I'm relatively new to electronics and I'm trying to understand how to use transistors, but I seem to be misunderstanding something.

Transistor Switch - HI = LED ON

I'm trying to implement a transistor switch as described in Electronic Formulas, Symbols & Circuits on p 118. The diagram above is based on my understanding of how the circuit should look if I want the LED on when the switch is high.

Transistor Switch - HI = LED ON

Based on my understanding of the first diagram, and the diagram shown in the book, I think this second schematic should turn the LED on when the switch is low. But I also see a basic LED driver with the battery, resistor and led, which seems odd to me. When I actually wire this up on a breadboard, I see that the LED is on for either position of the switch, and that it's actually brighter when I turn the switch on high.

Do these schematics actually function as I believe they should thus I am failing to implement it correctly, or am I misunderstanding something fundamental?

EDIT The resistors should read 1K in the diagrams, not 100.

  • 4
    \$\begingroup\$ Your battery is very wrong. \$\endgroup\$ Commented Sep 28, 2014 at 2:07
  • \$\begingroup\$ I'm not sure if I drew it correctly. The wire connecting to the battery is connected to the positive terminal. \$\endgroup\$
    – munk
    Commented Sep 28, 2014 at 2:11
  • \$\begingroup\$ I'd suggest leaving the transistor out of the circuit to begin with and try and make the switch, battery, 1k resistor, and the LED work. (get the LED to turn on.) Then you can work on the transistor. \$\endgroup\$ Commented Sep 28, 2014 at 12:48
  • \$\begingroup\$ If this is a sample of the circuits being proposed for learners I suggest you either get a better book or look through the questions and answers on this exchange. \$\endgroup\$ Commented Sep 28, 2014 at 15:09
  • \$\begingroup\$ I have no problem lighting an LED. My misunderstanding appears to be that there should be two batteries. I assumed that there was a single common battery just like there is a single common ground. The diagram represented a simplification based on my understanding. \$\endgroup\$
    – munk
    Commented Sep 29, 2014 at 14:15

1 Answer 1


There are many errors in the diagrams. How you connect these transistor, is called Common Emitter, and is one of the three ways in which you can connect a transistor.

For common emitter configuration, it is usually considered that the load is connected in the collector circuit. To light a led, the basic circuit would be:


simulate this circuit – Schematic created using CircuitLab

When the source \$V_b\$ is applied, the LED ligths on.

The LED is connected in the output circuit. The LKV for this:

$$ V_{CC} = I_C\,R_C + V_{D1} + V_{CEsat} $$

For example, if \$V_{D1} = 1.2\,\mathrm{V}\$ (like common red LED), \$V_{CEsat}= 0.8\,\mathrm{V}\$ (Collector-emmiter saturation voltage) and \$V_{CC}=9\,\mathrm{V}\$

$$ I_C\,R_C = 7\,\mathrm{V} $$

for a common red LED, we can suppose 15 mA, then

$$ R_C = \dfrac{7\,\mathrm{V}}{15\,\mathrm{mA}} = 466.66\,\Omega\approx 470\,\Omega $$

The input circuit is on the base terminal. For a current collector of 15 mA and a factor \$h_{FE}=100\$ (typical)

$$ I_B = \dfrac{I_C}{h_{FE}} = \dfrac{15\,\mathrm{mA}}{100} = 150\,\mu\mathrm{A} $$

If \$V_b = 3.3\,\mathrm{V}\$ (typical for a \$\mu\$C) and \$V_{BE}=0.7\,\mathrm{V}\$ (Base-emmiter voltage for a silicon transistor, typical)

$$ R_{b} = \dfrac{V_b - V_{BE}}{I_B} = \dfrac{3.3 - 0.7}{150\times 10^{-6}} = 17.333\,k\Omega\approx 18\,k\Omega $$


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