Common Emitter Amplifier

I have this circuit. In the data sheet, I found values for the DC current gain (beta.) Which value for beta do I use in this common emitter amplifier circuit? Is my method to get the base, emitter and collector voltages correct?

  1. calculated Rth by r1||r2
  2. calculated Vth by r2/r1+r2 * (vcc)
  3. calculated Ibq by vth-vbe / rth + (1+beta)R4
  4. calculated Icq by beta * Ibq
  5. calculated Ieq by (1+beta)Ibq
  6. calculated Vceq by vcc - ic(r4+r3)
  7. calculated emitter voltage by r4*Ieq
  8. calculated collector voltage by Vceq + emitter voltage
  9. Calculated base voltage by Vth - RthIbq

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  • \$\begingroup\$ I assume you are calculating the quiescent condition for this circuit. If the current gain is large enough, which it is, you can get accurate enough results without using β at all. Once you know the collector current, you can calculate the upper and lower limits of the base current. \$\endgroup\$
    – Bart
    Commented Oct 14, 2021 at 8:03
  • \$\begingroup\$ I did this circuit in a lab and want to create the circuit in software to try and get the values as close to my experimental values as possible, don't I need a value for beta to enter in the software? I am assuming that I must input 800 as beta in my calculations and the software. The aim of the task is uncertainty between measured results and expected results. \$\endgroup\$
    – james
    Commented Oct 14, 2021 at 8:07
  • \$\begingroup\$ You can calculate the voltage on the base as you said. Loading by the base current is small, so you can ignore that. Subtract 0.7V to get the emitter voltage. Dividing that by emitter resistance you get the emitter current, which is almost equal to the collector current, the base current being small. From the limits of β you can now calculate the limits of the base current. Note: a quick calculation shows this circuit is very close to saturation. \$\endgroup\$
    – Bart
    Commented Oct 14, 2021 at 8:11
  • \$\begingroup\$ If you need to use beta to calculate base resistors then use the minimum value for the amount of saturation you might encounter. That saturation level is based on how much p-p voltage your output needs to provide with your limited supply rail. Using a typical value gets you nowhere with a real design. \$\endgroup\$
    – Andy aka
    Commented Oct 14, 2021 at 8:18
  • 1
    \$\begingroup\$ Hint: To understand the main point of analyzing these circuits, try if it even matters that much if the Beta is 10, 100 or 1000. If the range is from 100 to 800 and you need to make say 1000 prototypes that operate within some tolerance you also don't know Beta or Vbe of each of the 1000 transistors you bought. \$\endgroup\$
    – Justme
    Commented Oct 14, 2021 at 8:31

2 Answers 2


Which value for beta do I use in this common emitter amplifier circuit?

Being "completely hooked up on \$\beta\$ or \$h_{FE}\$" is a very common thing among beginners. Fact is, as you can see in the dataheet, that \$\beta\$ varies a lot. And I mean really a lot. Here it's already a factor 4 between minimum and typical. They don't even mention the highest limit.

The reason for this is that \$\beta\$ is determined by production variables that are not easy to control. So transistors are made and then measured and sorted according to their \$\beta\$, that's why there are BC107A, BC108C etc. which are mostly the same except for \$\beta\$.

So how do circuit designers deal with this?

We use and design for the smallest \$\beta\$ we can expect. And add some margin as well. So let's say we have a bunch of BC108, it's lowest (guaranteed by datasheet) \$h_{FE}\$ is 110.

From this I subtract some margin like 20% (engineers add margins to prevent ever getting stuck on an "edge"). Then I round this off to say, \$h_{FE}\$ = 90. So I would use \$h_{FE}\$ = 90 to design the circuit.

Yes but what if \$h_{FE}\$ is much larger like \$h_{FE}\$ = 800? Then my circuit will stop working!

Will it? Do the excercise, design for \$h_{FE}\$ = 90.

Then, with the resistor values you got from that, re-calculate what happens when you use \$h_{FE}\$ = 800. Does that change a lot?

What happens when \$h_{FE}\$ increases is that the base current \$I_B\$ gets smaller, if we assume that the collector current \$I_C\$ stays the same.

So the NPN will draw less current from R1 and R2. In a proper design the current through R1 and R2 will be significantly (say a factor 10) higher than the highest \$I_B\$ we can expect.

When do we get that high \$I_B\$? When \$h_{FE}\$ is small!

So making \$h_{FE}\$ larger (800) will not change the voltage at the Base much. That means the voltage across R4 doesn't change much. What does that mean for \$I_C\$?


A well designed BJT circuit will cope with a large range of beta, it will not depend on an exact value.

There are two 'correct values' of beta to use.

The first is infinity. It allows you to understand what the circuit is doing without getting bogged down in the mathematics.

The second is the minimum beta guaranteed over your operating range. With the BC107/8/9 transistors you've posted, you'll notice a relatively large range being specified between min and typical, and even no minimum being specified for the 'A' grade devices at 10 uA. This means if you want to build a reliable circuit operating at 10 uA, you should use 'B' grade or higher, although typically, most 'A' grade devices will work.

So how can we design for a large beta range?

Generally we design for a collector current, and the base current is what it is. To design for the minimum beta, the maximum base current, we need to make sure that our bias circuit has low enough impedance to source that current with minimal voltage drop. If removing the base current (as beta goes to infinity) causes an excessive voltage shift, then we use feedback. This can be as simple as an emitter resistor (often used in audio amplifiers), or as complicated as an external opamp setting the bias level in response to a collector current sensor (often used in RF amplifiers).


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