Problem 43, chapter 6 in sedra and smith microelectronics.


simulate this circuit – Schematic created using CircuitLab

We are asked to find the current flowing in R1 and R2 knowing that VBE = 680 mV and Ic = 1 mA, β = 100.

if I compute Ie using (β+1)*Ic / β, I get 1.01 mA

if I first compute I(R1) using ohm's law (0.1 mA) and then use kirchoff current law: Ie+I(R1) = 1.1 mA, I get Ie = 1.1mA - 0.1 mA = 1mA

Not the same answer. Why?

Thank you very much.

  • \$\begingroup\$ Should the lower end of the current source be connected to ground? As shown, the circuit is nonsensical. \$\endgroup\$ – The Photon Aug 6 '13 at 18:00
  • \$\begingroup\$ \$I_C \neq I_E\$ in your Kirchhoff equation. \$\endgroup\$ – jippie Aug 6 '13 at 18:06
  • \$\begingroup\$ I'm not sure I understand. Kirchoff at node E gives: I(R1) + Ie = Is, so Ie = Is - I(R1) = 1.1 mA - 0.1 mA = 1 mA. The photon: that's how the circuit is given in the book. I think it's part of a circuit. \$\endgroup\$ – Douglas Edward Aug 6 '13 at 18:10

You did not read the problem's description carefully. It is a * problem (especially hard one), and there is a reason for this.

The problem states: "... current source \$I\$ is 1.1 mA, and at 25\$^{\circ}\$ C \$v_{BE}=680\$ mV at \$i_C=1\$ mA. At 25\$^{\circ}\$ C with \$beta =\$ 100, what currents flow in \$R_1\$ and \$R_2\$?"

You're asked to calculate currents through resistors, but you're not given constraints on \$v_{BE}\$ or \$i_C\$. You just told that, at room temperature, if \$v_{BE}=680\$ mV then \$i_C=1\$ mA.

Let us know if you need further help.

  • \$\begingroup\$ Thanks. Can we say that the collector current is the same as the current through R2? I am not sure because we don't know if something else is connected to ground. \$\endgroup\$ – Douglas Edward Aug 6 '13 at 19:27
  • \$\begingroup\$ What else could be connected to ground? The voltage at the other end of the current source is irrelevant in this setup. I don't see why would \$I_{R_2} = I_C\$ \$\endgroup\$ – Vasiliy Aug 6 '13 at 19:45
  • \$\begingroup\$ I was thinking that we can apply Kirchoff Current law at node C. But if something else is connected to this node then it doesn't hold. \$\endgroup\$ – Douglas Edward Aug 6 '13 at 19:52
  • \$\begingroup\$ @DouglasEdward, I'm not sure what you mean by "if something else is connected to this node". You have the schematic and there is nothing more to the circuit. \$\endgroup\$ – Vasiliy Aug 6 '13 at 19:57
  • \$\begingroup\$ I mean, since this circuit is probably extracted out of another bigger circuit, then there is probably something else connected to it in reality. But why IR2 is not = IC even if nothing else is connected? I'm confused. \$\endgroup\$ – Douglas Edward Aug 6 '13 at 20:04

The givens in your problem statement are contradictory.

If Vbe is 0.68 V, then, as you say, the current through R1 is 0.1 mA. Given the current source at 1.1 mA, then the emitter current must be 1.0 mA.

But you were also told that Ic is 1 mA. There's a variant of KCL that says that if we draw a closed curve across our circuit, then the sum of all the currents into the enclosed region must be 0. So if we draw a circle that encloses the BJT and no other elements of the circuit, the total current into that circle must be 0. This means

Ic + Ib + Ie = 0

if all of the currents are taken with a positive sign indicating current going in to the device.

Since Ic is 1 mA, and Ie is -1 mA, then we must have Ib = 0 mA.

Which violates the characteristic equations for the device (Ic = β Ib).

At least one of the "givens" in the problem statement must be rejected. Either Ic isn't really 1 mA, or R1 isn't really exactly 6.8 kOhm, or IS isn't really exactly 1.1 mA or ...

I suggest discarding the assumption that Ic is 1.0 mA and solve the problem from there.


the trans. is in saturation(since cathode of base collector diode is grounded) hence the drop across Tx is around .2v, 1.1mA is the emitter current viz Ie = Ic + Ib. therefore Ib comes out to be .1mA and the drop across R1 is .68V which is required to make BE diode forward. now we know the drop across R1 is .68V, we apply KVL from base to grnd through R2 completing the loop and find the current through R2 viz .68/68K = .01mA. one wrong info is beta which should be 10 not 100.

  • \$\begingroup\$ To improve readability of your answer, it is good practice to start your sentences with a Capital. \$\endgroup\$ – jippie Dec 8 '13 at 12:27

1ma is not correctly equal to 1.0ma.

1.1ma is not the same as 1.10ma.

1.00009ma is not the same as 1.00ma

I think you are being confused by significant figures.

Beware of assuming two things are equal.

Given a 10k 10% resistor that measures 10.1k would you refer to it as a 10k resistor or a 10.1k?

Are we splitting hairs or obeying the concept of significant figures when calculating numerical values? Your call.


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