# Operation of an astable multi-vibrator?

#### Allenph

Joined May 27, 2015
76
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?

#### Brownout

Joined Jan 10, 2012
2,390
Please expand comment block to see responses in red.

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 no, it goes to zero, due to the voltage drop across R1 . Because there is no path for current to flow through R2 to ground, C1's right plate starts to go negative since the + side of c1 has gone towards zero, current flows into the plate connected to R2, causing the voltage at that node to swing towards zero due to voltage drop across R2. . This negative voltage suppresses the current flowing through R3 and keeps Q2 off no, the zero-going voltage developed across R2 holds Q2 in cutoff (of condition).
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 sort of, R3 biases Q1 on, the voltage at the R3/Q1 base is equal to Q1's VBE ~.7V.

That's as far as I get.

Once C1 completes charging, current across R2 decreases, thus voltage drop across R2 decreases. Eventually Q2 turns on, causing C2+ side to drop towards zero, and the above cycle repeats in a complementary manner.

I also have a question in addition to understanding the operation of this circuit. Why does C1 charge when Q1 is on? Can you ascertain the reason from the discussion above?

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?

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#### Allenph

Joined May 27, 2015
76
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.

#### Brownout

Joined Jan 10, 2012
2,390
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 everything good up to here. Since C1's left plate is electrically common with ground, it cannot build up a positive voltage on its left plate Since C1's left plate has been pulled towards ground, current flows out of the left place, and INTO the right plate. Initially, the right plate moves towards zero volts, due to the current flowing into it, which also flows through R2, causing a voltage drop across R2 in accordance with Ohm's law (V = IR) As C1 charges up, current decreases, and the right place returns towards VCC. 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 Yes, this is consistant with what I wrote above..

#### Allenph

Joined May 27, 2015
76
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?)

#### Brownout

Joined Jan 10, 2012
2,390
If I understand correctly, then current is flowing in through the right plate, and out of the left plate. How is this accomplished? The controlling equation for capacitor terminals is: Ic = C*delta(V)/delta(T), where V is the capacitor's terminal voltage, and T is time. This is NOT an illusion, this is REAL current. If you don't understand this, then you MUST research capacitor theory. 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. The left plate quickly goes towards zero. Any charge in the left plate flows out towards ground. An equal amount of charge flows into the right plate, thus the total charge doesn't change. In a capacitor, the charge remains the same, when charge flows out of one plate, it flows into the other such that the TOTAL charge doesn't change.

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, when the charge in C1's right plate accumulates enough such that the voltage on that plate is more than Q2's threshold voltage ~.7v or when C1 is fully charged? If it is the former, isn't the RC constant essentially useless? Heck no it's not useless. The RC time constant determines how long Q1 stays on and how long Q2 stays off. But C1 need not completely charge up to VCC to turn on Q2, only enough such that Q2's base, connected to C1's right plate, is above ~.7v. This would lead to problems determining frequency down the road... no, the frequency is easily determined.

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

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#### upand_at_them

Joined May 15, 2010
766

#### Allenph

Joined May 27, 2015
76
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...

#### Allenph

Joined May 27, 2015
76
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.

#### Brownout

Joined Jan 10, 2012
2,390
I suppose my misunderstanding is inherent to my misunderstanding of the fundamentals of capacitors.

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...
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.

#### Allenph

Joined May 27, 2015
76
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?

#### Brownout

Joined Jan 10, 2012
2,390
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? No. If the left plate is common to ground, any positive charge will flow out of the left plate into ground. Just like positive charge flows from VCC to ground. If so, that means that C1's right plate will ALWAYS have a positive charge, right? As positive charge flows from the left plate to ground, an equal amount of positive charge flows into the right plate, such that the total amount of positive charge does not change. 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 Correct!.

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? Correct. Well almost. The right plate must only accumulate enough charge to turn on Q2 ~.7V. It doesn't need to fully charge.

#### Allenph

Joined May 27, 2015
76
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?

#### Brownout

Joined Jan 10, 2012
2,390
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... The left plate is alternatively charged and discharged via Q1/R1

Also, if C1's right plate never has a negative charge why doesn't Q2 turn on through R2? I already explained that as the voltage drop across R2 due to the charging current for C1 I would assume that it's because C1 is "leeching" all the current coming through R2 capacitors don't leech current, whatever that means. They accumulate charge on a plate, and lose an equal amount of charge on the opposite plate, as I have already stated many times in this discussion. until it gets to the point where the positive voltage on C1's right plate is sufficient to turn Q2 on? EDIT: Except for the vocabulary, which confused me, you're getting the right idea here.

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 .

During each RC time, the capacitor charges to ~70% if the final voltage (VCC) If, for example, VCC= 12V, then after 1 RC, Vc = 8.4v. So, somewhere around .1*RC, the voltage will be ~.84v. This is a VERY ROUGH estimate. It's not easy to write a simple equation because the voltage curve is not linear. If you want an exact equation,use

t = RC*ln(V/VCC) where 'ln' is the natural log function.

Then...

1/Time to .7V = Frequency in Hz

1 complete cycle requires a charge cycle for c1 and a cycle for c2 (remember, Q1 and then Q2 must operate for a complete cycle) So the complete period is twice the charge time for c1 (or c2) if R1C1 = R2C2.

Is that correct?

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#### Brownout

Joined Jan 10, 2012
2,390
Please see the edit in blue in the above post.

#### Allenph

Joined May 27, 2015
76
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?

#### Allenph

Joined May 27, 2015
76
Please see the edit in blue in the above post.
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...

#### Brownout

Joined Jan 10, 2012
2,390
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?
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.

Meaning that there is no current flow to Q2's base until C1 is charged? No, absolute not. I've explained when Q1 turns on several times now. Please refrain from asking the same question over and over. Please instead review what I've written. If that's true then why does Q2 turn on when C1's right plate reaches .7V...

#### Brownout

Joined Jan 10, 2012
2,390
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.

#### Brownout

Joined Jan 10, 2012
2,390
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.