# Class A amplifier - Transistor Q-Point calculation

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.

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.

• 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? – user3728501 Jun 12 at 14:50
• You should take some time to learn about Thevenin Equivalent Circuits. This is a very broad topic; too broad to address in these comments. – Elliot Alderson Jun 12 at 16:24

You treat Vcc as ground during resistance calculations, and determining the resistance of the circuit. (Short independent voltage sources and open independent current sources).

• This is the DC analysis part – user3728501 Jun 12 at 14:49
• Oh yeah sorry, you also short Vcc during resistance calculations. – Jaywalk Jun 12 at 14:49
• The equivalent of "turning the device off" is shorting independent voltage sources and opening independent current sources. – Jaywalk Jun 12 at 14:50
• 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?) – user3728501 Jun 12 at 14:55
• 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. – Jaywalk Jun 12 at 15:01

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: -

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.

• 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? – user3728501 Jun 12 at 14:59
• I've added a link to Thevenin's theorem – Andy aka Jun 12 at 15:01
• 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. – Andy aka Jun 12 at 15:07
• Ok, and Ieq IS that current that the base takes, and it is called the "quiescent current"? – user3728501 Jun 12 at 15:12
• Ieq = emitter quiescent current – G36 Jun 12 at 15:39