NPN “Saturation” Mode and switching

So, one thing I've been reading about is Sparkfun's tutorial on transistors,focusing on BJT (specifically NPN, but a bit into PNP). I'm super new to transistors and such, so please bear with me.

https://learn.sparkfun.com/tutorials/transistors

Looking at the switch applications they all utilize an Arduino to provide control at the Base (LO/HI (0V/5V)).The first circuit concerns controlling an LED in the framework presented for Saturation mode ("On") and Cutoff mode ("Off"). Vc is at 5V.

VE > VB < VC; Drive the Arduino pin low, the transistor will be in Cutoff mode. VE < VB > VC; Drive the Arduino pin Hi, the transistor will be in Sat mode.

However, if VB = 5v, and Vc is 5V, wouldn't that violate the condition that Vb > Vc? Ve < Vb checks out, because Ve is on the GND side of the circuit.

They only mention that Vb has to be above a certain Vth to saturate the transistor (this part makes sense in that Vb - Ve > Vth), but leave out Vb > Vc.

Is it because Vc isn't really at 5V then due to voltage drops from the resistor and LED BEFORE the collector terminal???

However, if VB = 5v, and Vc is 5V, wouldn't that violate the condition that Vb > Vc? Ve < 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 – 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.

• Phil, I posted a comment earlier than realized it probably wasn't worded correctly or didn't get the gist of my misunderstanding. Let's say you start out with cutoff mode (Vc = 5V) and want to saturate. Thus you build up Vb until it is now, .65V, at some moment in time Vb and Vc (still at 5V because of the open circuit existing) are directly compared for saturation (in my mind). Maybe what I'm trying to get at is a transient profile (i.e. does it go from cutoff-active-saturation?). With the second circuit (the one with the motor), will the PNP work if there was a large resistor between Vcc/Ve? – BlkbdyStarship Jan 27 '15 at 2:05
• @BlkbdyStarship Yes, to get to saturation from cutoff you must pass through active, but otherwise I don't really understand what you are asking. – Phil Frost Jan 27 '15 at 12:18

The voltage seen at the collector is not 5V, since there is a voltage drop across both the LED and the resistor. So Vb will be higher than Vc, and all the conditions are satisfied.

For instance lets say the LED is 20mA, you would see a 2V drop across the resistor (V=IR) and a voltage drop across the LED equal to its forward voltage (easily over 1V). Vc would then be 5V minus 2V minus the LED forward voltage.

I've always thought the Ve,Vb,Vc comparison is hard to use.

Look at the base-emiter junction as a diode. You send some current through it, probably using a voltage source and a resistor. In Sparkfun's example it is a 5V control signal and a 1kOhm resistor. This leads to (5-0.7)/1k = 4.3mA.

Once you know the base current the transistor will allow up to a current of $Ib*\beta$ to flow into the collector. If the circuit on the collector tries to put more than this, the transistor limits it and is in forward-active mode. If the circuit on the collector limits the current the voltage on the collector will be about 0.2V above the emitter and the transistor will be in saturation.

Suppose the transistor in Sparkfun's examples has $\beta=100$, then the most current allowed into the collector is 430mA. Since the LED/resistor leg only allows about 35mA to flow, the transistor will saturate, i.e. be in saturation. Another way of looking at it is that it is a "on" as the external circuit will allow it to be.

If there is no current going into the base there is no current going into the collector and the transistor is in cutoff. Don't worry about reverse-active, it's not really used.

• This probably is the clearest picture of what happens thus far. I think I'm starting to understand it. Maybe I'm also slightly confused by the definitions of "fully-on", "saturated" from a current perspective. You mentioned that 35 mA is what's actually passing, with a maximum of 430 mA allowed set by a certain Ib. Thus, if you were below this number, would you not be "saturated/fully on?" – BlkbdyStarship Jan 27 '15 at 18:33