Logic Gate Astable Multivibrator — How It Works and Example Circuits

The logic gate astable multivibrator is a simple oscillator circuit built using digital logic gates, such as NOT (inverter) gates. By adding a feedback network that comprises a resistor and a capacitor to cascaded inverters, the circuit generates a continuous square-wave output.

Logic Gate Astable Multivibrator — How It Works and Example Circuits

In this article, we’ll examine how an inverter-based astable multivibrator operates step by step, derive its frequency expression, and explore example circuits implemented with CMOS devices.

What are the Multivibrators?

Multivibrators are simple groups of electronic circuits or systems that have two states. They can switch between the two states. However, the condition for switching and the function after the switches depend on the type of multivibrators. Here are three of the multivibrator types:

※ Astable Multivibrator
An astable multivibrator has no stable states (“a-“stable). What it means is that, when powered, an astable multivibrator continuously switches between the two states. The constant switching creates an oscillation, where the oscillating frequency is the switching speed.

It’s often used as an oscillator to generate nonsinusoidal signals such as square or triangular waves. For instance, a relaxation oscillator used in vehicle flashing indicators is a type of astable multivibrator.

※ Monostable Multivibrator
A monostable multivibrator only has one stable state (“mono-“stable). When triggered by an external source, it switches from the stable state to the unstable state. It will then maintain the unstable state for a certain period of time before switching back to the stable state.

It’s sometimes used to produce a constant time delay when triggered. It’s also used in a single-pulse generator in the lab or for testing purposes.

※ Bistable Multivibrator
A bistable multivibrator has two fully stable states (“bi-“stable). As a result, if left alone, it will always maintain its latest state. When triggered externally, it switches to and holds on to the other state, unless triggered again.

Also known as a latch or flip-flop, it’s used mainly in sequential logic applications.

How Does an Inverter Gate Astable Multivibrator Work?

In this article, we’re mainly concerned with the astable multivibrator and how to create it using common logic gates (inverter, or NOT gate). That said, let’s look at a typical inverter astable multivibrator with a resistor and capacitor, and try to understand how it works.

Fig. 1 – A typical inverter astable multivibrator circuit with a resistor and capacitor outputs a square-wave signal.

The circuit has two inverters (NOT gates) connected in cascade. The resistor is connected from the first inverter’s output (G1) to the capacitor that is itself connected to the second inverter’s output (G2). The node (point A) where the resistor and capacitor join is also fed back to G1’s input.

The circuit output is pulled from the second inverter’s output (G2). The multivibrator should output a square-wave signal with frequency roughly determined by the following formula.

$$f=\frac{1}{2\, \ln(2) \, RC}$$

This form of expression is common in RC-based astable multivibrators. However, it could only provide an approximate value in practical implementations. Still, this gives us insight into how resistance (R) and capacitance (C) affect the output frequency.

Step-By-Step Operation

inverter astable multivibrator during the "1" output state, showing current flow from G2 to G1
Fig. 2.1 – A typical inverter astable multivibrator during the “1” output state, showing current flow from G2 to G1.

Figure 2.1: Suppose that the starting condition is G1’s output = “0” and G2’s output = “1.” Because G2’s output has a higher potential, current begins to charge the capacitor negatively and flows across the resistor to get to G1’s output. G2’s output has source current, while G1’s output has sink current. At this moment, a voltage would already have developed across the resistor.

Since G1’s output is “0,” we can assume that its input voltage (A point) is more than the threshold level (A.K.A. trigger point or logic threshold) for “0.” For simplicity’s sake, let’s assume the threshold levels for both logics are about half the supply voltage (VDD). I.e., the gate counts more than half of VDD as “1”, while counts lower than that as “0”.

Therefore, the voltage at point A—and in effect, across the resistor, since G1’s output acts as the ground—is currently more than ½ VDD. The circuit output is from the same node as G2’s output, which is at “1,” or equal to VDD.


inverter astable multivibrator, showing the output state switching from "1" to "0" logic
Fig. 2.2 – A typical inverter astable multivibrator, showing the output state switching from “1” to “0” logic.

Figure 2.2: As the capacitor is being charged, the charging current begins to drop. As a result, less current flows over the resistor, causing the voltage at point A to fall below the threshold level, or less than half the supply voltage, which causes G1 to switch its output logic from “0” to “1.”

Because it’s second in the logic chain, G2 also switches its output logic from “1” to “0.” This is effectively the inverse of the starting condition. In addition, the change in current direction also causes the capacitor to switch the polarity of its charge later.

The voltage across the resistor can briefly go negative as the switch happens. However, the chip protection should prevent G1’s input from going below 0V.


inverter astable multivibrator during the "0" output state, showing current flow from G1 to G2
Fig. 3.1 – A typical inverter astable multivibrator during the “0” output state, showing current flow from G1 to G2.

Figure 3.1: Now, since G2’s output has lower potential, the capacitor begins to discharge to it and becomes positively charged. The source current now flows from Q1’s output to Q2’s output as sink current. Current now flows in the opposite direction compared to the previous state. As current flows over the resistor, voltage develops across it, and the voltage at point A begins to rise yet again.

The circuit output is from the same node as G2’s output, which is currently at “0,” or equal to GND.


inverter astable multivibrator, showing the output state switching from "0" to "1" logic
Fig. 3.2 – A typical inverter astable multivibrator, showing the output state switching from “0” to “1” logic.

Figure 3.2: Again, as the capacitor discharges, its current begins to drop. However, before it came to a standstill, the voltage across would go above the threshold level, which, as we have established, is half the supply voltage. In other words, the voltage at point A rises above ½ VDD. When this happens, G1 switches its logic again, causing G2 to follow right after.

As before, the capacitor would soon switch the polarity of its charge as the current switches direction. After the switch, the voltage at point A abruptly changes to positive, potentially higher than the supply voltage. However, the chip protection shouldn’t let the actual gate input voltage exceed the supply. 

This brings us back to the starting condition of G1’s output = “0” and G2’s output = “1.” The cycle will start over again from here, repeating indefinitely.

As we can see, the multivibrator output switches back and forth between “1” and “0” depending on the current direction and voltage at point A. We could also graph the output voltage relative to the capacitor voltage and the point A voltage, which is shown below.

Voltage Waveforms and Relationships

three graphs showing output voltage, capacitor voltage, and point A voltage.
Fig. 4 – Three graphs showing the waveforms and the relationship between output voltage, capacitor voltage, and point A voltage.

In the first graph, we can see that the output voltage swings between VDD (“1”) and ground or 0V (“0”). This creates a symmetrical square wave signal. When driving a load, the G2’s output sources current during “1” and sinks current during “0.”

Moving on, next we have the capacitor voltage drop across. The waveform shows the capacitor reversing its charge polarity between each output state. The graph itself resembles a sawtooth wave with slight tapering near the peaks due to the decreasing current as the voltage approaches the switching point.

When the output is “0,” current from G1’s output charges the capacitor positively; thus, its voltage rises. Likewise, when the output is “1,” current from G2’s output charges the capacitor negatively; as a result, its voltage falls.

Note that the capacitor voltage swings by approximately ½ VDD in each direction. This is because the capacitor drop across voltage is related to G1’s input voltage. Once the input reaches the gate’s switching threshold of ½ VDD, G1 changes state, reversing the charging direction. As a result, the capacitor voltage cannot continue increasing in that direction beyond the point required to trigger the gate.

Lastly, we have the point A voltage. It shows an alternating ramp-like triangle wave centered at the threshold level, which is ½ VDD. The voltage jumps abruptly after the state transitions, then gradually decreases nonlinearly. 

Its rise and fall mirror that of the capacitor voltage, which is rising during the “0” output and falling during the “1.” However, it will always vary toward ½ VDD, as that’s what triggers the state change. As the state transitions while the capacitor still has a voltage drop across it, the input node can momentarily rise above the supply or fall below 0V. However, in practice, the input protection diodes clamp these excursions to slightly beyond the supply rails.

Example Basic Astable Multivibrator Circuits

Using what we learned from the theoretical circuit above, let’s now look at a couple of real, practical gate-based astable multivibrator circuits. These circuits fall under the relaxation oscillator category. Note that these circuits are just a basic configuration using different ICs and logic gates.

Basic 1kHz Clock Generator Circuit Using CD4049 NOT Gates

With this RC configuration, this relaxation oscillator can generate a 1kHz square-wave signal. The supply voltage (VDD) is given at 4.5V to 6V. VDD exceeding this range may be possible as long as it’s within the chip supply range; however, it may noticeably shift the frequency.

circuit schematic of basic 1kHz clock generator circuit using a CD4049 NOT gate chip
Fig. 5 – Basic 1kHz clock generator circuit using a CD4049 NOT gate chip.

The IC used is the CD4049 hex inverting buffer gate chip. An equivalent CD4069 should also work. In theory, the unbuffered version of the CD4069 works best for this usage. Note: CD4049 and CD4069 have different pinouts.

Using the frequency formula from before, a resistor value of 680kΩ and a capacitor value of 0.001μF (1nF) should result in approximately a 1.06kHz signal.

$$\frac{1}{2\, \ln(3) \times 680\text{k}\Omega \times 1\text{n}F} \approx 1060\text{Hz}$$

1KHz Oscillator Circuit Using NAND or NOR Gates

Here is another version of the same 1kHz square-wave RC relaxation oscillator configuration using a more popular CD4011BE quad NAND gate chip. The NAND has both of its inputs tied together, creating an equivalent inverter. Like the previous circuit, we use 5V VDD here as well.

circuit schematic of basic 1kHz clock generator circuit using a CD4011 NAND chip
Fig. 6 – Basic 1kHz clock generator circuit using a CD4011 NAND chip as a NOT gate substitute.

Likewise, we can also use a NOR gate from the CD4001BE quad NOR gate chip to construct an equivalent inverter. We’re using the same RC timing, so the result should still be the same. 

circuit schematic of basic 1kHz clock generator circuit using a CD4001 NOR chip
Fig. 7 – Basic 1kHz clock generator circuit using a CD4001 NOR chip as a NOT gate substitute.

1KHz Oscillator Using Schmitt Trigger

We could also use an inverting Schmitt trigger gate to create an astable multivibrator. This relaxation oscillator circuit uses a different frequency formula than the two-inverter version. The IC used is a CD40106 hex Schmitt trigger. However, we’d also tried with 7413 TLL, but it wouldn’t work for us.

circuit diagram of 1kHz clock generator circuit using a CD40106 Schmitt trigger inverter chip
Fig. 8.1 – 1kHz clock generator circuit using a CD40106 Schmitt trigger inverter chip.

The operating principle is similar to the two-gate solution: feedback is provided through an RC network that determines the timing. However, unlike a standard inverter, the inverting Schmitt trigger has two switching thresholds — a lower and an upper level.

The lower threshold is approximately ⅓ VDD, while the upper threshold is roughly ⅔ VDD. As a result, the capacitor charges and discharges between these two levels, which determines the charge and discharge timing. The simplified frequency formula for this circuit is as follows.

$$f=\frac{1}{\ln(3) \, RC}$$

It strictly assumes that the lower and higher thresholds are ⅓ VDD and ⅔ VDD, respectively. According to this formula, with R = 180kΩ and C = 0.01μF (10nF), the output frequency is approximately 506Hz.

However, in practice, this circuit produces a 1.1kHz square-wave signal with a 49% duty cycle at a 5V supply, significantly higher than the 506Hz approximated by the simplified formula. This demonstrates that the actual switching thresholds of the CD40106 at 5V deviate from the ideal ⅓ and ⅔ VDD values, which is the main limitation of these formulas.

real output reading of CD40106 1kHz clock generator circuit
Fig. 8.2 – 1kHz clock generator circuit using a CD40106 output oscilloscope readings.

Of course, there’s also a more accurate way of determining the frequency. That’s to use the chip’s actual measured upper and lower threshold voltages. However, that would be too complex for these basic circuits. Plus, in practice, we could easily tune the resistor value until we get our desired frequency.

In terms of waveform, the Schmitt trigger version generates a more symmetrical signal compared to the normal inverter gate version. The frequency is marginally higher than the 1kHz target, but it can be adjusted down by changing the resistor. In addition to a more compact circuit—needing only one Schmitt trigger gate—this circuit looks to be an ideal choice.

1kHz clock generator circuit using a CD40106 tested on a breadboard
Fig. 8.3 – 1kHz clock generator circuit using a CD40106 tested on a breadboard.

A Related Circuit: The Ring Oscillator

A ring oscillator is technically not a multivibrator, although it is still an astable digital circuit that continuously switches between logical “0” and “1.” The reason we want to bring it up here is that it looks similar to the astable multivibrator and relies heavily on logic gates in its operation.

example inverter gates ring oscillator circuit
Fig. 9 – Simple inverter gates ring oscillator circuit, showing three inverters (NOT gates) in a feedback loop.

A ring oscillator essentially consists of an odd number of inverter gates connected in a feedback loop, which is where the ‘ring’ part of its name comes from. Without additional components, it can generate a square-wave signal at very high frequencies, often in the MHz range. The frequency depends mainly on the propagation delay of each gate.

It’s a simple and interesting solution for producing non-sinusoidal signals. However, it’s outside the scope of this article, so we’ll discuss it in more detail in a separate article.

Check out these related articles, too:

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15 thoughts on “Logic Gate Astable Multivibrator — How It Works and Example Circuits”

    • IN MY OPINION ENGLISH IS VERY UNDERSTANDABLE, BUT IT DOESN’T SEEM VERY KIND TO REMARK IT, CONSIDERING THE EFFORTS MADE BY THE AUTHOR! MOREOVER YOU TOO MADE SOME MISTAKES ON YOUR WRITTEN!

      Reply
      • Hello,
        As for my English writing, it is undeniably bad—some parts were even considered unreadable. But in the last few months or so, we have been trying our best to improve our writing in our new posts or even rewrite some of our old posts on the website.

        But perhaps if you were to be so kind as to give us a piece of advice or two or even point out some of the things that we might have done wrong, that would help us tremendously improve our understanding of English writing even further.

        Thanks,

        Reply
  1. Keep the good work rolling.
    But I want to ask as electronic hubbist but who could not get chance or know how to start carry out electronic projects posted on this platform or see somewhere,where to start from?
    Thanks.

    Reply
    • I very much appreciate your question. It might be a great idea. I and my kids will compile how-to and ideas to create simple electronic projects for hobbyists and beginners. Please be patient.

      Reply
    • Oh! it is a great idea. I will teach my daughter to learn about AM radios. But now she loves LED flasher and LED running circuit. Never mind, electronics has a lot of information to do (fun).

      Reply
  2. The problem is this:
    The part of the cycle he is describing is the DISCHARGING of the capacitor. And it does not discharge via the resistor. But via the input of the first inverter.
    But he knows nothing about the diodes on the input of the gate.
    All the rest of his discussion is completely incorrect.
    Also he does not understand the fact that a gate changes state when the input voltage is slightly higher and slightly lower than mid-rail voltage. It does not have to rise to 100%. Maybe something like 30%, 70%.
    I will take up where he has started. The capacitor is fully charged and the output rises. As it rises, it delivers a voltage to the input that is higher than the supply. A protection diode on the input of the first gate will not allow the input to rise higher than the supply, so, as the output rises, the top of the capacitor “hits a ceiling” and will not rise. So it gets discharged via the protection diode. Now the output is HIGH and the input is HIGH.
    It would stay like that all day if it were not for the resistor. The resistor CHARGES the capacitor, because the bottom lead of the capacitor is HIGH and the bottom of the resistor is LOW.
    As the capacitor charges, the voltage on the input of the first gate reduces and when it fall to about 30-40% of rail voltage, point B goes HIGH. This makes the output go LOW and the whole capacitor drops by a voltage equal to rail voltage.
    The input voltage had about 30% rail voltage on it and now the voltage drops by full rail voltage.
    This means the input voltage will drop below 0v, but a protection diode on the input prevents it dropping below 0v, so the input just goes to 0v and the rest of the voltage on the capacitor is quickly removed.
    Now the output of the first gate is HIGH and it begins to charge the capacitor via the resistor and when the voltages reaches about 60% of rail voltage, the two gates change state to start the next cycle.
    Note: the capacitor has times when it is partially charged and discharged very quickly via the protection diodes.
    This has never been mentioned in any discussion. Everyone thinks it is a “smooth cycle.”

    Reply
    • First of all, I feel truly honored that Colin Mitchell (https://www.talkingelectronics.com
      ) took the time to share you knowledge with us.
      I sincerely acknowledge the mistakes in my content and am ready to learn and improve.
      It’s truly remarkable to understand how the circuit works correctly.

      I kindly ask for some time to review and reorganize the content to ensure it is accurate.

      Thank you very much again for your valuable input.

      Apichet Garaipoom

      Reply

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