Better explanation on BJT transistor regions?

[MarkusAnd]

Hello!

For the last day I've been trying to learn more about BJT transistors and how they work. I'm having a hard time understanding the theory behind cutoff, active and saturated region.
I understand that in the cutoff region the transistor is "off", in the active region the collector current is adjustable and in the saturated region the collector current is maxed out(Vce = 0).

For hours I've been looking at graphs displaying the different regions but I don't understand how the values correlate to eachother.

So if someone could explain how the different values make up the curves in the following graph I would be really thankful!



// Markus

[krayvonk]

Ive got a prob with BJTs,  and I heard somewhere that 2 back to back diodes makes a BJT  (p - n - n - p) but its not true,  I had a good search around again, after I prematurely posted and deleted my post, but I came up at the end, no its not true,  theres no configuration that makes it true.    Thats my issue with them...

[ferdieCX]

A BJT is not 2 diodes back to back, although it will behave like that, if you measure it with a VOM or DMM.
In the picture, you can see the internal structure of a BJT and the external and internal currents.
Be beware, that the middle layer is the Base and it is "VERY thin", much thinner than the Emitter and the Collector.

I would recommend you the book Electronic Devices and Circuits from Miilman and Halkias.

In a BJT, the collector current Ic is a function of the base current IB and the collector-emitter voltage VCE -->  Ic = f (IB, VCE),
ideally Ic = hFE * IB, but it increases also with VCE.

Using the attached circuit, you can feed the BJT base with a constant current, for example  IB = 50 uA, and by varying VCE you get one of the curves.
Then, you feed IB = 100 uA and and by varying VCE you get another curve and so on.

[Jwillis]

Hello!

For the last day I've been trying to learn more about BJT transistors and how they work. I'm having a hard time understanding the theory behind cutoff, active and saturated region.
I understand that in the cutoff region the transistor is "off", in the active region the collector current is adjustable and in the saturated region the collector current is maxed out(Vce = 0).

For hours I've been looking at graphs displaying the different regions but I don't understand how the values correlate to eachother.

So if someone could explain how the different values make up the curves in the following graph I would be really thankful!





// Markus

There are three modes of a Transistor. Cut off region is when a transistor is in the fully-off state . Ib is zero because theirs no forward voltage across Vbe . Vbe< 0.7V. 
Saturation is when the transistor is fully on . Maximum current is passing from base to emitter there by allowing maximum current to pass across emitter to collector.
In the Active region the current at the base is between the cut off and saturated conditions allowing variable amounts of current to pass from emitter to collector based on the amount of current passing from base to emitter.

[Jwillis]

The Q point (bias point, quiescent point) is when the transistor is in a steady state with a DC voltage and current cross at a mid point when no signal is applied. This is a sweet spot so when a maximum  AC signal is superimposed over the DC   it won't  cause the transistor to saturate causing distortion or clipping when it falls below the cutoff point.
The  Q-point does not necessarily remain stable if a transistor’s junction temperature is allowed to increase . Thermal run away will shift the Q point into the saturation region causing signal distortion and possible damage to the transistor.

[Ice-Tea]

Seems the diagram you've chosen is faulty? There's not supposed to be collector current (at least not this magnitude) when Ib=  0

[MarkusAnd]

So based on that circuit I would get the curve if I feed the base with a constanct current and increase the voltage(vcc) and by that increse Vce(and Ic)?

And is saturation something I want to avoid or why is the q-point considered av "sweet spot"?

[Jwillis]

If your using the transistor as a switch then saturation is fine . Provided you don't exceed the maximum voltage or current rating of the transistor.

By keeping a constant current at the base but only changing the voltage you change the Q point but the current remains the same.
 
Now if use the transistor as an amplifier. Lets say we  set a DC voltage at the Q point . The AC signal will "ride " that DC voltage as it's base line or zero point.If the AC signal peak goes above the saturation point you get distortion. If the AC signal peak goes below the cut off you get clipping. So you want to have the Q point right at the mid point of the active region to maximize the AC peak to peak signal.
The mid point of the active region is the "Sweet spot" but the Q point can change. That's what I meant. 

[ferdieCX]

So based on that circuit I would get the curve if I feed the base with a constanct current and increase the voltage(vcc) and by that increse Vce(and Ic)?

And is saturation something I want to avoid or why is the q-point considered av "sweet spot"?

Yes, you could draw the curve family for any transistor using this procedure. Usually students do it as a Lab assignment.

If you are going to use the BJT as an amplifier, you usually want to avoid saturation and cut-off regions.
You calculate Rc and RB to get a Q point in the active region.

Do you remember that you can solve a two equation system graphically ?
Well, in this case one of the equations is the collector mesh equation:  Vcc = Ic * Rc + VCE and it is called "static load line"
The other equation is represented by the different collector curves.
In your figure, the Q point is the crossing point of the curve for the selected IB and the static load line, and the solution is VCE = 3 V and Ic = 40 mA
When you apply a signal, it varies IB and consequently the operating point moves producing an output signal.

If you are going to use the BJT as a switch, you will operate it in saturation for a closed switch and in the cut-off region for open switch

[Cubdriver]

So based on that circuit I would get the curve if I feed the base with a constanct current and increase the voltage(vcc) and by that increse Vce(and Ic)?

And is saturation something I want to avoid or why is the q-point considered av "sweet spot"?

Yes, each of the curves shown in your diagram is the Ic/Vce curve for that particular base current.

The answers to your second and third questions depend on the intended use of the transistor.  If you're using it as a switch, then yes, you DO want to saturate it when it's on - it keeps the 'on' resistance to a minimum and reduces dissipation (wasted energy in this application) across the transistor.

If using it as an amplifier, then you want to stay out of saturation because if you saturate it then the response to the drive signal becomes distorted - saturation is the point where the transistor is all the way on (meaning that you can't turn it FURTHER on).  If for example, the circuit  design causes the transistor to saturate when the input signal reaches 75% of its maximum value, then the top 25% of the signal is cut off and lost.  The output hits its maximum at 75% in, then flatlines at max until the input again drops below 75% - this is known as 'clipping', as you are effectively clipping the top of your input signal off.

The 'sweet spot' you mention is the point you'd set the transistor to if you were operating it as a class A amplifier (one device does the amplification, and must handle both the positive and negative excursions of the input signal).  Imagine that you are controlling a faucet with a sine wave - you want to be able to vary the water flow from a minimum (at the negative peak if the sine wave) to a maximum (at the positive peak of the wave), but never turning the flow either all the way off or all the way on, as that would distort the fidelity of the signal.  This would result in you having the faucet about half way open (the sweet spot) when no signal is present.  As the input went more positive you open the valve further, and as it goes negative you close it to less than halfway.

There are other amplifier circuits that use two devices, one to handle each half of the input signal - one amplifying the positive half and the other the negative.  In that application, the transistors would be set to be off (biased closer to point B on your graph when no signal is present).  Each half of the input signal would then turn its respective transistor on while the other one would be completely off during that period.  (There are more subtleties involved with the actual bias point in the real world, but this is the gist of how it works.)

I hope the above has helped and didn't instead further muddy the waters for you.

TLDR for the last two questions is "It depends..."

-Pat

[Warhawk]

Markus,
your confusion is right. I found many books misleading. I'll describe it with my own words. I will focus only to the standard engineering conditions, let's skip the inverse mode.

BJTs have the active region and the saturation.
Saturation - the base-emitter junction is fully saturated with charge carriers. There are more charge carriers than needed. The transistor acts more as a resistor of a small value. (=therefore as a switch)

Active region - the limited number of free charge carriers limits the current from the collector to emitter node. No matter what voltage you apply, the current stays nearly same. The transistor acts as a current source. You want to operate a BJT in this region for analog circuits, such as A-B class amplifier etc. 

FET transistors have the linear and saturation region.
Linear region - the transistor operates more as a variable resistor of a small value therefore similar to a switch. This is an equivalent to the saturation region of a BJT.

Saturation region - The drain-source channel is completely saturated and there aren't more free charge carriers for the drain-source channel. The transistor acts as a current source. This is an equivalent of the active region of a BJT.

The most important information I am trying to say is that the saturation region for BJTs does not correspond to the saturation region of FETs. That's where a lot of confusion come from. It's been a while since I went to school but knowing

BJT
saturation - switch
active - current source

MOS
linear - switch
saturation - current source
 
is good enough for most engineering jobs.

I hope my post is understandable. I have had a drink or two.
Regards, Jiri

[Jwillis]

[MarkusAnd]

So I had to try this for myself. This is what my circuit looked like:




So I'm adjusting the RL(R2) and by that also adjusting Ic and Vce. I started measuring Ic when RL = 9.5k and did 10 measurements down to 0.5kOhm. This is the graph that I got:




So if I understand it correctly:

1. The area where the Ic rises when i decrease RL is the active area.

2.  The area where the Ic doesn't rise when i decrease RL is the saturated area.

3. The curve is determined by the base current(Ib)

So on the graph, the transistor is saturated in the "horizontal" area and active when the collector current still rises?


// Markus

[ferdieCX]

Hi Markus, it is the opposite. In a BJT:

1. The area where the Ic rises when you decrease RL is the saturated area  (Ic mainly depends from the value of RL and not from the value of Ib )

2. The area where the Ic doesn't rise when you decrease RL is the active area. In this area, Ic depends from Ib and it is ideally Ic = Beta * Ib
    This is what makes the BJT to amplify: An input signal produces a small variation in Ib, that as a consequence produces a large variation in Ic.

   The active area is the horizontal area.

3. You are right, the curve is determined by the base current (Ib)

Note: The definition of the saturated area is different in a FET.

[Siwastaja]

Saturation is where the transistor is fully on. Being fully on, by definition, can only mean there is the smallest possible voltage drop over the transistor, so that the load gets most of the voltage. It's like a closed switch.

Once the Vce starts rising from the minimum possible (near-zero) value, you are out of saturation.

With MOSFETs, I decline to use the word "saturation" at all, because I have noticed it confuses the shit out of people, because it has the opposite meaning than with BJT's, formally, but nearly not everyone knows that; and clearly there is disagreement even within the industry. I use made-up terms "fully on", "fully off", and "partially conducting" instead.

Of course, the exact points where the operation regions change, are, in the end, arbitrary.

[ferdieCX]

With MOSFETs, I decline to use the word "saturation" at all, because I have noticed it confuses the shit out of people, because it has the opposite meaning than with BJT's, formally, but nearly not everyone knows that; and clearly there is disagreement even within the industry. I use made-up terms "fully on", "fully off", and "partially conducting" instead.

Of course, the exact points where the operation regions change, are, in the end, arbitrary.

I agree, it is very misleading. When I teach FETs, I call it "constant current zone"

[ferdieCX]

Of course, the exact points where the operation regions change, are, in the end, arbitrary.

There is a formal definition for that:

Active region: The base-emitter junction is directly polarized and the base-collector junction is inversely polarized

Saturation region: Both junctions are directly polarized

Cut-off region: Both junctions are inversely polarized  (in practice, the base-emitter junction can just have 0 V applied)

[MarkusAnd]

Lets say I feed the base with a 1mA current and that the gain of the transistor is 100. Therefore the max Ic is 100mA(correct me if I'm wrong).
If I then connect a load that draws 50mA would the transistor be in the saturated region?
And if the load draws 150mA it would be limited by the 100mA and the Ic(max) could only increase with a larger Ib, therefore the transistor is in the active region?

[ferdieCX]

yes, you are right

[MarkusAnd]

Okay so if I'm using it as a switch the transistor should be in the saturated region because then it allows the load to draw enough current?

But if I'm for example feeding the base with a PWM signal to control the Ic the transistor should be in the active region so that it reacts to the smallest changes in the Ib?

[Siwastaja]

I see this question is more conceptual and applies to all kinds of semiconductor switches, not just BJT.

Yes, you want to switch fully on (to saturation, in BJT terms) to minimize the power loss and heating in the transistor, and provide as much of the power to the load.

If you want to use pulse-width modulated drive to control average power to the load, the idea is the same: always drive the transistor either fully on, or fully off, to minimize wasted power. You can't avoid a transition region where the transistor goes through the partially conducting mode (active region in BJT terms) - to reduce power dissipation, do that as quickly as possible.

However, in some cases, you want to drive the load with a nice, smooth voltage, not forcing it on/off alternatively. In this case you need to use the active region, with some sort of feedback. The downside is, power is wasted as heat in the transistor. Examples of this are linear voltage regulators and classical class A, B or AB audio amplifiers.

In the modern days, using transistors as switches (fully on / fully off) has become much more typical than driving them in partially conducting. Obvious example is digital logic; additionally, switch-mode power supplies have largely replaced linear power supplies. These circuits, however, tend to heavily use MOSFETs, not BJTs. With BJTs, you see partially conducting active region quite a lot, and will keep seeing it. Using BJTs as simple switches (100% on / 100% off) is becoming quite rare, to be fair.

The fact that transition to "fully on" depends on the load current is the same for MOSFETs, as well. You have the idea correctly; if you have just enough oomph driving the transistor to make it fully on at a load of 100mA, you need more oomph for it to be still fully on at a load of 200mA. For BJT, this oomph is base current; for MOSFET, this oomph is gate voltage.

[MarkusAnd]

I think I understand it now. Thank you for all the answers! Now I can continue digging deeper.

Until next time!

// Markus