Explanation of this Electronic circuit

[Sindhu]

 This circuit has been used between an IR sensor and microcontroller (AT89C51). Can anyone explain this desgin?? why these particular values of resistors have been used? The circuit’s purpose is to give correct HIGH and LOW voltage to the microcontroller because the output from IR sensor is around 3.2V and the 8051 might be confused whether this is HIGH or LOW signal.

 

The switching circuit makes sure that the HIGH= 5V and LOW=0v. But why only these values of resistors?

 

Thanks

[Mehtab Ali]

this circuit is simply a switch. when the sensor senses the ir radiations it sends a current to the base of microcontroller, thus turning the transistor on and hence 

their is zero volts at the ‘o/p to 8051’ pin. similarly when their is no current coming from the sensor then the transistor is off and the o/p is high at 5v.

  The 10k resistor is choosen in such a way to keep the transistor in the saturation region if a current is applied at the base i-e Vce<0.7v.

because for transistor to be operated as a switch, it should be operated in cutoff and saturation region.

The other resistor is used for proper biasing of transistor.

[Mehtab Ali]

this circuit is simply a switch. when the sensor senses the ir radiations it sends a current to the base of microcontroller, thus turning the transistor on and hence 

their is zero volts at the ‘o/p to 8051’ pin. similarly when their is no current coming from the sensor then the transistor is off and the o/p is high at 5v.

  The 10k resistor is choosen in such a way to keep the transistor in the saturation region if a current is applied at the base i-e Vce<0.7v.

because for transistor to be operated as a switch, it should be operated in cutoff and saturation region.

The other resistor is used for proper biasing of transistor.

[AJISH ALFRED]

I assume your IR sensor gives digital output. Here the transistor is operated as a switch only.

 

Suppose the transistor is off, you will get 5v ( maximum possible voltage ) at the output and it is independent of the value of collector resistor.

 

But suppose the transistor is ON, you need the minimum possible voltage at the output ( 0.7V ), which is the saturation voltage of the transistor. At that time an input current flows through the transistor depends on the input voltage and given 4.7 K resistor. For that input current an output current flows depends on the hfe of the transistor. That output current should generate a voltage not less than ( 5 – 0.7 ) V in the collector resistor, and the value of that resitor is designed for the same purpose using V = IR equation.

 

Simply for on condition of transistor,

 

we select an input resistance first

calculate the input current

multiply that current with hfe of transistor to get the output current

now we know the output current ( I )

The output voltage ( V ) should be 0.7 V, so 5 – 0.7 V should drop across collector resistor

now we know the current ( I ) through collector resistor and also the voltage drop across it ( V )

get the required value of resistor ( R ) using the equation R = V/I

 

 

[AJISH ALFRED]

I assume your IR sensor gives digital output. Here the transistor is operated as a switch only.

 

Suppose the transistor is off, you will get 5v ( maximum possible voltage ) at the output and it is independent of the value of collector resistor.

 

But suppose the transistor is ON, you need the minimum possible voltage at the output ( 0.7V ), which is the saturation voltage of the transistor. At that time an input current flows through the transistor depends on the input voltage and given 4.7 K resistor. For that input current an output current flows depends on the hfe of the transistor. That output current should generate a voltage not less than ( 5 – 0.7 ) V in the collector resistor, and the value of that resitor is designed for the same purpose using V = IR equation.

 

Simply for on condition of transistor,

 

we select an input resistance first

calculate the input current

multiply that current with hfe of transistor to get the output current

now we know the output current ( I )

The output voltage ( V ) should be 0.7 V, so 5 – 0.7 V should drop across collector resistor

now we know the current ( I ) through collector resistor and also the voltage drop across it ( V )

get the required value of resistor ( R ) using the equation R = V/I

 

 

[wajid aziz]

Hi Sindhu
     you are using ir sensors. if you calculate the voltage of ir sensor circuit
only then it gives you only for 0.3-0.9V and controller read as low logic. it is not
enough voltage to read controller as high voltage.therefore this circuit is used.
when sensor gives minimum of 1.8V then it give logic as high=5v
     The purpose of this circuit is to increase the voltage. the question is how?
first, start with diode, diode is working in that place where it block negative
voltage only. if you dont want to connect the diode, then there is no effect on
the circuit but you must give pure dc supply.
     Then the resistor is connect with BJT. you know BJT is depend on current to operate
therefore, resistor in series must must be connect. I tell you in first paragraph
ir sensors gives only 0.3-09 V only. here is use npn transistor, and sensor wire
is connect with base which is positive and to increase the positive voltage. you see
emitter is ground and positive base voltage voltage go to the controller with logic high and controller
take positive voltage only and now it won’t be confused.Bjt use for voltage increase from 0.9-
3.2V approximately. 
    high resistance about 4.7k is used because you want to increase voltage not current
v=ir when resistance is high, current reduce and voltage increase.    

[Author]

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