I came across a section of a tutorial regarding BJT DC analysis as follows:

enter image description here

The first two equations for Ic and KVL for the BE loop seems correct.

But I'm stuck at the last equation inside the box for the Ic.

It seems to me the factor for the Ico must be (β+1) but it is something different in the equation above. I checked it ten times and couldn't figure out and be sure whether I am or the material is wrong.

Here is how I write down:

Ic = β×Ib + (β+1)×Ico

Ib = (Vbb -Vbe) / (Rbb+(β+1)×Re))

These yield:

Ic = (Vbb -Vbe) × β/(Rbb+(β+1)×Re)) + (β+1)×Ico

I find the factor for the Ico as (β+1), but the material's result for this factor is:

(β+1)×(Rbb+Re) / (Rbb+(β+1)×Re) which is not equal to (β+1).

Is the original derivation or am I making a mistake?


It's been enough days, I think.

Two ways to think about the problem. Let's start with the way the problem was presented at the outset:

$$\begin{align*} I_\text{C}&=\beta\: I_\text{B}+\left(\beta+1\right)\:I_\text{CO}\\\\ V_\text{BB}&=I_\text{B}\: R_\text{BB}+V_{\text{BE}_\text{ON}}+\left(I_\text{B}+I_\text{C}\right)\:R_\text{E} \end{align*}$$

Take note that this is two equations and two unknowns: \$I_\text{B}\$ and \$I_\text{C}\$. So it's already solvable as a simultaneous pair of equations.

Yet you decided to introduce this KVL on your own:


That was completely unnecessary. You already had two equations and two unknowns and this "over-specified" the problem and did so, incorrectly.

Do you notice that your introduced equation for \$I_\text{B}\$ doesn't include the term, \$I_\text{CO}\$, at all? The problem is in the fact the equation you introduced for \$I_\text{B}\$ makes the assumption that \$I_\text{E}=\left(\beta+1\right)\:I_\text{B}\$. However, the reality is:

$$\begin{align*} I_\text{E}&=I_\text{B}+I_\text{C}\\\\ &=I_\text{B}+\beta\: I_\text{B}+\left(\beta+1\right)\:I_\text{CO}\\\\ &=\left(\beta+1\right)\:I_\text{B}+\left(\beta+1\right)\:I_\text{CO}\\\\ &=\left(\beta+1\right)\left(I_\text{B}+I_\text{CO}\right) \end{align*}$$

That's where you made your mistake.

So, you made two errors:

  1. You didn't just do a simultaneous solution of the two equations, when you should have done so.
  2. You introduced a new KVL equation that was incorrect, because it did not account for \$I_\text{CO}\$. (It would be correct if \$I_\text{CO}=0\$.)

The remedy is to either solve the two equations simultaneously, as you probably should have done; or else you should apply the corrected equation for \$I_\text{E}\$ when solving your added KVL equation:

$$\begin{align*} V_\text{BB} - I_\text{B}\: R_\text{BB} - V_{\text{BE}_\text{ON}} - I_\text{E}\:R_\text{E}&=0\:\text{V}\\\\ V_\text{BB}&= I_\text{B}\: R_\text{BB} + V_{\text{BE}_\text{ON}} \left(\beta+1\right)\left(I_\text{B}+I_\text{CO}\right)\:R_\text{E}\\\\ V_\text{BB}&= I_\text{B}\: R_\text{BB} + V_{\text{BE}_\text{ON}} + \left(\beta+1\right)\:I_\text{B}\:R_\text{E}+\left(\beta+1\right)\:I_\text{CO}\:R_\text{E}\\\\ I_\text{B}\: \left(R_\text{BB} + \left(\beta+1\right)\:R_\text{E}\right)&=V_\text{BB}- V_{\text{BE}_\text{ON}}-\left(\beta+1\right)\:I_\text{CO}\:R_\text{E} \\\\ I_\text{B}&=\frac{V_\text{BB}- V_{\text{BE}_\text{ON}}-\left(\beta+1\right)\:I_\text{CO}\:R_\text{E}}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}} \end{align*}$$

Do you note the difference, now? You cobbled up an equation for \$I_\text{B}\$ out of thin air, without taking into account the newly introduced term, \$I_\text{CO}\$. That works fine if you discount \$I_\text{CO}\approx 0\$. But if you have to include it, then you need to include it. And you didn't.

If you use this corrected value for \$I_\text{B}\$, perhaps things will work out okay.

$$\begin{align*} I_\text{C}&=\beta\: I_\text{B}+\left(\beta+1\right)\:I_\text{CO}\\\\ &=\beta\: \frac{V_\text{BB}- V_{\text{BE}_\text{ON}}-\left(\beta+1\right)\:I_\text{CO}\:R_\text{E}}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}}+\left(\beta+1\right)\:I_\text{CO}\\\\ &=\frac{\beta \left(V_\text{BB}- V_{\text{BE}_\text{ON}}-\left(\beta+1\right)\:I_\text{CO}\:R_\text{E}\right)+\left(\beta+1\right)\:I_\text{CO}\left(R_\text{BB} + \left(\beta+1\right)\:R_\text{E}\right)}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}}\\\\ &=\beta\frac{V_\text{BB}-V_{\text{BE}_\text{ON}}}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}}+\left(\beta+1\right)\frac{\:R_\text{BB}+\left(\beta+1\right)\:R_\text{E}-\beta\:R_\text{E}}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}}\:I_\text{CO}\\\\ &=\beta\frac{V_\text{BB}-V_{\text{BE}_\text{ON}}}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}}+\left(\beta+1\right)\frac{\:R_\text{BB}+R_\text{E}}{R_\text{BB} + \left(\beta+1\right)\:R_\text{E}}\:I_\text{CO} \end{align*}$$

That's it. You have the equation, done correctly.

Of course, you could also have just solved it as a pair of simultaneous equations in the first place.

| improve this answer | |

To simplify things and since you're only trying to check the second term put Vbb and Vbe equal to zero and find Ib in terms of Ic from your second equation. This is the third equation. Then replace Ib from the first equation with the third equation and solve for Ic. You get exactly the second term.

You are only going to verify the second term so you can put Vbb=Vbe=0. Then solve the second equation for Ib,

$$I_B=\frac{-R_E}{R_{BB}+R_E}I_C$$ Now if you put the above equation into your equation for Ic you get the 2nd term in Ic,

$$I_C=\beta I_B+(\beta+1)I_{co}$$ $$I_C= \beta \frac{-R_E}{R_{BB}+R_E}I_C+(\beta+1)I_{co}$$ $$(1+\frac{\beta R_E}{R_{BB}+R_E})I_C=(\beta+1)I_{co}$$ $$I_C=(\beta+1)\frac{R_{BB}+R_E}{R_{BB}+R_E+\beta R_E}I_{co}$$ $$I_C=(\beta+1)\frac{R_{BB}+R_E}{R_{BB}+R_E(\beta +1)}I_{co}$$

This only gives you the equation for Ic in terms of Ico, just to make sure that the second term is alright.

| improve this answer | |
  • \$\begingroup\$ If I set them zero Ib becomes zero. I dont get what you mean. Vbe cannot be zero in active mode as well. Do you think the original eq. is wrong? \$\endgroup\$ – pnatk Jun 13 '18 at 17:26
  • \$\begingroup\$ So the original is wrong? Please say your conclusion \$\endgroup\$ – pnatk Jun 13 '18 at 19:11
  • \$\begingroup\$ No, it's alright. The equation you've written for Ib is wrong. See if you can spot your mistake. \$\endgroup\$ – dirac16 Jun 13 '18 at 19:18

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