IR sensor supply in the diagram

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 ?

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...

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

why 270 ohms is in 0.5 watts

Click to expand...

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?

Click to expand...

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.

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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).

Genovator said:

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...

Click to expand...

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

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

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

Genovator said:

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

- - - Updated - - -

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

Click to expand...

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