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Operation of the monostable pulse generating circuit using OPAMP

September 27th, 2011 Remo No comments

monostable_using_opamp

monostable_signal-waveforms

Monostable generation

Here in this circuit the circuit generates a quasi-stable state transition from the stable state.Say the output of the V0= L+ I mean HIGH or Vsat.Now feedback fraction=b=R1/(R1+R2).Now as the output voltage is L+ then voltage at the C terminal is bL+.And as the diode D1 is getting High voltage at the P side so diode D1 is ON.So the voltage at the terminal B is just one dide drop say VD1.So the output remains at L+(I mean HIGH).So thisnis the stable state.BUT diode D2 is OFF as R4 is high.So this is the stable state of the monostable multivibrator.

How we go from stable state to quasi-stable state?

We apply a pulse at the trigger terminal. So then diode D2 will now conduct and the voltage at the C terminal will drop. Now when the voltage at the C terminal drops below voltage at B terminal then the output changes to L- or LOW state so this way HIGH to LOW(quasi-stable) state transition occur.

When back to stable state again from quasi-stable state.

Answer is when the output is at Low voltage (L-) then the voltage at terminal C is bL-.Now as the trigger is removed (because trigger is a very short duration pulse).So now as the diode D1 p-sde has low voltage so DIO D1 will be OFF and the capacitor C1 will try to charge towards L-.So a time will come when the voltage at the B terminal will be less than bL- and then output of the OPAMP go to L+ state again. But still this is not the stable state. Stable state will be achieved when the diode D1 will be On again.

What is recovery time

When the capacitor C1 will discharged as output (point A ) is L+. So C1 will discharge first and then when voltage at point B is more than diode potential barrier then diode D1 will be ON and then current will be flown via A to D1 and voltage at point B is again one diode drop.So stable state achieved. The time required to discharge the C1 is called recovery time.

Mono stable Multivibrator using IC 555:Details

September 27th, 2011 Remo No comments

untitled

fig2

See the figure and I will explain that to you.

First initially the SR Flip flop is in RESET state. Those who donot know what a flip flop is just keep remembering that it has 2 inputs named S and R and two outputs named as Q and Q_bar.If Q is HIGh then Q_BAR is low and vice versa. Note. Whan S=1 R=0 Then Q=HIGH and when S=0,R=1 Then Flip flop is in reset state. And When S=0,R=0 Then it is in no change state and Q and Q_bar value remains at it was earlier for other values of S and R. Here S-R kind of flip flop is being used.

So initially the flip flop is in reset state that means Q=LOW and the output of the Flip flop may be thought as HIGH ,another thing need to remember is that The trigger PIN is initially kept at a voltage higher than the VTL. So when the situation is like this then as VTH=2/3VCC and VTL=1/3VCC(because of the voltage divider network –voltage VCC is divided in three parts by using 3 equal registers).So What is the output of comparator1 :It is LOW.Why ?Because as Q is LOW so the Q_Bar is HIGH and this turns ON the transistor and as the point C is connected with the collector of the transistor so the voltage is zero there. Now consider the comparator 1 output ,it has Voltage LOW at +ve terminal than the voltage at the –ve terminal so output is LOW. For comparator 2 as the –ve terminal is maintained at a voltage more than VTL so the output is also LOW. So what we got –the outputs of the both comparators are low-So SR flip flop state remains unchanged –that means output will be at stable stage LOW.

What happens when we apply a trigger:

When we apply a trigger at the trigger terminal having pulse height lesser than VTL then output of comparator 2 is HIGH and output of comparator 1 is LOW, So this combination of S and R will set the flip-flop and so the output is HIGH now. So Q_bar is LOW and then and transistor goes OFF. So this way we reached at Quasi-stable state from the stable state.

How long to stay at Quasi-stable state ?

Now when it goes to quasi-stable state then as point C is no longer connected to ground via the transistor as transistor is OFF NOW so the capacitor begins charging so voltage at point C go on increasing ,as the capacitor will try to charge towards VCC so a time will come when the voltage at point C will exceed the VTH , Then the output of the comparator1 will go HIGH and the comparator 2 has LOW output because triggering pulse gone. So this is the RESET condition and the Flip flop will reset and the Q=LOW again. Q_bar is HIGH and transistor will be ON and then voltage at point C will now go towards zero after discharge of the charge capacitor stored the voltage at C is completely zero and so recovery done and multivibrator is aging at the stable state LOW.

Motor control using simple Schmitt trigger.

September 27th, 2011 Remo No comments

motor

Hello here we will see how we can control the speed of a DC motor by using a Schmitt trigger IC.Note the speed of the motor can be controlled by various ways ,may be by reducing the DC voltage across the motor etc .

But here we will follow the most conventional way of reducing/increasing the speed of the motor.The basic principle is if say the motor operates at speed 5 V DC then if this is done by sending a pulse of period say x with x/2 ti,e it has 5V and x/2 time it has 0 V. That means we are sending a square wave with 50% duty cycle.Because 50% of the time period we have HIGH voltage across the Motor(5V) and 50% of the time period we have 0 V across the motor.So what will be the result?

The answer is motor will rotate no doubt.But if we increase duty cycle more ? then ? then the speed of the motor will increase because the motor will now have the DC voltage across it more time,because we increased the duty cycle.How we increased te duty cycle ?simple just now we made a square wave such that 75% of time now there is HGH voltage across the motor and 25% of time there is 0V across the motor.So on the average the motor is getting more voltage in the sense that within one time period of the pulse that is being applied then motor is getting more average voltage when the dutry cycle is HIGH.

Actually what we do is that we configure an astable mutivibrartor such a way that its duty cycle can be varied using a potentiometer. So as the duty cycle can be varied then how this is used with the motor? The answer is the motor is connected with a transistor and the astable output is given at the base of transistor. So the transistor will be made on/off based on the duty cycle that way the Motor will be connected or unconnected with the supply at the same rate as the transistor will be ON/OFF. So this way the speed of the motor can be increased and decreased.

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