How does a capacitor in an astable multivibrator turn off a BJT?

Question

I've recently started learning about astable multivibrators. I understand that the capacitors in a circuit like the one below create the alternating current by charging up one at a time, and once they are charged up and the BJT next to them is closed, they get a negative potential which turns off the BJT on the other side by stopping current from reaching them. What I don't get, is how a capacitor getting a negative charge will stop current from the battery from going to the BJT all together. If, say, the transistor on the left is closed and C1 is charging, I feel that current from R2 should be able to split off at the junction to go left through C1 and right to the BJT on the right.

Answer 1

When the left transistor turns on, the RHS of C1 will rapidly drop to a negative voltage below ground while its LHS is held at close to ground by the turned on transistor. C1 now charges through R2 and the voltage on the RHS of C1 slowly rises but the right transistor can't turn on until the voltage on the RHS of C1 rises to about 0.7 V above 0 V which is the approximate turn on voltage of the transistor. When the right transistor turns on it rapidly takes the LHS of C2 to a negative voltage relative to ground which switches off the left hand transistor and the process repeats.

Answer 2

A capacitor gets charged or discharged if current flows through the capacitor. In this circuit: If the transistor on the left is not conducting, there is a current path from the positive supply through R1, the left LED, C1 and the base-emitter path of the right transistor. This path will charge C1 until the voltage of C1 reached a level that prevents further current from flowing through that path. In this circuit, the LED requires around 2V, the base-to-emitter path in the right capacitor requires around 0.7V, so if the capacitor accumulated enough charge to reach a level of 3.3V, the sum of the LED voltage, the capacitor voltage and the transistor voltage is 6V, which is equal to the supply voltage, so no further current will flow.

You should consider the voltage at a capacitor a secondary effect. The primary value to discuss at a capacitor is the amount of charge stored in it. Charge is created by current over time, and represents the number of electrons displaced in the capacitor. If you charged a capacitor of 10µF with an average current of 10µA for 3 seconds, you placed a charge of 30 µA*s = 30µC (C is the unit symbol for Coulomb, which is the same as Ampere multiplied by seconds) into it. The capacitor equation tells you that the voltage at the capacitor is 30µC / 10µF = 3V. While you can use the equation the other way to find out the charge of a capacitor if you know the voltage it has been charged to, you should keep in mind that physically, it's the charge that causes the voltage to appear and not the voltage that causes some charge to be in the capacitor.

When the capacitor is "fully charged", i.e. it reached 3.3V, the collector of the left transistor is at 3.3V (on the capacitor) + 0.7V (on the right transistor) = 4.0V. Now, if for some reason (to be explained later), the left transistor starts conducting, it will pull down the collector to around 0.2V, but this does not immediately change the amount of charge stored in C1, so the right end of C1 will still be 3.3V lower than the left end. If the left end is pulled to 0.2V, the right end is at -3.1V, which will turn off the right transistor.

Now, the primary current path through C1 changed: As the right end of C1 is below 0.7V, the base-to-emitter path in the right transistor does not conduct anymore, but some current can be sourced through R2. On the left side, the voltage is fixed by the left transistor being turned on. So now C1 is in the current path of R2 - C1 - left transistor. This current will discharge C1, and even slightly charge it in the "wrong" direction. As soon as C1 is charged to around -0.5V (measured from right to left, as we did all the time), the voltage at the base of the right transistor is enough to turn that transistor back on, which will then turn off the left transistor in the same way as we just discussed the turn-off process for the right transistor.

Answer 3

Late answer, but consider this: Both inductors and capacitors store energy, and there always is a time element to their charging and discharging. It's the "dt" in the equations. Because of this -

1. The voltage across a capacitor cannot change instantaneously. To do so would require infinite current. This is the central element of many timing circuits such as a monostable, relaxation oscillator circuits, and power circuits such as a charge pump.

2. The current through an inductor cannot change instantaneously. To do so would require infinite voltage. This is the central element of many switching power supplies such as the many variations of a flyback converter.

Because of #1, when you yank rapidly on one end of a capacitor, the other end has to move in the same direction.

Answer 4

Introduction

I am thrilled to answer this question, as it brings back fond memories from my high school days in the 60s. It was one of the first circuits I built, using PNP transistors and lamps in the collectors. While my teachers and professors explained the technicalities, no one ever really addressed the 'why' behind it. I have always been curious about the underlying concept. After many years of contemplation, I have finally grasped the underlying principles of this brilliant circuit design and am eager to share my insights.

Basic idea

The circuit consists of two timers in a loop, one for each half-cycle, which operate sequentially in time. They are subject to positive feedback, operating much like a latch. This reinforces their state and significantly enhances the switching speed. The timing capacitor is constantly recharging (reverse-charging). When its voltage drops close to zero, the circuit switches, improving the overall accuracy. This exact moment is sensed by the base-emitter junction which is reverse-biased during the charging process and does not allow any current flow. The base-emitter junction plays another important role since the capacitor is charged through it (this is not possible in the case of a FET implementation).

How to explore the circuit

The timer is made up of two components: a circuit that converts time into a voltage (capacitor) and a circuit that compares voltages (base-emitter junction). I will delve into the first part in detail. For the purpose of this intuitive explanation, I will replace the capacitor with a voltage source (a rechargeable battery) with the same voltage; this allows us to exclude time and examine the circuit using DC meters and DC Live Simulation. Thus, we will explore only some typical points of the exponential curve that are important for understanding. Individual schematics are like snapshots of circuit operation at different points in time.

How to convert time to voltage...

We can do this using a capacitor by changing its charge (voltage) in three ways.

... by charging a capacitor

Vc = 0 V: If we connect a 10 V voltage source through a 1 kΩ resistor (an ammeter with 1 kΩ internal resistance), the voltage across the capacitor is zero, and 10 mA current flows.

^{simulate this circuit – Schematic created using CircuitLab}

Vc = 10 V: Then the voltage increases exponentially and finally (almost) reaches 10 V; the current actually stops flowing.

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The problem with this method is that at the end of the process the voltage changes very slowly and the comparator threshold must be stable. Also, the range of voltage change is only 10 V.

... by discharging a capacitor

Vc = 10 V: Now the capacitor acts as a voltage source with exponentially decreasing voltage starting from 10 V.

^{simulate this circuit}

Vc = 0 V: Finally, both voltage and current become zero. The range of voltage change is 10 V as above.

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They have the same shortcomings as above - a sloping curve at the end and only a 10 V range.

... by recharging a capacitor

Vc = -10 V: This is our method where we first charge the capacitor back to -10 V and then start (re)charging it to 10 V. The result is impressive: The voltage range doubles (20 V)...

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Vc = 0 V: ... Since the current is doubled, the curve becomes steep in the middle (0 V) where the comparator switches.

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Vc = 10 V: The voltage should reach 10 V but in our case the circuit switches around 0.7 V.

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Conceptual bridge circuit

The question is how to recharge the capacitor. A classic way is through an H-bridge circuit. Something similar is in our multivibrator circuit. Let's consider the left part of the circuit.

Charging the capacitor

When Q1 is off and Q2 on, the capacitor quickly charges through the 1 kΩ Rc1 and the Q2's base-emitter junction. The charging current path is Vcc+ -> Rc1 -> C1 -> BE2 -> Vcc-. Initially, the capacitor voltage is 0 V...

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... and finally is close to 10 V.

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Recharging the capacitor

What I don't get, is how a capacitor getting a negative charge will stop current from the battery from going to the BJT all together. If, say, the transistor on the left is closed and C1 is charging, I feel that current from R2 should be able to split off at the junction to go left through C1 and right to the BJT on the right.

Electric current, similar to human behavior, always follows the easiest path. On the right side, we have a backward-biased base-emitter junction, while on the left side, a -10 V (better than 0 V) voltage source is present. So all the current will flow through the capacitor towards the negative voltage; the recharging current path is Vcc+ -> Rb2 -> C1 -> CE1 -> Vcc-. The capacitor slowly charges through the 100 kΩ Rb2 resistor.

^{simulate this circuit}

The voltage across the capacitor does not manage to reach 10 V since the base-emitter junction becomes forward-biased around 0.6 V and the circuit switches.

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Conclusions

• The circuit consists of two timers in a loop, one for each half-cycle, which operate sequentially in time.

• A positive feedback maintains their state and significantly enhances the switching speed.

• The timing capacitor is constantly recharging (reverse-charging).

• During this time, the base-emitter junction is reverse-biased and does not consume any current.

• When its voltage drops close to zero, the circuit switches; this moment is sensed by the base-emitter junction.

• In this BJT implementation, the base-emitter junction plays another important role since the capacitor is charged through it (this is not possible in the case of a FET implementation).