Correct.
VBE can be any value from -V to +V.
If the transistor is operating in the linear region with moderate Ic then we can assume that VBE is somewhere between 0.65V and 0.7V give or take a few.

 

Well I couldn't be more confused.
Can we go back to my circuit example
How can we find the current through Q1 given that the current through Q1 also draws into the base of Q2
That is my main question

 

Well I couldn't be more confused.
Can we go back to my circuit example
How can we find the current through Q1 given that the current through Q1 also draws into the base of Q2
That is my main question

Do you mean the circuit in your post#11 ?
there is no power supply and the base of Q1 is open....
Currents cannot be calculated.

But, why are you confused? What is confusing you?
It is a general rule that voltage determines current - and not vice versa.
Are you able to formulate any specific question?

 

Post 15 I posted an updated schematic

 

Do you mean the circuit in your post#11 ?
there is no power supply and the base of Q1 is open....
Currents cannot be calculated.

But, why are you confused? What is confusing you?
It is a general rule that voltage determines current - and not vice versa.
Are you able to formulate any specific question?

I guess the best specific question I have is. Back to my theoretical circuit in post 15
How to calculate current through Q1
Perhaps seeing it worked will aid me in a better understanding

 

Grr, I screwed up above.
LT changed my ref des on me
What I want to see is how to approximate the current through Q2 not Q1
sorry

 

Without knowing what you are doing you are creating impossible conditions for the device to satisfy. You CANNOT drive the base of a transistor to an arbitrary voltage level. This is because the transistor is a current device. That means it takes an input current and it produces a collector current. It is the same thing for an LED. You CANNOT impose an arbitrary voltage drop across an LED. You must use a resistor to limit the current. Let me modify your diagram to show you what needs to happen.

 

Mike - you know the potential of the base node - and with a realistic assumption vor Vbe you know the voltage against ground for the emitter node. Hence, you know the voltage across the resistor R3....not too complicated....

Correction: This answer is not valid anymore ; the BJT cannot work in the "quasi-linear" amplification region.

 

................
This is because the transistor is a current device. That means it takes an input current and it produces a collector current. ...

Papabravo - this is NOT CORRECT.
Did you read my contributions?
If you think, I am wrong - please correct me and tell me why and where I would be wrong!

 

Mike - you know the potential of the base node - and with a realistic assumption vor Vbe (as discussed) you know the voltage against ground for the emitter node. Hence, you know the voltage across the resistor R3....not too complicated....

and approximations are just fine

but can you explain how you know what you just said?

how do you know the potential of the base node
you cant just say its .7
the sim says its not .7

 

and approximations are just fine

but can you explain how you know what you just said?

how do you know the potential of the base node
you cant just say its .7
the sim says its not .7

Yes ...you are right. I have made an error. I have assumed that the BJT would be in its linear region.
But this is not the case.
I think, calculation by hand is rather complicated because of non-linear characteristics (saturation region).

 

so how do I get the current through q2

 

so how do I get the current through q2

The problem is that the transistor cannot work in the region where we have formulas which are sufficiently exact.
This is because the voltage Vce will not be large enough to allow "normal" operation. As you know, Vce must be large enough to keep the base-collector junction reverse biased.
For what reason did you select a circuit which looks rather "exotic" and has less practical meaning?

 

Let's take a step back and look at an actual circuit with it's associated simulation. I've eliminated Q2 for the time being and changed some of the resistor values to create a useful circuit called an emitter follower. It is called that because the signal at the emitter "follows" the signal at the base. So a small amount of base current produces a copy of itself on the emitter, but at a much higher level. In this case about 10 μA on the base produces about 1 mA on the emitter.



The voltages in blue are the inital voltages defined by computing the DC operating point. Resistors R1 & R3 form a voltage divider which establishes a voltage on one end of RB that is about one-half the supply voltage -- about 6 Volts DC. some small amount of current flows through Resistor RB (missing from your original diagram) into the base of the transistor. How much current is that?
It is approximately (5.988V - 5.986V)/220Ω ≈ 9.1 μA. this you will have to admit is a pretty small current. It is more than enough to turn the transistor on. The Vce Across the transistor is (12V - 5.21) ≈ 6.79 Volts. The VE voltage will now give us the emitter current which is about 1.11 mA. This means the generic NPN transistor in these conditions has a gain of 1.11e-3/9.09e-6 ≈ 122.

NOTICE Vbe can be seen to be 5.986 - 5.209 = 0.777 Volts

Why is this different than the typical value of 0.6 to 0.7? It is exactly because we are operating at such low currents. If we were to pick different resistors and voltage we would get different results. Once you understand how a single device works you are in a much better position to understand multiple devices strung together. These circuits are not thrown together at random but actually constructed to meet specific requirements. You need to learn what the basic forms look like, there are not that many of them. You can count them on the fingers of one hand.

 

I cropped your huge schematic so that it fits into my neighbourhood and added voltages on it.

 

...........It is called that because the signal at the emitter "follows" the signal at the base. So a small amount of base current produces a copy of itself on the emitter, but at a much higher level. In this case about 10 μA on the base produces about 1 mA on the emitter.

What kind of "signal"..?
It is the signal voltage at the emitter which follows the signal voltage at the base (voltage amplication is practically "1").
Therefore, the name "emitter follower".
The base current plays no role for this amplification. It only influences the input resistance.

Papabravo - may I suggest something?
The questioner MikeJacobs has some problems in understanding how the transistor really works.
I'm afraid that we won't help him much, but even increase his confusion if two different (contradictory) explanations are offered in this thread.
If you are of the opinion that my explanation should be wrong, please tell me where the mistake is.
However, I am in the position to offer sufficient evidence and explanations for voltage control - do you have evidence for current control? More than a pure assertion?
Of course, I don't expect to be believed blindly here, but I am in line with all serious textbooks and publications of leading US universities. (I have already mentioned the internationally renowned experts Barrie Gilbert and Robert A. Pease).
Thank you and regards
LvW

 

I cropped your huge schematic so that it fits into my neighbourhood and added voltages on it.

I have simulated the left part of the circuit only (BC177):
Collector voltage: 0.597 volts
Emitter voltage: 0.604 volts
Vce=7 mV only (unrealistic)
Emitter current: 1.140mA
Collector current: 59.7µA
Base current: 1.08mA
(The majority of the emitter current goes through the base node)

 

@MikeJacobs if this is all to much to wrap your head around at the moment you may need to take a step back and concentrate on actual circuits with a single device. There is no problem with learning to crawl and walk before you try to run a marathon. Just take it one step at a time.

 

Reliving my circuit days and came across something I forgot

Say you had a massive nearly infinite current source. Maybe its a giant battery or similar.

You can this giant battery to the base of a simple npn but with no base resistor.
You connect the emitter with say a 500 ohm resistor to ground.

With no base resistor how do you determine what the base current is? AKA what is the base terminal input impedance?
Trying to understand what would dictate how much base current can be sunk if there is no base resistor to limit it.

Thanks

Let's take a step back and put some things into perspective.

The precise behavior of a transistor is a quite complicated interplay between the conditions at its three terminals as well as other factors such as temperature, time (the same transistor under the same conditions will behave differently ten years from now than it does today even if it has been sitting on a shelf in the original package the entire time), and history (if you overheat the transistor today it will behave differently tomorrow than if you hadn't).

We can model some of these things fairly well and come with with equations that relate them, while other things have to just be treated as effectively random influences. Our equations are always just approximation and we try to strike a balance between having equations that are simple and easy to use and having equations that are sufficiently good approximations for our needs. As a result, we end up with several different equations for the same thing and need to develop the ability to know when we can use a particular set of equations and when they just aren't good enough.

As circuit designers, one of our primary goals is to design circuits such that we can use the simpler equations and get results that are good enough to achieve the desired purpose of the circuit. This is not just to make our lives easier, but also because circuits designed that way tend to be far less sensitive to variations that we can't control, such as the temperature in the room or what year our transistor was manufactured in, than circuits requiring more the complex equations -- it's that insensitivity that makes it so that we can use the simpler equations.

For many circuits involving BJTs, we design them so that we can assume that Vbe = 0.7 V (or something close, such as 0.6 V) and that β = ∞ (which makes the base current zero). We know that the base current is not zero and that Vbe is not exactly 0.7 V, but the design is such that the impact of the degree to which either of those two assumptions is incorrect is acceptable. In fact, it is usually well within the uncertainty associated with other component values in the circuit.

With this in mind, let's consider a healthier way to view equations such as

Ic = β·Ib

Do not think of this as Ib somehow causing a particular value of Ic. Instead, think of it as the following:

β = Ic / Ib

We have a transistor that, at a given operating point, has a particular Ic and a particular Ib and we define the present value of β to be the ratio of the two. It turns out that one useful property of transistors is that, for a particular transistor operating over a fairly wide range of operating conditions, that the value of β tends to be fairly constant. It is not a matter of one causing the other -- both could be caused by something entirely different (such as the base-emitter voltage), but merely an observation that whatever causes both of them tends to do so in a way that results in their ratio being fairly constant.

The same is true for Ohm's Law. Don't try to think of it dogmatically as a voltage causing a current versus a current causing a voltage. Think of it as merely as a device in which the ratio of the voltage across it to the current,through it, however they happen to come about, exhibits the property that it's a value that we call the resistance.

So let's look at that circuit you proposed:



IF the transistor Q2 is in the active or saturation regions, then we assume that Veb is about 0.7 V. Since the base is at 0 V, that means the emitter is at about 0.7 V. Since the other end of the resistor is at 12 V, that puts about 11.3 V across 10 kΩ yielding about 7.87 mA of emitter current.

Now, if the transistor were in the active region, most of that would flow out the collector, but that would raise the voltage on R1 to a level that is too high because it would make Vec negative. So we conclude that the transistor is in saturation and that Vec is about Vecsat, which is another value that we often approximate as somewhere around 0.2 V for most small signal transistors but realize that it is almost never exactly that. So, assuming that it is 0.2 V, that means that the collector voltage is about 0.5 V and that the current in R1 is about 50 μA.

With 0.5 V at the base of Q1 and the emitter voltage being greater than 0 V if there is any emitter current, we can see that Q1 is going to be essentially shut off. So there will be no appreciable base current in Q1 and no appreciable current in either R2 or R4, placing the emitter of Q1 at 0 V and the collector at 12 V.

So the total collector current of Q2 is about 50 μA which means that the base current is about 7.82 mA and the value of β at this operating point is about 0.006, meaning that it is in extremely deep saturation. In fact, it is so deeply saturated that the Vecsat at this point is probably nearly 0 V (probably just a few millivolts) and the Veb is likely elevated above 0.7 V by a noticeable amount. A good rule of thumb for silicon BJT devices near room temperature is that Vbe changes by about 60 mV for every order of magnitude change in the collector or base currents. If Q2 is a typical small signal PNP transistor, it likely has a Veb of 0.7 V at somewhere around a collector current of 10 mA. With a β of around 100, that means the base current is about 0.01 mA. To get to nearly 10 mA of base current we would need to increase the base current by three orders of magnitude, so Veb is probably in the range of 0.88 V, making the collector voltage also about 0.88 V.

This makes the current in the 10 kΩ resistor about 88 μA and places the base voltage of Q1 at 0.88 V. So Q1 won't actually be in cutoff, but how much will it actually be conducting. Well, let's start off with assuming it's Vbe is 0.7 V. That means that the emitter voltage would be at 0.18 V and the emitter current would be about 18 μA, which would put the collector voltage at about 11.82 V. We can probably do a bit better by noting that a collector current of 18 μA is about three orders of magnitude less than the 10 mA collector current that yields our nominal 0.7 V, so the actual Vbe is probably closer to 0.52 V, making the emitter voltage closer to 0.36 V and the emitter and collector currents closer to 36 μA, yielding a collector voltage of perhaps 11.64 V.

Notice how much effort was involved in getting a result that differs very little from the 12 V value we got almost by inspection.

 

Thanks to everyone for the effort. This circuit is just trying to help understand what parts of a bjt constrain the others. That was the whole point. I saw some questions about what kind of signal and such. There is no signal. Its just a dc circuit for the purpose of learning. Thats all. It serves no actual purpose