Examination of pseudo-monostable ordered by a downward face   Examination of a true monostable circuit   
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Created it, 06/10/19

Update it, 06/10/25

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Reception

This practice enables you to study the monostable circuits. The latter are rather often used in the design of digital circuits. These monostable concerns at the same time numerical electronics and analogical electronics.

The rockers can have two stable states and place from one state to the other thanks to an external order. On the other hand, the monostable one is characterized by only one state of stability.

This monostable circuit can place in an unstable state under the effect of an external order. It will remain in this state of instability for one function length of time of the characteristics of the circuit, then will return in a stable starting state.

The integrated monostable circuits are often used to increase the duration of a short signal, or to create a temporization. Their principle is based on a time-constant carried out by means of a cell R.C. (resistance and condenser).

1. - PREPARATION OF THE MATERIAL

In this practice, you will use the following components, which you will take among the material in your possession :

1 integrated circuit MM 74C08

1 integrated circuit MM 74C04

1 integrated circuit MM 74C74

1 integrated circuit MM 74C221

1 integrated circuit CD 4528

1 integrated circuit MM 74C85

2 diodes 1N 4148

1 electrolytique capacitor with tantalum 0,33 µF - 10 V

1 electrolytique capacitor with tantalum 0,47 µF - 10 V

2 electrolytique capacitors with tantalum 1 µF - 10 V

1 electrolytique capacitor with tantalum 10 µF - 10 V

2 resistances of 1 MW - 1 / 4 W tolerance ± 5 % (chestnut - black - green - but)

1 resistance of 2,2 M - 1 / 4 W tolerance ± 5 % (red - red - green - but)

2 resistances of 470 k - 1 / 4 W tolerance ± 5 % (yellow - purple - yellow - but)

2 resistances of 330 k - 1 / 4 W tolerance ± 5 % (orange - orange - yellow - but)

2 resistances of 220 k - 1 / 4 W tolerance ± 5 % (red - red - yellow - but)

2 braids of insulated canned wire (a red, black).

2. - FIRST EXPERIMENT : USE OF A PSEUDO-MONOSTABLE TO GENERATE IMPULSES OF DURATION GIVEN

You saw that the control signal of a digital circuit can be consisted the impulse generated by the push-buttons assembled on the desk.

The duration of this control signal corresponds to the time during which the push-button is inserted.

This type of control signal is not always suitable. Indeed, certain digital circuits require a control signal of well defined duration.

The monostable integrated circuits are used to this end.

You now will examine simplest of the monostable circuits called pseudo-monostable because it is an imperfect monostable circuit.

Begin the experiment now by carrying out the assembly of the circuit while following the indications hereafter.

2. 1. - REALIZATION OF THE CIRCUIT

a) You ensure that the food is disconnected and remove matrix and group of connectors the whole of the connections established previously as well as the integrated circuit into ICX.

b) Carry out the assembly such as it is indicated to the figure 1-a, while following the following indications :

Pseudo_monostable.jpgSchema_electrique_du_pseudo_monostable.gif

      take care of the correct orientation of the integrated circuit MM 74C04. All the integrated circuits are located by a notch or a point as represented on figure 2.

Identification_des_broches_d_un_CI.gif

     The pin 14 is connected to the positive tension.

      Pin 7 is connected to the negative tension (mass).

      R1 resistance of 1 MW 1 / 4 W tolerance 5 % (chestnut - black - green - but) is connected between pin 1 and masses it.

      Pins 2 and 3 are connected between them.

      Pin 4 is connected to the L0 contact of the group of connectors.

      Insert on the matrix the electrolytique capacitor into C1 tantalum of 1 µF - 10 V between pin 1 of the integrated circuit and a contact located on the left integrated circuit.

Do not insert for the moment the second condenser illustrated in dotted line.

You must respect the polarity of C1. The positive terminal is located by the sign (+) as indicated on figure 3.

Condensateur_electrolytique_au_tantale.gif

      The positive connection of C1 is connected to SW0 by a wire of connection, while the negative connection of C1 is connected to pin 1 of the integrated circuit MM 74C04.

The finished assembly, check the exactitude of the connections carried out (figures 1-a and 1-b).

2. 2. - OPERATIONAL TEST

a) Connect the food. The LED L0 is extinct.

If it is not the case, check with attention the whole of the assembly. Note however that with the powering, the LED can ignite a short moment.

b) Put switch SW0 on position 1.

The LED ignites during 1 second approximately to die out then, switch SW0 being always in position 1. Repeat the experiment while always leaving position 0 for SW0. The LED always ignites during 1 second approximately.

c) Commutate SW0 on position 0. Put SW0 on position 1 and quickly bring back SW0 on position 0. The LED dies out as soon as SW0 is brought back on position 0, even if it did not ignite during 1 second.

This is the principal limit of use of the circuit. Indeed, it is necessary that the impulse of order generated by SW0 lasts at least as a long time as the duration of the impulse of exit.

In the contrary case, the impulse of exit of the circuit will last only the time during which the switch is in position 1.

In a similar way, after having returned in position 0 for SW0 by reducing the duration of the impulse of exit, you give SW0 on position 1. You notice whereas the impulse of exit will be still lower than 1 second.

In theory 6 digital electronics you will find the explanations relating to this phenomenon.

d) Now take another electrolytique capacitor with the tantalum of 1 µF - 10 V (C2) and insert it by respecting the polarities of the terminals in parallel with the first C1 condenser, as illustrated with the figure 1-a in dotted line.

The equivalent capacity of the two condensers is 2 µF.

e) Actuate switch SW0. You note that the impulse of exit lasts approximately twice longer than the impulse of exit of the preceding test.

Thus the duration of the impulse increases with the capacity.

f) Remove C2, inserted in the last. There remains only C1, of value of 1 µF and replace the resistance of 1 MW (R1) by another resistance of 2,2 MW.

Remade the test. You also note that the LED ignites approximately 2 seconds.

An approximate value of this duration of impulse called T is given by product RC.

T = RC        (T in second, R in Ohm and C in Farad)

With a condenser having a capacity C = 1 µF = 1 x 10-6 F and a resistance R = 1 MW = 1 x 106 W time T is equal to :

T = 1 x 106 x 1 x 10-6 = 1 s

This value is that obtained in experiments with the assembly. In fact, you do not obtain this value exactly, because much of the characteristics of the components (resistance, capacity, reverser) are not taken into account.

In the assembly, you used two reversers whereas only one would have been enough for the test to themonostable one.

In fact, the second reverser enables us to light the LED during the impulse of exit. The forms of waves deferred on the diagram of the figure 1-b represent the voltage waveform in various points of the circuit, such as it appears with the oscilloscope.

The commutation of SW0 in position 1 delivers a positive impulse on the C1 condenser. At entry 1 of the diagram, there is also a positive impulse which decrease gradually. At exit 2, you have the negative impulse which lasts 1 second (or 2 seconds) and at exit 4, the same reversed impulse.

HIGH OF PAGE 3. - SECOND EXPERIMENT : EXAMINATION OF A PSEUDO-MONOSTABLE ORDERED BY A DOWNWARD FACE

The circuit relating to this experiment is very similar to that carried out before, as you note it on figure 4.

Schema_electrique_du_pseudo_monostable (1) .gif

It remotely only by R1 resistance, connected here to the positive tension, by the absence of the second reverser and finally by the inversion of the terminals of the C1 condenser.

You will carry out the assembly of the circuit using the electric diagram and then check operation while following the same procedure as in the preceding experiment.

Disconnect the food. Withdraw the terminal of R1 connected to the negative tension and connect it to the positive tension. You connect exit 2 of the first reverser to the L0 contact.

Do not forget to reverse the C1 condenser.

In this experiment, SW0 will be in initial position on 1. After having checked the exactitude of wiring, connect the food and commutate SW0 in position 0. The LED L0 ignites approximately 1 second with R1 = 1 MW and C1 = 1 µF and approximately 2 seconds with R1 = 2,2 MW and C1 = 1 µF.

The only difference compared to the preceding experiment is the fact that you apply a negative impulse to entry 1 of the reverser. This assembly avoids the use of a second reverser.

HIGH OF PAGE 4. - THIRD EXPERIMENT : EXAMINATION OF A TRUE MONOSTABLE CIRCUIT

The pseudo-monostable circuits seen previously are not able to provide impulses of exits longer than the impulse of order.

By using a rocker D and the cell including/understanding a resistance, a condenser and a diode, you experienced now a true monostable circuit, the duration of the impulse at exit not being a function of the duration of the impulse of order.

 4. 1. - REALIZATION OF THE CIRCUIT

a) Disconnect the food.

b) Remove all the components and the connections relating to the preceding experiment.

c) Carry out the assembly indicated to the figure 5-a, using an integrated circuit MM 74C74 (including/understanding two rockers D), a resistance of 1 MW, a condenser of 1 µF and a diode of the type 1N 4148.

Vrai_circuit_monostable.jpgSchema_electrique_d_un_vrai_circuit_monostable.gif

Do not omit the integrated circuit MM 74C00 in support IC1 (debouncing circuit).

You thus carried out the circuit schematized with the figure 5-b.

Entry CLOCK is connected to the contact P0Front_Montant.gif.

The exit Q is connected to the L0 contact.

4. 2. - OPERATIONAL TEST

a) Connect the food. L0 is extinct or remains lit a short moment before dying out.

b) Press on the P0 button. The LED L0 ignites during 1 second. The exit Q thus passed to the state H during 1 second.

c) Support and slacken immediately P0. The LED L0 always ignites 1 second.

It is not thus necessary any more to support on P0 beyond second, but a short impulse is sufficient to order the circuit.

Now let us examine the operation of this circuit using the figure 5-b.

The face going up of the clock signal allows the transfer of the data DATED to 1 at exit Q.

Q_barre.gif Thus pass on the level L. the condenser C thus takes care through resistance R. the tension present at entry CLEAR decreases and when this tension reached a rather low threshold, entry CLEAR is activated.

The exit Q passes to 0 and Q_barre.gif 1. Time necessary to activate CLEAR is a function of the time-constant RC. When Q_barre.gif is at the high level, the diode D is polarized in the direct direction and allows the fast discharge of the condenser C. entry CLEAR finds the initial potential in a high state.

The circuit is found under the initial conditions of rest.

You will try to change the value of resistance and the condenser (put in parallel of condenser and also of resistance).

The time of illumination of the L0 diode is always equal to approximately the time-constant RC.

In conclusion, the true monostable circuit shows the following characteristics :

     it generates an impulse of given duration each time a face going up is applied to entry CLOCK.

     duration of the impulse at exit east function only of the values of R and C.

     the control signal can fall down immediately on the level L without modifying the shape and the duration of the output signal. 

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Daniel