No IC's allowed, part II: determining active vs. saturation state in BJT transistors

Back for more punishment!  

Quick one this time.

Last time I discussed BJT basics...if you are a "BJT cookbook" tech as I was--find a bipolar junction transistor circuit fragment, put it in your design, and hope it works--you might want to skim the last post before reading this one.

Maybe simulate it using Falstad?

If you are a BJT superstar--you taught 1970's electronics at a community college--by all means skip this post--instead, watch the video here.

but really? this is all for me.

SATURATED VS. ACTIVE

You can get Falstad simulations for these "No IC" posts from github, here.

So...how the heck do you figure out if a transistor is in its active or saturation region?  Should be easy, but this one drove me nuts...

Cheat code: Plug your BJT circuit fragment into Falstad, hover over the transistor and see what it says, right?  

Nope, too easy. 

We start by guessing we are in active region, and go with these assumptions:

• B-C junction is forward biased; its voltage drop across B and C is .7V
• current at emitter and current at collector are the same
• B-C is forward biased; B-E is reverse biased.(normal for active)

Let's do some simple alegbra and use ohm's law to see what's really going on.

• Base is 2V less .7V to ground, so the voltage at the BJT emitter is 1.3V.  
• Using ohm's law: current at BJT emitter junction is 1.3/1000 or 1.3mA  
• Assuming active region for the transistor, Ie and Ic are about the same, so let's assume Ic is also 1.3mA
• Again using ohm's law, for emitter to be at 1.3mA, the voltage drop across a 10k resistor would need to be 13V
• But, the the collector power is only 5V!!
• The transistor lowers its Ve to Vc resistance as much as it can to satisfy the need to satify the 13V drop
• But fails--even if C to E is a short, the transistor can't make the Vc junction 13V relative to ground.
• Transistor is saturated; B-C and B-E are both forward biased.

Let's try again:

• BJT collector current is 1.3mA; that hasn't changed.
• Assuming active, Ic = Ie, so Ie is also 1.3mA
• 1.3mA * 1000 = 1.3V  (drop across collector resistor)
• 5V - 1.3V is 3.7V0--that's the collector voltage
• B-C is forward biased; B-E is reverse biased (see active assumptions above)
• Transistor is active.

OK with this framework, any simple fragment can be analyzed. for PNP, all the same ideas, but all polarities are reversed.  

Notes for this post....

OH NO IT'S THE PROMO

Quick word from this blog's sponsor! Once you have your design working in Falstad, run, don't want to PCBWAY and get some prototype PCB's made....They also do great work with 3D printing, assembly and can even act as your OEM.  Check them out!

OUTTRO

OK no more this time, it's Juneteenth, I have to get to the bench and mess around with some IC's.