# Understanding Transistors

transistors

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

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)

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