What you seem to have a problem with, is any kind of intuitive feel for Ohm's law and how it works, even though you say you understand it, so you blindly keep asking questions where the answers should be obvious from the other answers we have given you, and we thus keep going in circles.
That's why your questions seem like you are trolling us.
And unfortunately I don't think intuition is something we can help you get.

Ohm's law isn't enough to solve some problems in some circuits.
I am more likely confused at calculating some stuff here.
Making some assumptions and calculating the range. I was assuming of using Ic_max as a saturation point.

I am also confused with the moment when I have Rc, Re and Rc in one transistor because Vb and Vc changes there is no "stiff" point like Re, Rc combination or Rb,Rc combination.

Smoller voltage at the base means that the Re will also see a smaller voltage. Thus, Ie current will drop too. So, for an unchanged RC value, the voltage drop across RC resistor drops too.
So, we are moving away from saturation.

I do not understand this part. I mean normally the divider at active region should make the divider "stiff"? At saturation it should change the voltage from 2,5V to for example 1,8V ?
If I had divider with 10 k ohm resistors then if the voltage changes in 200 ohm resistor does it mean that the transistor is in saturation ?

 

I do not understand this part. I mean normally the divider at active region should make the divider "stiff"? At saturation it should change the voltage from 2,5V to for example 1,8V ?
If I had divider with 10 k ohm resistors then if the voltage changes in 200 ohm resistor does it mean that the transistor is in saturation ?

Repeat this exercise.
And this time examine how Vb (Vb = 1V and VB = 3.7V ) changes affect the allowed voltage drop across Rc
https://forum.allaboutcircuits.com/threads/transistors-why-is-it-here-saturated.188858/post-1758844

 

Repeat this exercise.
And this time examine how Vb (Vb = 1V and VB = 3.7V ) changes affect the allowed voltage drop across Rc
https://forum.allaboutcircuits.com/threads/transistors-why-is-it-here-saturated.188858/post-1758844

Here it is, but I meant something different.

1. For voltage divider with 800 ohms resistors and with 10k resistors divider will have 2,5V on each resistor, so does the voltage drop from 2,5V to 1,9V will indicate that the transistor is in saturation? If not then how to usually calculate the saturation while the voltage on divider can change ?

2. Also why if I want to calculate Rc_max I assume that Ie = Ic ? Maybe I missed something, I know that Ie = Ib+Ic.

 

Assumed Ie = 45mA and the "real" one is Ie = (2.5V - 0.7V)/(40Ω + 400Ω/101) ≈ 40.9mA. For me personally, this is not a HORRIBLE assumption. But I understood your point here.

As far as I remember this is app. the value of Ic which was mentioned already in post#2.,

 

If not then how to usually calculate the saturation while the voltage on divider can change ?

Exactly the same way as you did in the exercise that I have given you. Simply use a Thevenin's to find Vb or even better- Ie current.
https://forum.allaboutcircuits.com/threads/transistors-why-is-it-here-saturated.188858/#post-1758751

Also why if I want to calculate Rc_max I assume that Ie = Ic ? Maybe I missed something, I know that Ie = Ib+Ic.

This is another assumption we made very often. When working in an active region transistor beta is "large" thus, we can ignore the base current and assume that Ic = Ie *β/(β+1) ≈ Ie
Because for β = 50 we have Ic = 0.98*Ie and for β = 100 --> Ic = 0.99*Ie, Thus Ic ≈ Ie.

 

Exactly the same way as you did in the exercise that I have given you. Simply use a Thevenin's to find Vb or even better- Ie current.
https://forum.allaboutcircuits.com/threads/transistors-why-is-it-here-saturated.188858/#post-1758751

Okey so I understand that voltage can change on resistor from 2,5V to 1,5V or something like that and still be in active region ?


https://tinyurl.com/2mr4tu2s

When I am changing Rc the voltage on Rth and Re doesn't change. Hmmm.
But overall the voltage on R2 (the one with 2,5V) did change and it is constant with 1.49V.

I don't know how to interprete it. Because before we assumed that 2,5V is constant.

 

Assumed Ie = 45mA and the "real" one is Ie = (2.5V - 0.7V)/(40Ω + 400Ω/101) ≈ 40.9mA. For me personally, this is not a HORRIBLE assumption. But I understood your point here.

But that is not even CLOSE to the actual result!

You claim that Ie = 40.9 mA.

Okay, where is that 40.9 mA coming from?

It's not coming from the collector! Even if Vcesat = 0V, the collector voltage wouldn't go below 1.8 V. With a 1 kΩ resistor connecting it to 5 V, that's only 3.2 mA. So AT LEAST 37.7 mA is coming from the base current.

But even if ALL of the current in the top bias resistor was going into the base, then that would put 30.2 V across it, requiring that Vcc be at least 32.7 V, not 5 V.

The base voltage is NOT 2.5 V, it is barely 1 V. Nearly 75% of the current that flows into the top resistor becomes base current, leaving just 25% of it flowing in the bottom resistor.

The emitter current is not 40.9 mA, it is less than about 8.2 mA. The emitter voltage is not 1.8 V, it is just under 0.33 V.

For me, personally, that makes the assumption pretty horrible.

 

When I am changing Rc the voltage on Rth and Re doesn't change. Hmmm.

This should not be a surprise for you. As long as BJT is in the active region the "collector" is acting just like a constant current source. And this current is equal to Ie = Ve/Re.

 

But that is not even CLOSE to the actual result!

You claim that Ie = 40.9 mA.

Okay, where is that 40.9 mA coming from?

It's not coming from the collector! Even if Vcesat = 0V, the collector voltage wouldn't go below 1.8 V. With a 1 kΩ resistor connecting it to 5 V, that's only 3.2 mA. So AT LEAST 37.7 mA is coming from the base current.

But even if ALL of the current in the top bias resistor was going into the base, then that would put 30.2 V across it, requiring that Vcc be at least 32.7 V, not 5 V.

The base voltage is NOT 2.5 V, it is barely 1 V. Nearly 75% of the current that flows into the top resistor becomes base current, leaving just 25% of it flowing in the bottom resistor.

The emitter current is not 40.9 mA, it is less than about 8.2 mA. The emitter voltage is not 1.8 V, it is just under 0.33 V.

For me, personally, that makes the assumption pretty horrible.

You misunderstand me. I will use 45mA (or 40.9mA) to only check if the BJT is in saturation for a given Rc resistor value. Nothing more.

 

This should not be a surprise for you. As long as BJT is in the active region the "collector" is acting just like a constant current source. And this current is equal to Ie = Ve/Re.

I thought that changing Rc will change voltage on Vb and V_rth.
So usually in active region I can assume that Ic = Ie.

Calculating voltage range for Rc is also a good thing.
Usually I have to calculate the voltage range basing on Rc ? Or can it be also calculated for Vth, Rth and Re, this voltage range ?

I think this is a huge progress in my calculations. I slowly getting how it works. I just wonder what I don't know now.
Maybe how the additional load affects the Rc and Re depending on how it is connected.

I will try to calculate some other circuits (basic ones).

 

Or I can't calculate the range of Re, Rth and Vth. Because only Rc can be changed so other stuff won't change ?
Because I think that changing Rth,Re and Vth will change Ib,Ic and other voltages. But in Rc it won't ?

 

The discussion has been going on long enough that it is probably useful to work the original problem, from scratch. There are multiple ways to do this, I will show one and give the reasoning for each step.

Some of this was posted previously, but I'll repeat it here for completeness.

First, here is the original circuit with my annotations of the symbols I will be using and/or referring to.

View attachment 274950

Active vs Saturation

We often think of these as being distinct operating regions as if there is some fundamental dividing line between them at which point the behavior of the circuit instantly changes. This is far from reality.

As we move from active to saturation, the device characteristics change smoothly. So there is a significant region where the transistor is operating somewhere between active and saturation.

For silicon BJT transistors, particularly small signal transistors, device manufacturers specify their saturation characteristics at a beta of 10. This does NOT mean that a beta of 10 defines the transition between active and saturation, it merely means that, for historical reasons, device manufactures typically choose to force the transistor to operate at a beta of 10 when making their saturation performance measurements.

From a theoretical perspective, the start of the transition from active to saturation can be defined as when the collector-base junction stops being reverse biased. But, again, this does NOT result in a sharp change in behavior at this point. In fact, most devices continue behavior more like they are in the active region (e.g., beta hasn't changed much) until the collector-emitter voltage drops below about 400 mV (meaning the collector-base junction if forward biased by about 300 mV) and the major shift usually starts when Vce reaches about 300 mV. For many small signal transistors, the current gain does reach 10 until the Vce is down around 50 mV to 150 mV. hence, for small signal transistors, forcing the transistor to operate where beta will be 10 or less will safely put it into saturation.

The bottom line here is that sometimes your analysis will lead you to a conclusion that you are at/near the active/saturation boundary. More likely, you are asking questions like how big can the collector resistor be before the transistor saturates. Be a bit wary when working around this boundary, because it is NOT well defined. The takeaway, therefore, should be that well-designed circuits strive to avoid operating near this boundary precisely because the behavior is not well-defined. They either operate well into the active region (by staying well away from Vcesat) or they operate well into the active region (by forcing Ib to be well above Ibsat).

Note of caution -- the above discussion is primarily for small-signal NPN transistors. PNP transistors have typical numbers that are a bit different and power transistors can have significantly different values -- in particular, the active-region beta for power transistors may or may not be greater than ten to begin with. Check the data sheets to determine what are decent parameters to use.

Assumptions

As Jony130 and others have said, we often make simplifying assumptions in order to analyze circuits quickly. But, when we do, we need to check whether the results are compatible with the assumptions. If they turn out not to be, we need to go back and redo the analysis without making that assumption.

These are some common initial assumptions that are often made for silicon BJT transistor circuits:

The transistor is in the active region.
The base-emitter voltage is one diode drop (~0.7 V).
The collector current is about equal to the emitter current (equivalent to saying that beta >> 1).
The base current is small compared to the relevant currents in the base bias network.

If we believe (or conclude) that the transistor is in saturation. then these assumptions change:

The transistor is in the saturation region.
The base-emitter voltage is one diode drop (~0.7 V, though often ~1V is a better estimate).
The collector-emitter voltage is at Vcesat (~0.2 V typically, often taken to be 0 V for quick analysis).

A big difference is that we cannot assume that the collector current is about equal to the emitter current because the fact that we are in saturation means that beta is not very large. Device manufacturers typically spec their saturation characteristics at a point where Ic/Ib = 10, but you can't assume that this is the beta when doing the analysis. It can easily turn out that nearly all of the emitter current is coming from the base (i.e., beta < 1).

A consequence of this is that, for circuits in saturation, it is seldom safe to assume that the base current is small compared to the base bias network currents.

PASS #1

Let's assume that this circuit is operating in the active region and has been designed so that our initial assumptions are valid.

That means that our bias network is negligibly affected by the base current, and hence the base voltage is 2.5 V.

This, in turn, means that the emitter voltage is one diode-voltage-drop less, or about 1.8 V.

This, in turn, means that the emitter current is about 45 mA.

Our assumption that the emitter and collector currents are about the same means that the collector current is about 45 mA, which means that the collector voltage is about 45 V below Vcc, or -40 V.

This last result is completely at odds with our assumption that the device is in the active region because, in that region, the lowest we could go would be Vc = Ve + Vcesat, which even if we use Vcesat = 0V, would be 1.8 V.

So we know that at least one of our assumptions is bad, but we don't know which one. We made a few. The two big ones were that the device was in the active region and the other is that the base current is negligible. It's possible that only one of those assumptions is bad -- so we can't just conclude that BOTH must be bad.

Before stepping away from this, let's ask what the collector resistor would need to be in order for our assumption that the transistor is active to be reasonable.

Under the assumption that the collector and emitter currents are about equal in the active region, that means that we want to keep the collector voltage above the base voltage of 2.5 V (or at least above about 2.1 V). So with 2.5 V dropped across Rc with 45 mA flowing in it, that means that Rc can't be any larger than about

Rc_max ~= (5.0 V - 2.5 V) / (45 mA) ~= 56 Ω

If we make Rc no greater than this, how about our assumption that teh base current is negligible? The current flowing into the bias network is nominally 5 V / 1.6 kΩ = 3.13 mA. The base current, assuming a beta of 100, is about 0.45 mA. This is 14% of the bias network current, which is generally considered to be more than "negligible". So this assumption would still be at least shaky.

PASS #2

Let's next assume that the device is in saturation, but that the base current is still negligible compared to the base bias network currents. In this case, we would still have Vb = 2.5 V, Ve = 1.8 V and Ie = 45 mA. But now our collector voltage would be no lower than the emitter voltage plus Vcesat. Let's use Vcesat = 0.2 V, placing the collector votlage at 2.0 V even. That means that we have 3 mA in the collector, leaving 42 mA for the base current. This is clearly far, far larger than the ~3 mA of current available to the base bias network, so we cannot assume that the base current is negligible.

PASS #3

So now we know that we can assume that the transistor is in saturation but we cannot assume that the base current is negligible.

One thing that we can do to make our lives easier is to replace the base bias network with its Thevenin equivalent, which would be a 2.5 V source in series with 400 Ω. But I will leave the circuit as is in order to illustrate doing the analysis of the given circuit.

We will now keep only two assumptions: that Vbe is 0.7 V and that Vce is Vcesat. The first one is pretty shaky with these kinds of base currents, but without a specific diode part number we can't do much better. After we get a result using these assumptions, we can look at an actual data sheet to refine things a bit (assuming we were to use that transistor).

As you can see from the annotated diagram, we have eight unknowns (five currents and three voltages). But we only have three meshes, so let's use mesh current analysis (if we did the Thevenin replacement for the bias network, we would only have two meshes.

Summing the voltage drops round the three meshes, we have:

Ix: Ix*Rc + Vcb +(Ix-Iz)*R1 = 0

Iy: Vbe + Iy*Re + (Iy-Iz)*R2 = 0

Iz: -Ix*R1 - Iy*R2 + Iz*(R1+R2) = Vcc

Putting these in standard form:

Ix*(R1+Rc) - Iy*(0) - Iz*(R1) = -Vcb = -(Vce + Veb) = -(Vce - Vbe) = Vbe - Vce
Ix*(0) + Iy*(R2+Re) - Iz*(R2) = -Vbe
-Ix*R1 - Iy*R2 + Iz*(R1+R2) = Vcc

Solving these equations using whatever method you like, we get

Ix = 4.474 mA
Iy = 8.158 mA
Iz = 9.441 mA

These then yield:

I1 = 4.967 mA
I2 = 1.283 mA
Ib = 3.684 mA
Ie = 8.158 mA
Ic = 4.474 mA

Vb = 1026 mV
Ve = 326 mV
Vc = 526 mV

As you can see, the assumption that the base current was negligible was completely wrong. Nearly 3/4 of the I1 current was diverted into the base, thus starving R2 and drastically reducing the base voltage from 2.5 V down to 1.03 V.

The beta is 1.2, which confirms that this transistor is WAY into saturation.

2n3904 data sheet

https://www.onsemi.com/pdf/datasheet/2n3903-d.pdf

If we look at the data sheet for a very common small-signal NPN transistor, namely the 2n3904, we see the following:



At a collector current of about 4.5 mA and base current of 3.7 mA, we are probably looking at Vce of under 100 mV and likely closer to 50 mV.

The Vbe is a bit tricker to get a handle on, since we are operating well below a beta of 10. But the Vbe is driven more by the base current than the collector current, so we want to look for the typical Vbe when Ib is in the range for 3.7 V. The ON characteristics spec the maximum Vbe as 0.95 V when Ic is 5 mA. So we are probably safe assuming something in the 0.8 V to 0.9 V range, which would not be expected to change the results much.

 

You misunderstand me. I will use 45mA (or 40.9mA) to only check if the BJT is in saturation for a given Rc resistor value. Nothing more.

Okay. But with an Ie that is more than an order of magnitude too large, how well are you going to be able to determine if a given Rc is in saturation?

 

PASS #1

Let's assume that this circuit is operating in the active region and has been designed so that our initial assumptions are valid.

That means that our bias network is negligibly affected by the base current, and hence the base voltage is 2.5 V.

This, in turn, means that the emitter voltage is one diode-voltage-drop less, or about 1.8 V.

This, in turn, means that the emitter current is about 45 mA.

Our assumption that the emitter and collector currents are about the same means that the collector current is about 45 mA, which means that the collector voltage is about 45 V below Vcc, or -40 V.

This last result is completely at odds with our assumption that the device is in the active region because, in that region, the lowest we could go would be Vc = Ve + Vcesat, which even if we use Vcesat = 0V, would be 1.8 V.

So we know that at least one of our assumptions is bad, but we don't know which one. We made a few. The two big ones were that the device was in the active region and the other is that the base current is negligible. It's possible that only one of those assumptions is bad -- so we can't just conclude that BOTH must be bad.

Before stepping away from this, let's ask what the collector resistor would need to be in order for our assumption that the transistor is active to be reasonable.

Under the assumption that the collector and emitter currents are about equal in the active region, that means that we want to keep the collector voltage above the base voltage of 2.5 V (or at least above about 2.1 V). So with 2.5 V dropped across Rc with 45 mA flowing in it, that means that Rc can't be any larger than about

Rc_max ~= (5.0 V - 2.5 V) / (45 mA) ~= 56 Ω

If we make Rc no greater than this, how about our assumption that teh base current is negligible? The current flowing into the bias network is nominally 5 V / 1.6 kΩ = 3.13 mA. The base current, assuming a beta of 100, is about 0.45 mA. This is 14% of the bias network current, which is generally considered to be more than "negligible". So this assumption would still be at least shaky.

So as I thought the voltage on the divider can change and the transistor can be still active.
Like if I change the resistors from 800 ohm to 10k the drop I had was 1,6V from 2,5V.

PASS #2

Let's next assume that the device is in saturation, but that the base current is still negligible compared to the base bias network currents. In this case, we would still have Vb = 2.5 V, Ve = 1.8 V and Ie = 45 mA. But now our collector voltage would be no lower than the emitter voltage plus Vcesat. Let's use Vcesat = 0.2 V, placing the collector votlage at 2.0 V even. That means that we have 3 mA in the collector, leaving 42 mA for the base current. This is clearly far, far larger than the ~3 mA of current available to the base bias network, so we cannot assume that the base current is negligible.

I think I get it ?

Also for PASS #3 I thought I would see KCL or KVL, I am more trained in them. I know this is similar but I just don't know how to read it myself.

I thought to be honest that it is easier to calculate. But there are many assumptions, many stuff that I'm trying to "consume".

I also know how to calculate Rc_max. But I was also wondering if it works the same with Vth, Rth and Re.
I think no because Vth, Rth and Re changes the Ib and Ic. But changing Rc doesn't change alot. It changes Vce and V_RC.
but Ib and Ie doesn't change. So I thought of other changable components. Or how load affects Re or Rc.

 

So here is the LTSpice operating point simulation of the original circuit, using a 2n3904 transistor.



As you can see, the results are in very good agreement with the mesh current analysis done previously.

The Ve is 324 mV (just 2 mV less) and Vb is 1.13 V (which is about 100 mV more). The Vbe is 0.804 V, right at the low end of the range I suggested. The Vce is only 14 mV, so even less than the 50 mV I was throwing out as being very possible. The beta comes in at 1.36, which is close to the estimated 1.2,

 

So as I thought the voltage on the divider can change and the transistor can be still active.

I don't follow what you are claiming here. Yes, the voltage on the divider can change because it is being very heavily loaded by the base of the transistor. But what does that have to do with "the transistor can still be active". The transistor is not still active -- it never was active. The initial assumption that the transistor was active turned out to be very wrong.

Like if I change the resistors from 800 ohm to 10k the drop I had was 1,6V from 2,5V.

???
If you change the both resistors from 800 Ω to 10 kΩ, the base voltage will be about 900 mV and the emitter voltage will only be about 200 mV.

Also for PASS #3 I thought I would see KCL or KVL, I am more trained in them. I know this is similar but I just don't know how to read it myself.

By all means, use KCL and KVL and see if you get similar results.

But just because you are "more trained" (i.e., more comfortable and confident) with a more basic technique is not a reason to forego more powerful techniques. Quite the opposite. You need to get comfortable and confident with the more powerful techniques by forcing yourself to use them. These techniques have been developed and gained wide acceptance for a reason -- they make the engineer's life easier. Since engineers tend to be lazy by nature (it's why we went into engineering -- we thought we would design and make all these things to do our work for us, we just didn't count on how much work it takes to be lazy). that is a good indication that learning all of these techniques will actually do so.

 

???
If you change the both resistors from 800 Ω to 10 kΩ, the base voltage will be about 900 mV and the emitter voltage will only be about 200 mV.

https://tinyurl.com/2j7vgvhl

Like here, voltage on the resistor is not 2,5V but the transistor is still active, so I assume that because the resistor doesn't have 2,5V doesn't mean that transistor will be saturated.

So I wonder why we had to think that it must be 2,5V at the first place.

Tommorow I'll send my KVL/KCL results. I know there are better methods but I want to know if KVL or KCL can also work. Just for my own satisfaction.

 

https://tinyurl.com/2j7vgvhl

Like here, voltage on the resistor is not 2,5V but the transistor is still active, so I assume that because the resistor doesn't have 2,5V doesn't mean that transistor will be saturated.

So I wonder why we had to think that it must be 2,5V at the first place.

Tommorow I'll send my KVL/KCL results. I know there are better methods but I want to know if KVL or KCL can also work. Just for my own satisfaction.

These are completely (well, largely) unrelated things.

Whether the base of the transistor is at 2.5 V (because that is what the bare voltage divider results in) is determined by how significant the base current is relative to the current flowing in the divider.

Whether the transistor is active or not depends on whether the Vce can stay out of saturation when the collector current is what it needs to be to not be in saturation.

Now, as the transistor goes into saturation, many (most?) circuit topologies will result in emitter current falling slowly while the collector current shifts rapidly from collector to base. As the base current increases, that increases the load that the base presents to the bias network, which in turns accelerates the degree to which the base voltage gets pulled away from its no-load value.

 

These are completely (well, largely) unrelated things.

Whether the base of the transistor is at 2.5 V (because that is what the bare voltage divider results in) is determined by how significant the base current is relative to the current flowing in the divider.

Whether the transistor is active or not depends on whether the Vce can stay out of saturation when the collector current is what it needs to be to not be in saturation.

Oh I thought it was relevent because you mentioned it at PASS 1.
So I thought it must be 2,5V. But It seems it doesn't have to.

I do understand that Vce is the factor I could say that tell me whether it is in saturation or not. Vce = 0,2V or Vce =0,1V.

Now, as the transistor goes into saturation, many (most?) circuit topologies will result in emitter current falling slowly while the collector current shifts rapidly from collector to base. As the base current increases, that increases the load that the base presents to the bias network, which in turns accelerates the degree to which the base voltage gets pulled away from its no-load value.

I mean I can see the emitter current falling slowly down here :

https://tinyurl.com/2gaa4l2f

Also I see small difference. Can I call it "fully" saturated transistor when voltage on Rce is close to Vcc ?
For example Vce = 0,1 V and V_Re = 0.03 V, so V_Rc = 4.87 V.

By the way does it really need that much calculation ? Isn't there a shortcut ?
I thought that it is more simple.

Are there any exercises on the internet that I can use to practice ? I want to calculate more of those circuits to understand more about them. Because I know now something but I don't feel to confident. Especially when I add additional load like in common emiter or collector or other configurations.

 

I gave you an exercise but it seems that you ignored it.