I have a setup as shown below:


simulate this circuit – Schematic created using CircuitLab

I am using the transistor as a sort of a switch. I am ensuring about 0.7 V (approx) is provided from the mid-point of the resistors to the base of the transistor. Because as I recall from my school days the breakdown voltage of silicon is 0.7V. Which means it will start forward-biasing above this voltage.

The calculated midpoint value is coming to 0.757V however. Observed is also something around .7V.

However, in this setup the LED doesn't turn on. What am I doing wrong here?

PS: I have used a standard LEDs that you use for common breadboard prototyping. Not sure the one I mentioned in the diagram is the same...

  • \$\begingroup\$ Remove R2, and move the diode and R3 into the collector terminal, emitter to gnd. Or swap the places of a R1 with R2. \$\endgroup\$ – G36 Oct 13 '18 at 5:56
  • \$\begingroup\$ Because what you have made is an Emitter Follower. \$\endgroup\$ – G36 Oct 13 '18 at 6:00

The base-emitter volage Vbe is typically 0.7V maximum, but that is the voltage between base and emitter, not the absolute voltage relative to ground. With R1=56k and R2=10k, the voltage at the base is 0.75V, leaving only 0.05V between the emitter and ground. The diode D1 also needs a forward voltage to turn on. Assuming D1 is a typical red LED, it needs about 1.5V.

This circuit could be made to work if you change the values of R1/R2 to set the Q1 base voltage to the correct voltage, and reduce the value of R3.

Assuming the D1 forward voltage is 1.5V, and the desired current is 10mA through D1 and R3, then the Q1 emitter voltage Ve must be 1.5V + (10mA * R3). Note that with R3 = 330 ohm, Ve must be about 4.8V. If you're trying to run from a 5V source, use a smaller value for R3. Finally, select R1/R2 ratio for Vb = Ve + 0.7V.

  • \$\begingroup\$ So this means I should arrange for Vb = 5.5V? (4.8 +.7)? My option then becomes to go for a high volt power source. Is this correct? \$\endgroup\$ – deostroll Oct 15 '18 at 2:43

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