# DC analysis on this NPN BJT

Discussion in 'Homework Help' started by u-will-neva-no, May 14, 2011.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
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Hello everyone. Would it be possible to check my analysis for my attached circuit?

Here is my analysis:

The question states that B is 200 and that Vbe = 0.7v.

From the graph, vb = 0v, because it is grounded. Therefore:
Vbe = Vb - Ve
0.7 = 0-Ve
Ve = -0.7V

Now to find IE. using ohms law I have:
Ve - (-10v) = IE*RE
(-0.7)+10 = IE *10K
IE = 0.93mA

finding IC by using the formula IC = αIE
α=β/(β+1)IE
α = 200/201* 0.93mA
IC = 0.925mA

Finding Vc using omhs law again
Vcc -Vc = ICRC
10-Vc = 0.925mA * 10k
Vc = 0.75V

I would just like to know if my method is correct, because my teacher has completely different values! Thanks.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You forgot about RB resistor in you analysis and the base current.
VEE - Vbe = Ie*Re + Ib *Rb
And for me the BJT will be in saturation

EDIT
I do simply calculation
And I got
Ie = 886uA; Ic = 881.6uA;
And the BJT for sure is in active mode

Last edited: May 14, 2011
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3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
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I am having difficult trying to visualise how you obtained that equation. I accept that your correct though!

4. ### jegues Well-Known Member

Sep 13, 2010
735
43
You are confusing the where the voltage at the base is.

I've attached a figure to help clear up your confusion.

You can write a KVL in the BEJ loop and solve for Ic, Ie or Ib, whichever you prefer. (Exactly like what Jony130 has demonstrated for you)

If you find that they are nonzero then certainly VB is nonzero as well.

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5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I wrote KVL for this loop

• ###### 4.PNG
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6. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
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Ok I think I understand it now. I am going to try to explain what I think is going on, so that I can apply it to different examples in the future.

Just to explain some of my thoughts to different problems, say I wanted to calculate the current flowing through the collector, i would do something like Vcc - Vc = ic* Rc and rearrange. If I wanted to find the current ie, i would do something like Ve - VEE = ie*Re.

from these two approaches, i would do VEE - Vbe = Ib + Ie (1). Am i correct to say that there is a pattern, such that you must go in a clockwise direction for the Vcc - Vc,
Ve - VEE and VEE - Vbe part?

Now, im picturing that we are interested in the current Ib and current Ie, hence why we get the Ib + Ie part in equation (1)? Let me know if this is the correct way to visualise it

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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For the BJT we always need to find Ib or Ie first.
Then if we know Ib we can find Ic form
Ic = β*Ib and Ie = Ib + Ic

You equation are wrong
How can current be equal to voltage (mismatch in units) ??

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8. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Woops, I messed up...I meant to have IeRe +IbRb!

9. ### jegues Well-Known Member

Sep 13, 2010
735
43
I don't think this is true.

You can solve it in whatever order/manner you choose (I could solve for Ic first if I choose to do so) as long as your results are correct.

After doing so, you must check to see that any assumptions you've made during the initial analysis are valid. (i.e. assuming active mode)

10. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
@Jony130, using the formula VEE - Vbe = IeRe IbRb, and also, Ib = Ie/(B+1), i get:

VEE-Vbe = IeRe + (Ie/(B+1))*Rb

-10 -0.7 = Ie (Re + Rb/(B+1))

Ie = -1.02mA....Im clearly doing something wrong...

11. ### jegues Well-Known Member

Sep 13, 2010
735
43
When I write a KVL I find that,

$I_{b}R_{b} + V_{be} + I_{e}R_{e} + V_{ee} = 0$

$\Rightarrow I_{e} = \frac{-V_{ee} - V_{be}}{\frac{R_{b}}{\beta + 1} + R_{e}}$

But it is not clear how the Vee source is hooked up, so it could change by a negative sign.

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12. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
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jegues, your answer is correct. My problem is that the formula is different to Jony130's which is what I understood. Could you explain me how you obtained your KVL?

Jony130's formula would work if i took my Vee value (-10v) as just 10v, but i dont think i should be doing that...

13. ### jegues Well-Known Member

Sep 13, 2010
735
43
It was a simple application of KVL.

Why don't you make an attempt of drawing the voltages and their respective polarities around the BEJ loop.

If there is any problems, someone will correct you.

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14. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Oki doki, my picture is attached. Its practically exactly the same as Jony130's representation (Just a lot messier). Anyway, from my diagram, I have VEE in a clockwise direction whereas all the other voltages are in an anticlockwise direction. Therefore, i am not getting the same formula as Jegues..

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15. ### jegues Well-Known Member

Sep 13, 2010
735
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Here is my drawing of my KVL.

I cannot make it any clearer than this. This is nothing but a basic application of KVL.

• ###### VBKVL.JPG
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16. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
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Thank you to everyone for your help, much appreciated!

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Vee = -VRB + Vbe - Vbe

But you have VEE in a clockwise directions so Vee = 10V
You are measure the voltage across Vee withe respect to minus terminal of a Vee and Vbe = - 0.7V

Ie = Vee + Vbe/ ( Re + Rb/(β+1) = (10V + (-0.7V))/ (10K + 497.5Ω) = 9.3V/10.497K = 886uA

BJT is a current control current source, so to find Ic you need to know Ib or Ie first.

18. ### Jony130 AAC Fanatic!

Feb 17, 2009
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And let me remind you what negative voltage is in real life

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19. ### jegues Well-Known Member

Sep 13, 2010
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Yes I agree it acts as a current controlled current source, but as far as the analysis is concerned I could solve for Ic first if I choose to do so.

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In principle the same equation can be translated to immediately solve for any of the three interdependent values Ie, Ic or Ib. The original equation itself explicitly accounts for the functional controlled source.

So, per Jony130's notation

$I_e=\frac{V_{ee}+V_{be}}{R_e+\frac{R_b}{1+\beta}}$

$I_b=\frac{V_{ee}+V_{be}}{(1+\beta)R_e+R_b}$

$I_c=\beta \frac{V_{ee}+V_{be}}{(1+\beta)R_e+R_b}$

Certainly in that sense one may immediately solve for any of those values - as desired. Obviously, one must firstly form the correct algebraic equation for the solution to be valid.