1
\$\begingroup\$

Light Sensitive Buzzer circuit using transistors

https://www.eleccircuit.com/four-light-sensor-alarm-circuits/ This is the link to the website where the circuit is explained.

I have some trouble understanding this circuit, or more precisely, understanding the explanation provided in this website, where I found the circuit. In the website it is stated that when it's dark (High LDR resistance) Q1 turns OFF and Q2 AND Q3 turn ON, which activates the Buzzer, and that the Q1 and Q2 portion of the circuit forms a Schmitt-Trigger.

My question is whether the explanation provided is consistent with the actual workings of the circuit, because I have some doubts about it as a beginner in electronics, which could also mean I'm entirely wrong too.

So since we have a Schmitt-Trigger wouldn't that mean that when Q1 is On, Q2 would be Off and vice versa? If so when Q1 is off, because it's dark and LDR turns into an open circuit I guess, then that means Q2 would be ON, and since Q2 is on we have a current Ic that goes through it's collector, but that current also appears at R5 so the voltage drop across R5 would be Vr5 = (Ic + Ib)*R5, where Ib is the current for R5 and R6 when they are in series, which means when Q2 turns on the voltage drop across R5 increases so that means less Volts at the base for Q3, which would mean Q3 is off and the buzzer wouldn't activate. That was assuming LDR is open circuit when it's dark, realistically wouldn't it just have a high resistance? In that case we would have a voltage divider at the base of Q1 ,which really makes things different.

Am I missing something or is my explanation just wrong? I would love to hear suggestions and corrections regarding my question.

\$\endgroup\$

3 Answers 3

Reset to default
0
\$\begingroup\$

You are not right about the Q3. Note that it's PNP, not NPN, its emitter at VCC. So, R5 and R6 arent in series, they actually are parallel. When Q2 opens, its collector voltage goes lower and the current through both R5 and R6 increases. This opens Q3, as the current of R6 is the Q3's base current. So, the buzzer really activates when LDR is of high resistance or even open-circuit.

\$\endgroup\$
4
  • \$\begingroup\$ So because less volts at Q3 base because Q2 is on, and 9v at Q3 emitter, Q3 turns of because it's PNP? I haven't looked too much into PNPs. \$\endgroup\$ Commented Feb 18 at 19:00
  • \$\begingroup\$ Yes, the PNP is almost the same as NPN, but with polarity reversed. So, the "ground" or "0V" for it is at battery + (VCC), and when the Q2 opens, the voltage at its collector (voltage between collector and battery -) decreases, but the voltage between VCC and Q2 collector increases, increasing the current through R6 and Q3 emitter-base junction and opening the Q3. \$\endgroup\$ Commented Feb 18 at 19:31
  • \$\begingroup\$ Thanks, I meant to say Q3 turns ON because less volts at Q3 base and because Q2 is on. And by the voltage between VCC AND Q2 collector you refer to voltage drop across R5, correct? \$\endgroup\$ Commented Feb 18 at 19:36
  • \$\begingroup\$ Yes, exactly. And the voltage drop on R5 equals to voltage drop on R6 + base-emitter of Q3. The latter is usually small (below 0.7-0.8V for silicon BJTs), so most of the voltage is on R6, defining the Q3 base current, which opens it. Remember, BJT's are opened by base current, and FETs are opened by the gate voltage. \$\endgroup\$ Commented Feb 18 at 22:12
0
\$\begingroup\$

The explanation provided on the website is reasonable, albeit short.

Your explanation is incorrect.

If Q2 is on the Q3 is on, and if Q2 is off, Q3 is off.

\$\endgroup\$
0
\$\begingroup\$

Here are the "cyclograms" of the circuit.

Made with microcap v12

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.