1
\$\begingroup\$

I'm very, very new to EE, so this might be a really basic question. I am trying to make a very basic LED-switch using a bipolar NPN-transistor. The circuit I use can be seen below.

schematic

simulate this circuit – Schematic created using CircuitLab

Now, when I prepare this circuit on the breadboard, the LED will dimly emit some light even when the switch is open. When the switch is closed, the LED turns fully on, as desired.

Does anyone know why the LED is on, even though the switch is open?

\$\endgroup\$
  • 2
    \$\begingroup\$ Tie the base to ground with a resistor, of say 10k. The base left floating like this will be very susceptible to noise which will make the transistor conduct more or less. \$\endgroup\$ – Bart Nov 4 '19 at 9:03
  • 1
    \$\begingroup\$ Thank you for the advice, works like a charm! \$\endgroup\$ – M. Koopmans Nov 4 '19 at 9:16
1
\$\begingroup\$

The QED 123 is a infrared LED. Light is not visible. The 1k is to high. Calculate the correct value. The QED 123 = 1.7V 100mA says the datasheet. So 1.3V (3V-1.7V) over the resistor (when the transistor is fully open). The (If) 100mA 1.3/0.1= 13 ohm. 15ohm is ok. The switch circuit should work. But best to connect the base to ground with a resistor if the transitor opens slightly a wierd way when the switch is open.

\$\endgroup\$
  • 1
    \$\begingroup\$ I think the question says it is a LED that emits visible light because there is no mention of an IR detector. Probably OP picked the first type that was readily available in Circuitlab, which happens to be an IR led. You are right about the 1k resistor limiting the current too much. Though I think 15 ohm is a bit low (+100 mA in the LED), unless it is a power LED. I would recommend more like 100-150 ohm. \$\endgroup\$ – Bart Nov 4 '19 at 10:22
  • 2
    \$\begingroup\$ Okay. If your not using the QED 123. What should I say. My calculation is correct, but then use the correct values. If your using a regular led it will blow with 15 ohm. \$\endgroup\$ – seine Nov 4 '19 at 12:11

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.