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I have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

And I am trying to calculate Rc and R2. I started by adding the currents but I am not sure for I1 or I2:

  1. Is I1 the same with Ib or Ic?
  2. Is I2 the same with Ib or Ie?
  3. Does it matter? What I mean is even if I add more currents in the circuit will the result be same?

About transistor voltages:

  1. Does Veb equals to 0.7V?
  2. If yes, then Vbc equals to Vec - Veb (4.3V)?

And last question:

Is it true that Ie = Ic + Ib or is it true only when the transistor works on the linear region?

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(First thing to do is to make explicit note that the bottom node of your circuit is to be taken as ground, or \$0V\$, to simplify discussion. Any node can be arbitrarily made 0V. But you only get to do that with one node and no more. So that's my \$0V\$ node.)

For small signal transistors like the 2N3904, I mentally use a active region figure of about \$V_{be} = 700mV\$ when \$I_C=4mA\$. Your \$I_c\$ isn't far enough way to worry about the difference (only makes about \$5mV\$ change.) So yes, I'd assume the transistor is in the active region until later calculations suggest differently and I'd tentatively take \$V_{be} = 700mV\$.

(Take note of the fact that \$\beta = \frac{I_C}{I_B} \approx 167\$, which is typical for an active region, small signal BJT like this. Just a quick confirmation of an earlier assumption before going forward. A second directly explicit confirmation is the statement on the diagram that \$V_{ce} = 5V\$. Note, though, that they say this for \$V{ec}\$, which is probably inconsistent with the usual nomenclature used and so I'm assuming they don't really mean \$V_{ec}\$ but instead \$V_{ce}\$.)

The emitter current will be almost the exact same as the collector current, so for a first estimate it must then be the case that \$Q_1\$'s emitter voltage will be \$V_e = 0V + R_e\cdot I_c \approx 1.25V\$. (The other \$15\mu A\$ only adds \$7.5mV\$ to that.) So, now it is estimated that \$V_b = V_e + V_{be} = 1.95V\$. Also, you can now use the stated fact that \$V_{ce} = 5V\$ to realize that \$V_c = V_{e} + V_{ce} = 1.25V + 5V = 6.25V\$.

The value of the collector resistor can now be computed as \$R_c = \frac{V_{cc}-V_c}{2.5mA} = 1500\Omega\$. This is also a standard value for resistors. Nice.

Since \$V_b = 1.95V\$, we can now compute \$I_1 = \frac{V_{CC} - V_b}{R_1} \approx 1mA + 6\mu A\$. I added that last little bit there so that you can see that the indicated base current doesn't materially change the value of \$I_2 \approx 1mA\$. In short, the voltage divider at \$Q_1\$'s base is stiff enough so that reasonable variations (or incorrect assumptions about it due to \$\beta\$ being different than estimated) probably won't materially affect the design calculations. So now we can compute \$R_2 = \frac{1.95V - 0V}{1mA} \approx 1950\Omega\$. (Slightly less or more depending on if you decide to get all technical and account for some wayward \$\mu A\$ that I've been suggesting you ignore, for now.) That's not a standard value resistor and it appears that your problem hasn't been using entirely standard values, anyway. But the problem does use rounded values, so I'd put down \$R_2 = 2k\Omega\$ as the answer there.

SIDEBAR: From my notes here you may remember that I said \$V_{be}\$ might be \$5mV\$ less, but also that \$V_e\$ might be \$7.5mV\$ more, so the estimated value of \$1.95V\$ for the base voltage really is pretty close despite those second order comments, earlier. You could go back through all this, using some of these newly developed resistor values and make some re-estimations to narrow things down. But honestly, you don't really know the value of \$\beta\$ for the transistor, anyway. Nor, technically, a more accurate value of \$V_{be}\$. Those things will be buried into the model parameters for the part and, even then, actual parts do vary. So there's no point wasting further time on this.

The result is \$R_2 = 2k\Omega\$ and \$R_c = 1500\Omega\$ and \$I_1 \approx I_2 \approx 1mA\$.

P.S. No, I didn't do any simulation to check the work. I didn't feel a need in this case. Chances are that the base current will be even lower (as the \$\beta\$ of a typical 2N3904 that I have laying around here is closer to \$250\$ at these currents and at room temperature [15mW heating on the BJT in a TO92 will add perhaps \$3^{\circ}C\$ and isn't going to mess with this or anything else by much.]) Also, the \$V_{be}\$ is likely to be a little lower than the estimate I used, and this will slightly raise the collector current a bit. But there's no point in nailing it down further partly because real parts do vary and you need to make designs that work without depending on fine details you can't and don't control. The operating point of this circuit will be about where it is calculated to be. And good enough for most uses even with the variations of BJT parameters. So there's no point in my simulating it. It would be a looking for "foolish consistencies." (Or trying to make sure I didn't screw up. But this is just too easy for that, so I wasn't worried.)

Final EDIT: Yes, given that \$V_{ce} = 5V\$ and \$V_{be} = 700mV\$, you can reasonably say that \$V_{cb} = V_{ce} - V_{be} = 4.3V\$.

Final NOTE: The quiescent voltage drop across \$R_c\$ is indeed \$3.75V\$. There is nothing wrong with your calculation of \$1500\Omega\cdot 2.5mA = 3.75V\$. It's just that this is the voltage drop across the collector resistor. It's not the voltage at the collector. To get the voltage at the collector, using the voltage drop, you have to subtract it from the \$10V\$ rail, which is at the top end of the collector resistor. So you find that the collector voltage is \$V_c = 10V - 3.75V = 6.25V\$. Which is the value I found by instead added \$V_{ce} = 5V\$ to \$V_e = 1.25V\$ in order to get the same value. Two ways of doing the same thing. Same answer.

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  • \$\begingroup\$ Perfect and detailed answer but I am a bit confused about one thing. I did my own calculations and like you I found Rc = 1500 and R2 = 2000 and I1=I2=1mA so far so good. What I did is Vc+Vec+Ve=10V so Vc=3.75=1500*0.0025A but you did Vc=Ve+Vce=6.25V!=1500*0.0025A. \$\endgroup\$ – DimChtz Aug 20 '16 at 13:45
  • \$\begingroup\$ @DimChtz: See added note at the bottom of my answer. \$\endgroup\$ – jonk Aug 20 '16 at 15:48
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Normally Ib is much less than Ic, perhaps you are missing a decimal point. Besides that the 8k base R can not source more than 10V/8K =1.25mA if Vb was 0V.

Both these suggest your Ib is incorrect.

Normally you compute Ve from Ie then the other node voltages determine the R values.

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  • \$\begingroup\$ Ib is a typo. It's actually 15μΑ. \$\endgroup\$ – DimChtz Aug 19 '16 at 14:10
  • \$\begingroup\$ Now compute as Ve= Ie*Re then Vb then Vc then R2 and Rc \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 19 '16 at 14:19
  • \$\begingroup\$ Vbe for 15 uA is closer to 0.6 to 0.65 V range than 0.7 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 19 '16 at 14:20
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Answer to your questions:

1.) I1 is a current through the voltage divider which provides the necessary base voltage VB across R2. Normally, R1 and R2 are chosen to allow a current I1 that is at least I1=(10xIB). I think, this is ensured in your circuit.

2.) I2 is simply the difference I2=(I1-IB).

3.) I do not understand the question. Perhaps the answer is contained in 2) ?

Next two questions:

1.) During calculation, the voltage VBE is assumed to be app. (0.65...0.7) volts. This estimate is sufficient because you have an emitter resistor RE (emitter degeneration, negative DC feedback) which reduces the sensitivity of the circuit to VBE uncertainties. This is the classical method because the exact value of VBE (which allows the desired current IC) is not known.

This stabilization works good because you have provided a relatively stable (stiff) voltage VB at the base. Therefore, the voltage VB depends only a little on the base current IB (which has very large tolerances).

2.) VBC=(VB-VC) with VB=(I2xR2)-(ICxRC).

  • General comment:

The circuit and the currents look reasonable and realistic. I do not intend to calculate for you all resistors; instead I gave you some hints for doing it by yourself.

Additional comment:

Circuit simulation (PSpice) confirms the design. For IC=2.515 mA the base current is IB=16.21 µA. Your design confirms that the collector current IC depends on VBE only - the resistors R1 and R2 are nearly independent on the base current IB (due to the "stiff" voltage VB and RE providing DC feedback). When the additional base current through the voltage divider is assumed with IB=0 (unity-gain buffer between the base node and the voltage divider) the collector current is still IC=2.56 mA.

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  • \$\begingroup\$ Because I continuoiusly are trying to improve my knowledge I kindly ask the "downvoter" to tell me where I am wrong. Or is there any other reason ? I think, such a clarification is also important for the questioner because he could be confused if a correct (and helpful) answer is qualified by somebody as not helpful. \$\endgroup\$ – LvW Aug 20 '16 at 7:52
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  1. I1 is not the same with Ib or Ic. Actually I1 = I2 + Ib.
  2. I2 is also not same as Ic or Ib.

about Transistors:

Vbe is 0.7V and Veb is -0.7 (typically considered during analysis).

So voltage across R2 is 0.7V. Use voltage division rule between R1 and R2. You will find R2 out of which you can find I2. With Ib known you can find I1. With Ic know, Ib known you can find Ie = Ib + Ic. Apply KVL at the output and you can find Rc.

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  • \$\begingroup\$ The voltage across R2 is not 0.7V but the sum of VBE and the voltage drop across RE. \$\endgroup\$ – LvW Aug 20 '16 at 7:53
  • \$\begingroup\$ ohh sorry.. I missed the Re resistor by oversight. You are correct. \$\endgroup\$ – Sanat Mishra Aug 21 '16 at 14:38

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