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"this dc voltage is known as bias voltage and it's ...regardless of the polarity of the signal source" now I have this question : **1.**Do i manually need to choose bias voltage so that so that transistor remains on even when negative cycle kicks in? for example let us say my signal source has value of 5 volts peak to peak , so now if I choose 7 volts or 12 volts as a bias voltage it should work perfectly, this also means that the bias voltage is not unique , it depends on the signal we are applying, we can change bias voltage according to our source signal, is it correct? (assuming everything is sine wave)

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    \$\begingroup\$ I hate these diagrams. They are nothing but confusing. And they lead to questions, such as the ones you are asking, where any short answer will just add to your confusion. So please be aware that this whole thing is a mess, from start to finish, and that trying to directly answer will lead to more difficulties than it resolves. I'll need to think a little before composing my thoughts on how to approach your question in way that will actually help. (Until then, just imagine that the AC sine is constrained to a very very tiny peak-to-peak magnitude compared to the DC base bias.) \$\endgroup\$ – jonk Oct 24 at 15:53
  • \$\begingroup\$ thanks jonk, I hope after sometime you would like to clear my confusion \$\endgroup\$ – aud098 Oct 24 at 15:58
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    \$\begingroup\$ Question 1 is actually a very good question (and the answer will eventually be yes) but the rest of your speculation makes no sense for reasons that will become apparent a bit further into the course. So note the question and come back to it when you're ready. Meanwhile, suspend disbelief and stay the course As @jonk says, think "small signal" like 5 mV pk-pk for now. \$\endgroup\$ – Brian Drummond Oct 24 at 16:28
  • \$\begingroup\$ can somebody clear this confusion: let us assume the activation voltage of the transistor at base is 0.7 V , and assuming it can tolerate up to 20 volts in its base, and suppose my AC signal has pk-pk voltage 6 volts, can I use 13 volts as a bias voltage? because 13+6=19 v it is not going to destroy the device, and when negative cycle comes, 13-6=7v , which is sufficient to remain on the transistor. following similar logic , can i use 9 volts too as a bias voltage? \$\endgroup\$ – aud098 Oct 24 at 16:48
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    \$\begingroup\$ @aud098 I've added something to start. There's a lot more that could be written. But for a start, perhaps its enough? \$\endgroup\$ – jonk Oct 24 at 16:50

The circuit you have is a highly non-linear one, even if you assume that the AC voltage itself is exactly zero volts, peak to peak. Let's look at it without regard to the AC, for a moment. It'll be easier to show what I mean, then.


simulate this circuit – Schematic created using CircuitLab

I've removed the batteries, as they really just add "noise" to the diagram without helping at all in understanding it. It's enough to simply label the voltages, as they are implicitly referred to the ground reference.

Let's say that \$R_\text{C}=4.7\:\text{k}\Omega\$, \$V_\text{CC}=10\:\text{V}\$, \$V_\text{BB}=650\:\text{mV}\$ and that therefore \$I_\text{C}=1\:\text{mA}\$ (simplified Ebers-Moll model in active mode.) Then we'd expect to see a voltage drop of \$4.7\:\text{k}\Omega\cdot 1\:\text{mA}=4.7\:\text{V}\$ across \$R_\text{C}\$, meaning that the collector voltage sits at \$V_\text{C}=10\:\text{V}-4.7\:\text{V}=5.3\:\text{V}\$. All seems well, as the collector voltage is higher than the base voltage and so the BJT is really in active mode.

What happens if we change things so that \$V_\text{BB}=750\:\text{mV}\$? In this case, temporarily ignoring the collector resistor and assuming that the BJT stays in active mode, I'd expect the collector current to change by a factor of almost 50, to about \$I_\text{C}=50\:\text{mA}\$. (This is because there will be a factor of 10 change in the collector current for each and every increase of \$60\:\text{mV}\$ in the base-emitter voltage.) But now the voltage drop across \$R_\text{C}\$ should be nearly \$235\:\text{V}\$!! And that's just not possible. So already we know that even a very tiny adjustment of the base voltage leads to an impossible result -- if we assume the BJT is still in active mode.

Instead, what really happens is that the collector voltage will smash down close to the emitter voltage (about \$0\:\text{V}\$) and we will have a collector current nearing \$I_\text{C}=\frac{10\:\text{V}}{4.7\:\text{k}\Omega}\approx 2.13\:\text{mA}\$ and the BJT will be in saturation mode.

By now, I think you may be able to see why I hate that diagram so much. The BJT is a non-linear device. It responds wildly with respect to minor base-emitter voltage changes.

When your textbook says, "This d.c. voltage is known as bias voltage and its magnitude is such that it always keeps the emitter-base junction forward biased regardless of the signal source...", it is just begging you to do exactly what you thought about doing -- adjust \$V_\text{BB}\$ so that no matter what the input signal AC peak magnitude might be the base-emitter junction remains in forward bias.

If you take their comment "literally" and apply it for any-sized AC input signal, then the entire idea becomes impossible! This is why you are asking a good question, right now.

When I was first trying to understand BJTs more than 50 years ago, these same stupid examples and very similar writing about them caused me similar confusion for a time. And I'm not happy to see that nothing has changed in the education about BJTs in all that time.

Let's see why, using your own example of \$V_\text{AC}=5\:\text{V}_\text{PK}\$ for the AC source. You might then consider \$V_\text{BB}=6\:\text{V}\$, since that's large enough to keep the base-emitter junction forward-biased.

But that means that the base voltage is \$1\:\text{V} \le V_\text{B}\le 11\:\text{V}\$! But the collector current is a non-linear function of the base voltage (the emitter is grounded.) And I mean highly non-linear. For every \$60\:\text{mV}\$ change, the collector current is multiplied (or divided) by 10!! Look at that voltage swing, \$\Delta V_\text{B}=10\:\text{V}\$! How many orders of 10 in collector current does that imply??? About 170 orders of magnitude change!

Do you think there is any chance in this universe of that happening?

The impossible implications from taking their comment as being true under all circumstances you can imagine for the AC signal source means that you must let go of the idea that their model works for any imaginable AC signal magnitude. In short, you must instead imagine that it only applies for a very tiny AC signal magnitude. One that does NOT materially change the "bias-point" of the BJT. Or, put still another way, doesn't have a noticeable impact on the magnitude of \$V_\text{BB}\$.

In that case, the schematic they provide works.

But then you probably have another good question. Why did they bother to write, "This d.c. voltage is known as bias voltage and its magnitude is such that it always keeps the emitter-base junction forward biased regardless of the signal source..."? I mean, if the AC signal is assumed to be so tiny, what's the point of writing it that way?

Also, the way they write that almost makes you think that even if you keep the AC signal magnitude small, you might still be able to let the base-emitter junction swing widely if perhaps not as widely as you were imagining. For example, suppose \$V_\text{BB}=700\:\text{mV}\$. Then you might imagine that the peak AC voltage might be \$500\:\text{mV}\$. But that would NOT be valid for this circuit. Yet it meets with their statement, just fine, and you didn't even change \$V_\text{BB}\$!

So I think they chose poor writing.

Why not just say, "The AC peak voltage must not exceed a couple of millivolts in peak magnitude?" That's easy to read and directly addresses the situation and you'd still also know that the base-emitter junction is forward-biased, too.

I don't know why they didn't just do that. You'd then know that you aren't free to make much adjustment to the AC voltage and things would be a lot better for the example at hand. (It would keep you nearer the much simpler "linearized" small-signal model of the BJT and well away from its global, highly non-linear behavior.)

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Jonk has given you a good answer, but I want to point out something that may have contributed to the confusion.

The signal in the example is not a voltage signal.

It is a current signal.

Rather than trying to imagine what happens when you apply a 5V peak to peak sigal, you should be considering what happens as the current varies.

That the current source is in series with \$V_{BB}\$ is important.

Connecting \$V_{BB}\$ straight to the base and the emitter would cause a large current to flow through base, destroying the transistor.

With the current source in series with \$V_{BB}\$, the current through the base is limited to the current from the signal source.

In any real transistor amplifier circuit, the biasing is done via current. I suggest taking a look at the Wikipedia Bipolar transistor biasing page. It covers many more ways to bias a transistor - and the \$I_B\$ current is at the heart of every one.

The externally applied voltage signal is also converted to a current signal.

Look at the biasing in a simple amplifier like your example:


simulate this circuit – Schematic created using CircuitLab

You choose R2 (the biasing resistor) such that \$I_B \times \beta \times R1 \approx \frac {VCC}{2}\$. (There's a lot more to it than that because you have to pick \$R1\$, and that depends on the transistor load line and other stuff. Not going into that now.)

You've not picked a biasing voltage, you've picked a biasing current.

The voltage at the base ends up at (about) 0.7V because that's the forward voltage of the transistor's base-emitter junction - it acts like a diode.

Now look at the input:


simulate this circuit

Since V1 is a voltage source, you have to have R3 to limit the current. A voltage source applied directly to the base will try to raise the voltage on the base above 0.7 V, and will pour in as much current as it takes to make that happen - rapidly resulting in a dead transistor.

A voltage source in series with a resistor is (or approximates) a current source.

With R3, the voltage at the base will stay at (about) 0.7V, but the current through the base will vary as the input signal voltage varies. That current is multiplied with the \$\beta\$ of the transistor. The varying base current varies the collector current, which together with \$R1\$ varies the output voltage at \$VOut\$.

The real tricky thing about transistor amplifiers is that transistors only have current gain, but the typical transistor circuits have voltage gain.

The example from your text book mixes voltage and current sources, and makes it difficult to see what is going on.

Looking back, I'm not sure I've made things any clearer. It seems clear and correct to me, but it is written out of my self taught view of transistors.

The point I was trying make, though, is that the originally posted circuit has a current source as a signal source rather than the voltage source most folks expect.

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