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".

  • \$\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

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:


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.
| improve this answer | |
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.