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Monostable Circuits based on NAND gates

  Question 7.  What is meant by the term monostable?  ANSWER

How does the circuit work?  At first the circuit is stable.

• When switch S is open, the point L is at 1.

• The capacitor C is not charged and the voltage at M is 0.

• Therefore the output of Y is 1, and both inputs of X are 1.

• The output of X is 0.

•  Since the inputs of Z are 1, and the output of Z is 0.

  Now let us momentarily close the switch S.

• T becomes 0.

• Therefore the input L into gate X becomes 0.

• The output of X becomes 1.

• The flow of charge onto the capacitor C causes there to be a current through the resistor R.  This makes the voltage at M high.

• Therefore gate Y gives out a 0.

• Since the voltage at N is a zero, the output of gate Z is 1, and the LED lights up.

Question 8.  What does gate Z do and why?  ANSWER

This lasts a short time, determined by the RC time constant.

• The current through R decays exponentially as the capacitor charges up.  So the voltage at M decays exponentially as well.

• As the voltage at M drops, it passes below the threshold at which Y is triggered to change state.

• Gate Y gives out a 0 until the voltage passes below the threshold.

• Now gate Y changes to 1.

• Since L is at 1, this makes the output of X low again.

  If the switch is held closed, the output of X remains 1, but no current flows onto the plates of the capacitor, so the voltage at M remains low.  Therefore holding the switch has no effect on the behaviour of the circuit. We can show the behaviour of the circuit with timing diagrams for each of the points.

Depending on the threshold at which the gate triggers, it can be shown that the time period T at for which the output of Z is high is approximately RC.  

If the gate triggers at 0.5 Vs, then T  is approximately 0.7 RC

  Question 9. Why does the voltage at M show the shape shown?  ANSWER

NAND gate Astable

We can make an astable circuit the output of which oscillates at a frequency determined by the value of the time constant of a capacitor and a resistor. 

If you look carefully at the arrangements of the NAND gates, it does not take a genius to see that the two NAND gates are wired as NOT gates, so this set up is also called a NOT gate astable.  Let’s have a look at how the circuit works:

• Suppose the output of Y is high.

• This means that the input to Y is low.

• The capacitor will charge up.

•  A current flows through the resistor R which means there will be a voltage across it.

• This raises the input to Y to high and it will trigger to the output being low.

• Since X is connected to the feedback loop, its output will be low.

• The low output of X will cause the capacitor to discharge.

• This makes the input to Y low, hence the output to go to high.

• And so on…

  We can summarise this in the timing diagram:

We can show that the mark time is given by the relationship:

                                    t_H is approximately 1.1 RC

  Similarly the space time is given by:

                                    t_L is approximately 1.1 RC

  Therefore the period T = t_H + t_L = 2.2 RC

So the frequency f = 1/T = 1/2.2 RC

  Question 10. A NAND gate monostable has a capacitor of capacitance 20 mF with a resistor of resistance 150 kW.  What is the period of the monostable?  ANSWER

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