NPN "Saturation" Mode and switching
However, if VB = 5v, and Vc is 5V, wouldn't that violate the condition that Vb > Vc? Ve &lt; Vb checks out, because Ve is on the GND side of the circuit.


If \$V_B = 5\:\mathrm V\$, your transistor is probably on fire. Why? Because the base-emitter junction is a silicon diode, and current rises very rapidly after about 0.65V.

I think the mistake you are making is considering the voltages not at the transistor, but at the components attached to the transistor.







simulate this circuit &ndash; Schematic created using CircuitLab

\$V_B\$ is the voltage at the base of the transistor, not the voltage at the microcontroller output (\$V_{in}\$). \$V_C\$ is the voltage at the collector, not the power supply (\$V_{cc}\$). In this circuit, \$V_B\$ will be 0V (when off) or about 0.65V (when on). \$V_C\$ will be 5V (when off) or about 0.2V (when on). Remember that when current flows through a resistor, there will also be a voltage across that resistor, by Ohm's law.