2
\$\begingroup\$

I find the definitions of Q-point that I have been given in class to be a bit ambiguous, so please read on and confirm or correct my understanding.

As I understand from this SO answer, the Q-point or Quiescent Operation Point of a BJT are the values of the voltages and currents of its bias circuit (bias values) when no signal is present. In class-A amplification applications, it is also that combination of bias values that enable the BJT to work properly as a linear amplifier for any AC input signal.

Take the following bias circuit:

Bias circuit for a BJT

The input characteristic curve of a BJT is something like this:

Input characteristic curve of BJT

If we are looking for class-A linear amplification, then it is only logical that the Q-point (which must lie on the input characteristic curve) should be roughly midway inbetween its two extremes of operation with respect to it being 'fully ON' or 'fully OFF'. That is, they are where the curve is 'most linear', and we will get more faithful amplification. Also, we must ensure that in the output characteristic curve for the transistor, we will be far away from the saturation region:

Output characteristic curve of BJT

Here is where my confusion begins. I will try to explain what I understand. What I would like is if someone could clarify whether what I am explaining from now on (or previously, for that matter) is correct or not:

There is another line drawn in the previous two graphs: the load line of the input and output circuits. It is fixed by the values of \$V_{BB}\$ and \$R_{BB}\$ in the input circuit and by the values of \$V_{CE}\$ and \$R_C\$ in the output circuit. Here are their equations:

$$ \boxed{I_B} = \frac{V_{BB}-\boxed{V_{BE}}}{R_{BB}} $$

$$ \boxed{I_C} = \frac{V_{CC}-\boxed{V_{CE}}}{R_{C}} $$

where the boxed variables are varying throughout the lines.

So, in order for the biasing of our transistor to be appropriate for the purpose of class A linear amplification, we should regulate those voltages and resistances so that the load lines intersect the characteristic curves at a convenient place for the Q-point.

The typical exercise (like this one) in electronics courses about finding the Q-point of a BJT bias circuit is not really asking about finding the optimal voltages and resistances of the bias circuit for the transistor to operate as a class-A amplifier. In other words, it is not asking you to calibrate the load lines so that they intersect the characteristic curves at a convenient spot. Rather, they are asking to find the values of all the voltages and currents in the transistor given a specific set of bias circuit voltages and resistances. The circuit that they are providing is already biased to a given Q-point. What they ask of you is to find what that Q-point is.

In other words, 'find the Q-point of this transistor' does not mean 'find the optimal way of biasing this transistor', but rather, it means 'find how this transistor has been biased'.

Is that correct? Sorry for the lengthy question, but this to me was explained very ambiguously and I need confirmation (or correction) of what I understand.

\$\endgroup\$
4
  • \$\begingroup\$ You are correct, 'find the Q-point of this transistor' means 'find how this transistor has been biased'. \$\endgroup\$
    – G36
    Sep 18, 2022 at 10:14
  • \$\begingroup\$ For each way a bjt is biased, we can define what is called the 'bias stability factor'. The lower, the better. \$\endgroup\$
    – Antonio51
    Sep 18, 2022 at 10:37
  • 1
    \$\begingroup\$ I was taught to NEVER simply bias a transistor like that because a transistor has a range of current gain then some will be saturated and others will be cutoff. \$\endgroup\$
    – Audioguru
    Sep 18, 2022 at 17:28
  • \$\begingroup\$ True @Audioguru, this is just the simplest way of biasing a transistor and it is used for exemplification purposes. However, it is highly unstable with respect to variations in \$\beta\$ which may appear even in transistors from the same supplier. That is why it is not really advisable to use it in practice for many applications. \$\endgroup\$
    – Clerni
    Sep 18, 2022 at 17:53

2 Answers 2

3
\$\begingroup\$

It depends whether you are doing analysis or synthesis.

Analysis is easier, so it's what students are given as an exercise. With these components, work out what's happening.

What engineers really need to do to create circuits is synthesis. Find the values of components that will make this circuit work well. And before that, choose a topology for which good component values exist.

Analysis is the 'walk' before you do the synthesis 'run'.

In practice, many people do synthesis by iterative analysis. That is, throw a circuit together, and analyse what it does. If it appears to do exactly what you want, stop there. If it doesn't, change something, and re-analyse. With any luck or imagination, the difference between the old performance and the new performance gives you an insight into what the components are doing, and your third iteration might be even better. Computer based optimisers work a bit like this.

With a simple enough circuit, it's possible to write an expression for what it's doing in terms of its components. Then to find a maximum for one of its performance parameters, differentiate with respect to that parameter, set it to zero, and solve. This works, for very simple circuits.

For even slightly complicated circuits, and those where the performance parameters need a compromise between competing aims (for instance performance and cost), that simple mathematical synthesis route tends not to work, and you're back to modifying component values and repeated analysis. Or experience. You gain experience by doing repeated analysis!

\$\endgroup\$
5
\$\begingroup\$

In other words, 'find the Q-point of this transistor' does not mean 'find the optimal way of biasing this transistor', but rather, it means 'find how this transistor has been biased'.

Is that correct?

It depends on the context. If you are given a transistor amplifier complete with component values it means, "find how this transistor has been biased".

In a design context, where component values are being determined, design constraints like VCC and/or input-output signals are provided then, "find the optimal way of biasing this transistor" is probably expected, although the expectation should be clarified in words.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.