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Reference: http://www.learningaboutelectronics.com/Articles/How-to-find-the-q-point-of-a-transistor-circuit

I am trying to understand this q-point calculation. I recently asked a question about the notation used, however I now understand this.

Consider the line RB=(R1 R2)/(R1 + R2). I understand that this is the calculation of 2 parallel resistors, R1 and R2. However I do not understand what the relevance of this is in the calculation.

I have seen this from other sources, for example the last example at the bottom of this page:

https://www.electronics-tutorials.ws/amplifier/transistor-biasing.html

Again, in this example, I understand everything until the line RB=...

How is this equation derived? I thought it might be obtained from one of Kirchoff's Laws, but if this is the case I can't see which law is used and how.

I copy the relevant diagram and calculation for convenience below.

enter image description here

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  • \$\begingroup\$ Here you can find a bit of information on Thevenin Equivalent circuit. electronics.stackexchange.com/questions/377467/… and here forum.allaboutcircuits.com/threads/transistor-base.81886/… \$\endgroup\$
    – G36
    Jun 12 '19 at 15:37
  • \$\begingroup\$ Still looking for an approach here? Or are you done? \$\endgroup\$
    – jonk
    Feb 1 '20 at 5:06
  • \$\begingroup\$ @jonk I put this particular project on hold for a while, but yes at some point I will go back to it and I am still interested to understand it \$\endgroup\$ Feb 2 '20 at 15:44
  • \$\begingroup\$ @user3728501 I'll assume for now that the answers you already have weren't sufficient. So I may add something. Not sure when. \$\endgroup\$
    – jonk
    Feb 2 '20 at 19:25
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You really, really need to understand why a resistor divider can be reduced to a voltage source and a series resistor.

If you don't understand it, don't believe it, don't have a feel for it, then you won't apply it. Or, if you do, you'll always be unsure and nervous about it. That kind of worry has to be driven out of you. You need to have confidence in the Thevenin equivalent of a resistor divider. The idea needs to be so deeply buried inside your bones that you will never question it, again. You will just "know" and be confident in that fact.

Let's proceed...

Thevenin Voltage of a Resistor Divider

A resistor voltage divider looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

On the left side we have two resistors in series between a power supply. What I'd like to know is what is the voltage at \$+V_\text{TH}\$. But when I ask that question, I have to say, "relative to what other location in the circuit?" So I've labeled another spot (node) called \$-V_\text{TH}\$, which identifies the location I've picked as the "relative to" answer. So now I'm asking, "What is the voltage at \$+V_\text{TH}\$ with respect to the voltage at \$-V_\text{TH}\$?"

In your transistor circuit, \$-V_\text{TH}\$ is ground. So on the right side, I'm renaming \$-V_\text{TH}\$ to "GND." There's no harm in this and it doesn't change the circuit. It just renames a node. Which is harmless. (Well, you don't get to rename a node to the same node name given to a different node, of course. And so you only get to have one GND node.) So I think you can easily see that the right side is the same as the left. I also decided to remove the +-sign and therefore rename the mid-point voltage as \$V_\text{TH}\$. (It's just a variable, now.)

I'm taking this slowly for a reason. I think you already know how to calculate \$V_\text{TH}\$ since it is hard to imagine you haven't had to work that out, just yet. But I want to make sure that the ground-work is laid out carefully, too, and that you can follow along.

(NOTE: By convension, any node labeled "GND" is assumed to be the "default reference point," whenever anyone is talking about the voltage at some other place in the circuit. We just "assume" that's the "relative to" location. So now, I can just ask "What the voltage at \$V_\text{TH}\$?" and you are then supposed to insert in your head "with respect to GND" in your own head. It's just a "common" that is always inferred whenever anyone talks about voltages without explicitly saying what it is in reference to. (Voltages are always a "voltage here with respect to a voltage there," as they are always relative measurements and have no absolute meaning.)

Now, as I proceed below, I want you to temporarily "forget" the fact that your circuit above has a BJT base also connected up. For now, we don't want to know about this and don't want to think about it. We'll get back to it, soon enough. So just hold your horses, for now.

With that in mind, we can work out the voltage at \$V_\text{TH}\$. We know the current through the series circuit is \$I_\text{TOT}=\frac{V_\text{CC}}{R_{\text{B}_1}+R_{\text{B}_2}}\$. But \$I_\text{TOT}\$ through \$R_{\text{B}_2}\$ will cause a voltage difference from one end to the other end of the resistor of \$V_\text{TH}=I_\text{TOT}\cdot R_{\text{B}_2}=V_\text{CC}\cdot\frac{R_{\text{B}_2}}{R_{\text{B}_1}+R_{\text{B}_2}}\$. This is a classic form for computing the voltage in the middle of a resistor divider pair.

This is the often simply called the Thevenin voltage, \$V_\text{TH}\$, for a resistor divider (with that implied reference to GND, of course.)

So, we have:

$$V_\text{TH}=V_\text{CC}\cdot\frac{R_{\text{B}_2}}{R_{\text{B}_1}+R_{\text{B}_2}}$$

Thevenin Resistance of a Resistor Divider

There is also a Thevenin resistance. This is a little trickier to gather up. Many will just tell you about linearity and superposition. But then this is just two more things I have to explain, in detail. And they are abstract and probably require a calculus viewpoint, anyway.

Let's add a load to the above circuit, as follows:

schematic

simulate this circuit

Note that I've drawn a box around our resistor divider circuit on the left side. \$R_\text{LOAD}\$ doesn't "know" what's in there. All it knows is that there is a connection point and ground that it will attach to. But suppose we want to find out if the idea found on the right side could be used. We already know how to compute \$V_\text{TH}\$, so the only remaining question is if we can find an expression for \$R_\text{TH}\$ that doesn't depend upon \$R_\text{LOAD}\$. If it does depend upon \$R_\text{LOAD}\$, we are screwed. But if the expression cancels things out such that \$R_\text{LOAD}\$ magically disappears, then we may have something.

Let's start with the easier right side schematic. Find the right-side is \$I_\text{LOAD}=\frac{V_\text{TH}}{R_\text{TH}+R_\text{LOAD}}\$. The left side schematic is a little more complex. Here, will have the left-side is \$I_\text{LOAD}=\frac{V_\text{O}}{R_\text{LOAD}}\$. But we also have \$V_\text{O}=V_\text{CC}\cdot\frac{R_{\text{B}_2}\mid\mid R_\text{LOAD}}{R_{\text{B}_1}+R_{\text{B}_2}\mid\mid R_\text{LOAD}}\$. Clearly, both these left-side and right-side calculations for \$I_\text{LOAD}\$ should be equal to each other. So we can say:

$$\begin{align*} \frac{V_\text{TH}}{R_\text{TH}+R_\text{LOAD}}&=\frac{V_\text{CC}\cdot\frac{R_{\text{B}_2}\mid\mid R_\text{LOAD}}{R_{\text{B}_1}+R_{\text{B}_2}\mid\mid R_\text{LOAD}}}{R_\text{LOAD}}\\\\ V_\text{CC}\cdot\frac{R_{\text{B}_2}}{R_{\text{B}_1}+R_{\text{B}_2}}\cdot\frac{1}{R_\text{TH}+R_\text{LOAD}}&=\frac{V_\text{CC}\cdot\frac{R_{\text{B}_2}\mid\mid R_\text{LOAD}}{R_{\text{B}_1}+R_{\text{B}_2}\mid\mid R_\text{LOAD}}}{R_\text{LOAD}} \end{align*}$$

You can see, at least for a start, that \$V_\text{CC}\$ cancels out. But I'll save you the trouble of the algebra and simply say that if you re-work the above equation, solving for \$R_\text{TH}\$, you will find:

$$R_\text{TH}=\frac{R_{\text{B}_1}\cdot R_{\text{B}_2}}{R_{\text{B}_1}+R_{\text{B}_2}}$$

Using calculus (I'll avoid the details here), I'd instead just solve the following (the negation is due to the fact that I know if the load current increases that the output voltage decreases):

$$\begin{align*} R_\text{TH}&=-\frac{\text{d}\, V_\text{O}}{\text{d}\,I_\text{LOAD}}=\frac{R_{\text{B}_1}\cdot R_{\text{B}_2}}{R_{\text{B}_1}+R_{\text{B}_2}} \end{align*}$$

Experimental Validation

Suppose \$V_\text{CC}=5\:\text{V}\$, \$R_{\text{B}_1}=1\:\text{k}\Omega\$, and \$R_{\text{B}_2}=4\:\text{k}\Omega\$. We can now compute \$I_\text{TOT}=\frac{5\:\text{V}}{1\:\text{k}\Omega+4\:\text{k}\Omega}=1\:\text{mA}\$ and therefore that \$V_\text{TH}=4\:\text{k}\Omega\cdot 1\:\text{mA}=4\:\text{V}\$. From the above work, we would also now say that \$R_\text{TH}=1\:\text{k}\Omega\mid\mid 4\:\text{k}\Omega=800\:\Omega\$.

So this is what our above theory says is the Thevenin equivalent circuit:

schematic

simulate this circuit

Let's consider two different load resistor values that we'll place between the \$+V_\text{TH}\$ output wire and the GND wire. Suppose we use \$R_\text{LOAD}=800\:\Omega\$ and \$R_\text{LOAD}=1200\:\Omega\$. We'll analyze the first circuit and then we'll analyze the "Thevenin equivalent" circuit for both cases. So we'll have four results and we'll compare them.

schematic

simulate this circuit

On the upper-left, we have \$800\:\Omega\mid\mid 4\:\text{k}\Omega=\frac23\:\text{k}\Omega\$ that is in-series with \$1\:\text{k}\Omega\$. So the total current from the power supply will be \$\frac{5\:\text{V}}{1\:\text{k}\Omega+\frac23\:\text{k}\Omega}=3\:\text{mA}\$. This means that \$R_1\$ will drop \$1\:\text{k}\Omega\cdot 3\:\text{mA}=3\:\text{V}\$, leaving \$+V_\text{TH}=5\:\text{V}-3\:\text{V}=2\:\text{V}\$. From this, we find that \$I_\text{LOAD}=\frac{2\:\text{V}}{800\:\Omega}=2.5\:\text{mA}\$.

On the upper-right, we have a total current of \$\frac{4\:\text{V}}{800\:\Omega+800\:\Omega}=2.5\:\text{mA}\$. Note that all of the total current is flowing through \$R_\text{LOAD}\$. So this matches up with what we just calculated for the upper-left circuit.

On the lower-right, we have \$1.2\:\text{k}\Omega\mid\mid 4\:\text{k}\Omega=923 \frac1{13}\:\Omega\$ that is in-series with \$1\:\text{k}\Omega\$. So the total current from the power supply will be \$\frac{5\:\text{V}}{1\:\text{k}\Omega+923 \frac1{13}\:\Omega}=2.6\:\text{mA}\$. This means that \$R_1\$ will drop \$1\:\text{k}\Omega\cdot 2.6\:\text{mA}=2.6\:\text{V}\$, leaving \$+V_\text{TH}=5\:\text{V}-2.6\:\text{V}=2.4\:\text{V}\$. From this, we find that \$I_\text{LOAD}=\frac{2.4\:\text{V}}{1.2\:\text{k}\Omega}=2\:\text{mA}\$.

On the lower-right, we have a total current of \$\frac{4\:\text{V}}{800\:\Omega+1.2\:\text{k}\Omega}=2\:\text{mA}\$. Note that all of the total current is flowing through \$R_\text{LOAD}\$. So this matches up with what we just calculated for the lower-left circuit.

I think you can see, at least from these examples anyway, that it appears this "trick" works.

Applying the Thevenin Equivalent to your Circuit

If you think closely about your circuit, you'll see the following as true:

schematic

simulate this circuit

Just applying KVL now, we can find:

$$V_\text{TH}-I_\text{B}\cdot R_\text{TH}-V_\text{BE}-I_\text{E}\cdot R_\text{E}=0\:\text{V}$$

But as \$I_\text{B}=\frac{I_\text{E}}{\beta+1}\$, this can be re-written as:

$$V_\text{TH}-\frac{I_\text{E}}{\beta+1}\cdot R_\text{TH}-V_\text{BE}-I_\text{E}\cdot R_\text{E}=0\:\text{V}$$

And solved for \$I_\text{E}\$ as:

$$I_\text{E}=\frac{V_\text{TH}-V_\text{BE}}{R_\text{E}+\frac{R_\text{TH}}{\beta+1}}$$

Or, alternately, for \$I_\text{B}\$ as:

$$I_\text{B}=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\cdot R_\text{E}}$$

The above assumes that the BJT is in active mode and isn't saturated by the circuit that surrounds it. (But this is easy to test. If you compute \$V_\text{C}=V_\text{CC}-\beta\cdot I_\text{B}\cdot R_\text{C}\lt V_\text{B}\$ then it is in some level of saturation.)

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  • \$\begingroup\$ This is probably one of the most detailed answers on SE... I'm following you up until Find the right-side is I_{LOAD}=V_{TH} R_{LOAD}/(R_{TH}+R{LOAD}). I don't yet fully understand this - but it might be because I'm not yet fully awake. Sorry if it's a stupid question. \$\endgroup\$ Feb 13 '20 at 11:00
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The equation for RB comes from finding the Thevenin Equivalent for \$V_{IN}\$. If you assume that \$V_{CC}\$ is deactivated (set to zero volts) then R1 and R2 are effectively in parallel. So, the Thevenin Equivalent resistance from \$V_{IN}\$ to the base is equal to R1||R2.

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  • \$\begingroup\$ Ok, I understand what that is, but what I don't understand is "what does RB actually mean" - is it something like the effective resistance as seen from the base of the transistor? Even than, I don't fully understand what that means, physically so to speak. Why do you say "set Vcc to zero" also? \$\endgroup\$ Jun 12 '19 at 14:50
  • \$\begingroup\$ You should take some time to learn about Thevenin Equivalent Circuits. This is a very broad topic; too broad to address in these comments. \$\endgroup\$ Jun 12 '19 at 16:24
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You treat Vcc as ground during resistance calculations, and determining the resistance of the circuit. (Short independent voltage sources and open independent current sources).

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  • \$\begingroup\$ This is the DC analysis part \$\endgroup\$ Jun 12 '19 at 14:49
  • \$\begingroup\$ Oh yeah sorry, you also short Vcc during resistance calculations. \$\endgroup\$
    – Jaywalk
    Jun 12 '19 at 14:49
  • \$\begingroup\$ The equivalent of "turning the device off" is shorting independent voltage sources and opening independent current sources. \$\endgroup\$
    – Jaywalk
    Jun 12 '19 at 14:50
  • \$\begingroup\$ Perhaps I should try and rephrase the question: "Why are we calculating RB and what does it correspond to in the real world?" Is RB the input impedance of this entire amplifier circuit? If so I don't understand where that impedance is seen from, or why we are calculating it, since this is a DC analysis, and the input impedance is only relevant when doing the AC analysis part. (Perhaps I'm wrong about that?) \$\endgroup\$ Jun 12 '19 at 14:55
  • \$\begingroup\$ Oh, Rb is required to calculating the KVL around the base to determine base current. You can see the KVL in the first link you posted. \$\endgroup\$
    – Jaywalk
    Jun 12 '19 at 15:01
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The equivalent DC source (and its effective series resistance) applied to the base is what they are calculating so, in the example in the first reference they calculate the voltage at the base under the assumption that the base takes no current (3 volts) and, the effective series resistance of that 3 volts. That effective series resistance is the parallel combination of R1 and R2. If you don't understand why that last part is so then that's a different question. Go study Norton's and Thevenin's theorem.

However I do not understand what the relevance of this is in the calculation

So, armed with the open circuit voltage of 3 volts and the effective series resistance of that source, you plug those numbers into the formula to calculate q-point: -

enter image description here

The above formula uses \$R_B\$ and \$V_{BB}\$ as previously discussed to yield \$I_{EQ}\$ and that then allows you to calculate the voltage q-point.

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  • \$\begingroup\$ I followed everything there up to a point: I think no I don't know what effective series resistance is exactly, although I do know how to calculate it just not "why". You raise a new question however which is where does that formula for Ieq come from? And what does Ieq actually mean? Is it the quiescent current? \$\endgroup\$ Jun 12 '19 at 14:59
  • \$\begingroup\$ I've added a link to Thevenin's theorem \$\endgroup\$
    – Andy aka
    Jun 12 '19 at 15:01
  • \$\begingroup\$ You need to calculate the ESR because the small amount of base current turns what should be a perfect 3 volts into something closer to 2.9 volts (or thereabouts). The base takes current because \$\beta\$ is not infinity. If beta were infinity (or the resistors R1 and R2 were very much lower in value), exactly 3 volts would be applied to the base and the formula would be much simpler. \$\endgroup\$
    – Andy aka
    Jun 12 '19 at 15:07
  • \$\begingroup\$ Ok, and Ieq IS that current that the base takes, and it is called the "quiescent current"? \$\endgroup\$ Jun 12 '19 at 15:12
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    \$\begingroup\$ Ieq = emitter quiescent current \$\endgroup\$
    – G36
    Jun 12 '19 at 15:39

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