2 builds up on the right plate, attracted to the "holding power" of the ground-equivalent left-plate. When C1's right plate is completely saturated (I.E. the capacitor is fully charged) the illusion of "current flow" from the right plate to the left plate collapses.

If this is correct, I'm confused as to WHEN Q2 turns on. Is it when C1's right plate builds up enough positive charge to saturate Q2's base, or when C1 is fully charged? If it is the former, isn't the RC constant essentially useless? This would lead to problems determining frequency down the road...

In addition, now I'm having a hard time understanding what would turn Q1 off. (I.E. how would C2 desaturate Q1's base?)

6. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998

Last edited: May 27, 2015

May 15, 2010
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8. ### Allenph Thread Starter Member

May 27, 2015
76
2
I suppose my misunderstanding is inherent to my misunderstanding of the fundamentals of capacitors.

I thought that a capacitor was two plates separated by an insulator. Current can't flow THROUGH the dielectric?

What I Think I Know About Capacitors:

Capacitors are devices which store electrical charge. They consist of two plates separated by an insulator called a dielectric. Dielectric material, plate material, dielectric thickness, and plate surface area all have an effect on the capacitance of the capacitor which is measured in farads. Current is held in a capacitor if there is a positive charge on one plate while the other plate has a negative charge. The charge is "held together" because the positive and negative charges attract each other. The excess electrons in the negatively charged plate are attracted to the "holes" in the positively charged plate. However, the excess electrons in the negatively charged plate cannot fill the holes in the positively charged plate because their path is blocked by the dielectric. When you provide a path for the electrons to flow, you discharge the capacitor, utilizing the current through the path along the way.

If I was wrong, I would appreciate it being pointed out and explained. I can't be completely right, because this would mean that current can't flow through the capacitor...

9. ### Allenph Thread Starter Member

May 27, 2015
76
2
I appreciate that Upand_At_Them, but I've viewed several animations, several tutorials, several video tutorials, and invested quite a few hours in theory of the components involved. Somewhere I'm misunderstanding what all of them are saying, and thus this circuit is beyond my reach. As per my conversation with Brownout, I think that my misunderstanding lies in capacitance.

10. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
When charge flows out of one plate (current) an equal amount of charge flows into the opposite plate (current) No charge flows through the insulator. Think about this: If you bring two charged objects close together, charge will move to the opposite side of each object, although no charge flows in the space between the objects. This is due to electro-static force, which repels like charges (attracts opposite charges) Same for a capacitor. When you remove charge from one plate, an equal amount of charge will flow into the opposite plate, and no charge flows in the insulator.

11. ### Allenph Thread Starter Member

May 27, 2015
76
2
Sigh...the complexity of what is supposed to be one of the most bread-and-butter circuits is vexing to me.

I think I understood your response, but I'm not sure. It's brought on a new barrage of questions.

If the left hand of C1 is electrically common to ground, then the left plate of C1 is perpetually positively charged, yes? If so, that means that C1's right plate will ALWAYS have a positive charge, right? If that is so, then it will try to acquire a positive charge through R2. Because R2 will limit the current, it will take time to charge up the capacitor's right plate to the maximum capacity of the capacitor.

AFTER the capacitor is full then R2's current will have to find another avenue to go to ground...and that would be Q2's base. Is this correct?

Jan 10, 2012
2,375
998

13. ### Allenph Thread Starter Member

May 27, 2015
76
2
Hmm. If the capacitors right plate only acquires the positive charge that is proportional to the loss of positive charge on the left plate...shouldn't nothing happen? The left plate never acquired any positive charge...

Also, if C1's right plate never has a negative charge why doesn't Q2 turn on through R2? I would assume that it's because C1 is "leeching" all the current coming through R2 until it gets to the point where the positive voltage on C1's right plate is sufficient to turn Q2 on?

In addition, if the right plate of C1 only needs to attain the forward-bias voltage of Q2's base-emitter junction, then the formula for the time to get to .7V should be the proportion...

Rail Voltage/(R2(C1)(5)) = .7/Time To .7V

Then...

1/Time to .7V = Frequency in Hz

Is that correct?

14. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998

Last edited: May 28, 2015
15. ### Brownout Well-Known Member

Jan 10, 2012
2,375
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Please see the edit in blue in the above post.

16. ### Allenph Thread Starter Member

May 27, 2015
76
2
Let me try again. Hopefully I'll be more successful.

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. This negative voltage on C1's right means that Q2 is off. C1's right plate will begin to charge through R2 displacing charge on the left plate. C1's right plate continues to charge until it's at .7V?

17. ### Allenph Thread Starter Member

May 27, 2015
76
2
Meaning that there is no current flow to Q2's base until C1 is charged? If that's true then why does Q2 turn on when C1's right plate reaches .7V...

18. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
You're getting the idea, and maybe I should leave it alone here. However, it's not correct to say that because C1's left plate is positive that means the right plate is negative. Depend on how the capacitor is charges, one plate can have a positive voltage, and the other plate can have a different voltage. However, the voltage may be negative, but that depends on charge, current and circuit conditions. You must evaluate the capacitor voltage during charging and during static conditions.

19. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
I think you need to go over the discussion and give the circuit careful thought. You seem to be slowly getting the idea, but you require me to answer the same questions several times. If after thinking more about it, you still have questions,then I'll try to answer. For now, I'm heading to bed.

20. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
PS: I had to change something in post #14. A capacitor charges to ~ 70% of VCC for each RC time, not 7% as I had written.