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https://www.electronicshub.org/wp-content/uploads/2013/10/infrared-control-remote-switch-Receiver-Circuit-1024x425.jpg

In this circuit, resistor R4 (just before the transistor) is used to stop loading the transistor.

I do not understand the part "resistor used to stop loading the transistor".

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  • \$\begingroup\$ "to stop damaging the transistor" would be more accurate. Not to mention damaging the 4018 and the LED. \$\endgroup\$ – Finbarr Jun 26 at 15:51
  • \$\begingroup\$ It is to limit current through the transistor. \$\endgroup\$ – evildemonic Jun 26 at 16:21
  • \$\begingroup\$ It's weird, inaccurate wording. \$\endgroup\$ – DKNguyen Jun 26 at 16:39
  • \$\begingroup\$ it probably means ... resistor used to prevent overloading the transistor \$\endgroup\$ – jsotola Jun 26 at 23:28
  • \$\begingroup\$ It's not a good sentence. Is it translated from some other language? \$\endgroup\$ – Pete Becker Jun 27 at 12:10
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The short answer is, as other commenters have stated, "to limit the current through LED and the transistor". Here's the equivalent circuit of that section:

schematic

simulate this circuit – Schematic created using CircuitLab

!Q1 output of U1 can be at the level of U1's supply voltage, 12V.

From KVL, \$\mathrm{12V = V_{D7} + V_{R4} + V_{BE-Q7}}\$

So, the current flowing through these components will be \$\mathrm{(12V-2\cdot(0.6V))/660\Omega = 16mA}\$ which is quite acceptable for illuminating an indicator LED and saturating a transistor which drives a relay.

If R4 is shorted then D7 and the BE junction of the transistor Q1 will see the full 12V across them. Here's what can happen next:

  • The !Q1 output of U1 will see very low resistance. So a high current will flow through the diode and the transistor's BE junction even for a short time.
  • BE junction of Q1 (and thus the transistor) and/or D7 will break down which (most likely) results in a short-circuit.
  • Finally, the !Q1 output of U1 will see even lower resistance which may result in destroying the pin's output driver if there's no on-chip current limiting mechanism.
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  • \$\begingroup\$ Thank you very much now I understand clearly. Thanks for detailed answer. \$\endgroup\$ – Ayushprime Jun 27 at 4:32
  • \$\begingroup\$ @Ayushprime you can accept my answer. \$\endgroup\$ – Rohat Kılıç Jun 27 at 7:10
  • \$\begingroup\$ How to accept as I am new to electrical stack \$\endgroup\$ – Ayushprime Jun 27 at 7:48

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