IR sensor supply in the diagram

[rajaram04]

hello sir

in the above diagram supply voltage to the circuit is in between 6 to 9 volts as shown above
then whats the use of 270 ohms & 150 ohms resistor actually ? & why 270 ohms is in 0.5 watts ?

 

[Genovator]

Because a zener diode is connected just after R7. That is, after 270R, potential on the line will be 5.1v. So max potential drop across R7 is 9-5.1 = 3.9v.
It is a huge potential drop. So drawing very low current will cause a huge power dissipation. max current can be drawn is 3.9v/270R = 14.4mA only...

And R5 seems a current limiting resistor, as 1738 needs just 5mA as supply current...

 

[rajaram04]

hmm i see , and here is one more circuit related to it containing a diode IN4148 with 100k resistor ?:?

 

[KerimF]

why 270 ohms is in 0.5 watts

In worst case, if the zener becomes faulty (shorted), the dissipation of the resistor (Rs):
P_max = 9 * 9 / 0.27 = 300 mW

In normal case, the minimum zener current is:
Iz = ( Vcc - Vz ) / Rs - I_load_max
Iz = ( 6 - 5.1) / 0.27 - 1.5 mA = 1.8 mA
(the maximum 1.5 mA from datasheet)
So the zener limits the voltage though Vz is in this case a bit lower than 5.1 V

whats the use of 150 ohms resistor actually?

It seems the designer of this circuit tried to supply the IR IC with as close as 5V.
Knowing (from datasheet) I_typical = 0.6 mA, the 150R drops 0.15*0.6 = 0.09 V.
Therefore Vc on the IR IC is 5.1 - 0.09 = 5.01 V.
In my opinion, it can be removed since Vc_max is 6 V.

=========

The output of 1738 is an open collector with a 80K pull-up resistor. With no load (open circuit) and no IR (pulsed) detection, the output voltage is Vc (about 5V).

 

[kam1787]

So max potential drop across R7 is 9-5.1 = 3.9v.
It is a huge potential drop. So drawing very low current will cause a huge power dissipation. max current can be drawn is 3.9v/270R = 14.4mA only...

Did you mis-type? 3V9 is more a normal potential drop IMO.

The power dissipation for the resistor is only 56mW - very small.

 

[Genovator]

The power supply is 6-9v. So I just said max PD = 3.9 considering 9v be the highest input..

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@rajaram
And about the 1N4148 diode, refer to this link....
http://www.doctronics.co.uk/555.htm

 

[rajaram04]

The power supply is 6-9v. So I just said max PD = 3.9 considering 9v be the highest input..

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@rajaram
And about the 1N4148 diode, refer to this link....
**broken link removed**

ohh ya okk sir i got it , thanks :smile: