Сan you help please with solving this problem?

The circuit shown below is used as an amplifier The value of the componets are Rb=470 kΩ om RC = 4.7 kΩ Power supply Vcc = 12V The transistor has Vbe = 0.6V β(Beta) = 80 Compute the value of the quiescent operating point:

Ib = ? μA
Vc = ? V

The transistor is working in ____________ region.


simulate this circuit – Schematic created using CircuitLab

So this my doubts and calculations, are they correct?

  1. Base current (Ib):
    Ib = (12V - 0.6V) / 470 kΩ
    Ib = 11.4V / 470 kΩ
    Ib ≈ 24.26 μA

  2. Collector-emitter voltage (Vc):
    Vc ≈ Vcc / 2
    Vc = 12V / 2
    Vc = 6V

    Therefore, the quiescent operating point of the amplifier is:
    Ib ≈ 24.26 μA
    Vc = 6 V

  3. The transistor is working in the active region (as an amplifier) because the collector-emitter voltage (Vc) is not saturated (close to Vcc) or cut-off (close to 0 V).

  • 1
    \$\begingroup\$ Vc is not correct. Hint: You'll need to use Beta to calculate Vc in this case. \$\endgroup\$
    – sai
    Commented May 1, 2023 at 17:43
  • \$\begingroup\$ @sai Vc = about 2.9 V. am i right? \$\endgroup\$
    – Mirmir
    Commented May 1, 2023 at 17:49
  • \$\begingroup\$ that is correct. \$\endgroup\$
    – sai
    Commented May 1, 2023 at 18:10

3 Answers 3


This is what I teach my students when they deal with such a circuit. It is a 3 steps algorithm:

  1. is the transistor cut-off? For it to be cut-off, the voltage on the base needs to be lower than the Base-Emitter diode voltage (noted Vbe or VD0). In your case, the Vbe is 0.6 and the VB is 12V

  2. is the transistor saturated? 2-1) Calculate satuartion current Ic(sat) 2-2) Calculate the minimum base current to achieve the saturation point Ib(min) 2-3) Calculate the real base current Ib (you got that part correctly in your question)

  3. If it is not in saturation or in cut-off, then it is simply in active/linear region and is neither fully closed or fully open.

In your question, you did not provide a value for Vce(sat) which is the voltage between the collector and emitter, but you can fairly assume values between 0v and 0.2.

Calculation become Ic(sat) = (Vcc- Vce(sat)) / Rc Ib(min) = Ic(sat)/beta Ib is what you previously calculated.

If the transistor was saturated, Vc would be equal to Vce(sat) or nearly 0V.

I am pretty sure that the transistor is in the linear/active region since the Rb resistor is pretty big.

In that case, the real Ic is obtained by multiplying Ib and beta together.

so 24.26 uA * 80 yield 1.94mA for Ic.

You can then apply Ohm's law accordingly: Vcc - Ic*Rc = Vc

12V - 1.94mA* 4.7k = 2.88V

Vc is then 2.88V

In this answer, when I refer to open and close it means that the transistor is free to let current through like a valve (open) or restrict it completely (closed). An open transistor (Saturated) yield a closed circuit and a cut-off transistor yield an open circuit.

As a side note, I was puzzled with your approximation that Vc is Vcc / 2. I never saw that before and I don't think it is a thing. Vc is equal to the voltage Vce and that voltage can take any value from Vcc when the transistor is full closed to Vce(Sat) when the transistor is fully open.

The more open the transistor is, the lower Vc will become. This is because the ground is slowly being exposed to the Vc point through the transistor. If the transistor was completely ideal, Vc would be able to reach 0V. However, it will never be the case in practice.

I hope this helped you a bit.

  • \$\begingroup\$ "The more open the transistor is, the lower Vc will become". An open circuit does not pass current. A closed circuit does. For switches an open switch does not pass current but a closed switch does. Contact designators like NO or NC mean normally open and normally closed respectively. Open contacts do not pass current, Closed contacts do.This is electronics. \$\endgroup\$
    – RussellH
    Commented May 2, 2023 at 2:30
  • \$\begingroup\$ Sorry, I will correct my answer to clarify that ambiguity. I used the word open in the meaning and analogy of a valve that you ''Open'' to let current pass through the transistor. \$\endgroup\$ Commented May 2, 2023 at 16:17
  • \$\begingroup\$ Furthermore, the expression cut-off (closed) is in opposition of the open state of the circuit which enhance even more the ambiguity in the vocabulary. I added a section to treat the difference between the transistor state and the circuit state. If you think this is still not clear, please advise so I can edit further. \$\endgroup\$ Commented May 2, 2023 at 16:25
  • \$\begingroup\$ I hope you also teach your students that this is a terrible circuit because it depends on a particular value of beta, which cannot be relied upon between component samples, or across temperature. Art of Electronic refers to this type of reliance as "sadness-inducing" (3rd edition, page 85). \$\endgroup\$
    – gwideman
    Commented Mar 23 at 3:04

Here is how I solve such a problem:

  1. Base potential is fixed. So we can determine approximate base voltage and base current: $$\begin{aligned} V_B &\approx 0.6{\rm\,V} \\ V_{R_B} &= V_{CC}-V_B = 12-0.6=11.4{\rm\,V} \\ I_{B} &= I_{R_B} = V_{R_B}/R_B = 11.4/470{\rm k}=24{\rm\,\mu A}. \end{aligned}$$

  2. Assuming linear region, we figure out collector current $$\begin{aligned} I_C &= I_B\cdot \beta \\ I_C &= I_{R_C} = 24\mu \cdot 80 = 1.9{\rm\,mA}. \end{aligned}$$

    There's no need to assume anything about collector voltage. The assumption should be about operating region, and only if the assumption proves wrong, we redo the parts of the problem that used that assumption.

  3. Assuming linear region, determine collector voltage. If higher than approximately base voltage, the linear operation assumption stands. Otherwise, we'd have to redo the problem assuming saturation. $$\begin{aligned} V_{R_C} &= I_{R_C}\cdot R_C \\ V_{R_C} &= 1.9{\rm m}\cdot4.7{\rm k} = 8.9{\rm\,V} \\ V_{C} &= V_{CC}-V_{R_C} = 12-8.9=3.1{\rm\,V}. \\ \end{aligned}$$ Since \$V_C>V_B\$, the transistor is indeed operating unsaturated.

Now, just to see what would happen in saturation, we can cut the base resistance in half:

  1. $$I_B' = I_{R_B}' = 11.4/240{\rm k}=48{\rm\,\mu A}.$$

  2. Assuming linear region, the collector current $$I_C' = I_B' = 48\mu\cdot 80 = 3.8{\rm\,mA}.$$

  3. Assuming linear region, the collector voltage $$V_C' = V_{CC}-V_{R_C}' = 12 - (3.8{\rm m}\cdot4.7{\rm k})= 12-18=-6{\rm\,V}.$$ Since this is well below \$V_B\$ of 0.65V, the transistor is saturated. We have to change assumptions.

  4. Assuming saturation, and \$V_{\rm CE\,sat} \approx 0.2{\rm\,V},\$ the collector current is independent of beta: $$ I_C'=(V_{CC}-V_C')/4.7{\rm k} = (12-0.2)/4.7{\rm k} = 2.5 {\rm\,mA}.$$

    We now also know the approximate collector current range before saturation occurs - it should be below about 2.5mA.

Since the transistor model we've used is simplified and assumes a fixed beta, all of those are of course rough approximations. They have more than 1 significant digit, but not much more. In real discrete transistors, beta will vary quite a bit, base voltage will vary by several tens of mV for a given base current, etc.


First, figure out the maximum possible collector current \$I_{C(SAT)}\$. This occurs when the transistor is saturated, \$V_{CE}\$ is at a minimum, and collector potential \$V_C\$ is as low as it will ever be. In other words, the voltage across \$R_C\$ is at a maximum of 12V:

$$ I_{C(SAT)} = \frac{12V}{R_C} = \frac{12V}{4.7k\Omega} = 2.6mA $$

Now figure out what base current \$I_{B(SAT)}\$ would be just enough to evoke this condition, just enough to saturate the transistor:

$$ I_{B(SAT)} = \frac{I_{C(SAT)}}{\beta} = 33\mu A $$

You already correctly established that base current is 24μA, so now ask yourself, is this sufficient to saturate the transistor?

If not, then you may then assume that the transistor is in its active region, and then continue to find \$V_C\$. Start with your base current \$I_B=24.3\mu A\$ to find out actual collector current:

$$ I_C = \beta I_B = 80 \times 24.3\mu A = 1.94mA $$

From there it's trivial to find collector potential.


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