# BJT Biasing - Simple easy 1 minute answer!

Discussion in 'Homework Help' started by aly34, Feb 3, 2012.

1. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
1. The problem statement, all variables and given/known data
Find IB, IC, IE, VCE, VBC, and VEB for the NPN transistor in the circuit below for β = 50, 150, 500. Let VCC = 10 V, R1 = 100 KΩ, R2 = 400 KΩ, RC=2 KΩ, and RE = 0 Ω. Assume VBE = 0.7V.

2. Relevant equations

V = iR
IE = Ib + Ic

3. The attempt at a solution

My question is that if RE = 0 ohms, then is Ie automatically 0 amps?

But the how does Ie = Ib + Ic apply? Because I got 92.25uA for Ib and 4.56mA for Ic?

Help! :]

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2. ### jimkeith Active Member

Oct 26, 2011
539
99
My question is that if RE = 0 ohms, then is Ie automatically 0 amps?
No--If Re = 0Ω, then Vre = 0V

But the how does Ie = Ib + Ic apply? Because I got 92.25uA for Ib and 4.56mA for Ic?
In this case, it does not apply because the 'resistance' has already been determined. However, it is not likely to work unless Re is adjusted correctly to bias the collector voltage midway.

Also, regarding thermal stability, this base divider /emitter swamping resistor bias topology goes back to the germanium transistor days. In low power applications, decent stability may be obtained without Re (Re = 0), provided that the top of R1 is tied to the collector and has its resistance adjusted correctly. However, in this case, it provides a negative feedback path that tends to decrease gain--and decrease distortion. I'll let you figure out the math.

There are many, many nuances to this stuff...

3. ### hgmjr Moderator

Jan 28, 2005
9,030
214
HINT:

Find the Thevenin's Equivalent voltage and resistance of the bias network R1 and R2. Then you can calculate Ib using Vbe.

hgmjr

4. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
I think I am even more confused than I started?

So if Ie is not 0 then I can proceed with using Ie = Ib + Ic?

Do not really understand how that works?! :[

5. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
I did that already...but I am not sure what to do with Re = 0 ohms.

I calculated Ib to be 91.25uA

And Vb to be 8V

And Ic to be 4.56mA

After the thevenins equalivant..I tried doing a KVL loop...and just get (Ie)*(Re) and Re = 0?

Ahhh!

6. ### jimkeith Active Member

Oct 26, 2011
539
99
Let Re = 0Ω because that is given
Calculate the required currents and voltages as if the circuit really works (even though it probably cannot)--it is essentially a hypothetical situation
Note that Vce will be all over the place
hint: Ib will be a constant

7. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
Yes - We do not have to construct this circuit or anything...

He just wants us to understand some of the concepts...

Is my Ib and Ic calculations correct?

Based on that Ie = Ib + Ic would be correct?

Thank you for your help so far! <3

(I am really really bad at circuits!) Lol!

8. ### jimkeith Active Member

Oct 26, 2011
539
99
Is my Ib and Ic calculations correct?

No, it is incorrect
Calculate the thevenin resistance & voltage of R1 & R2 with the base open.
Then calculate the current flowing into the base assuming the provided Vbe.
hint: it has to be less than the current through R2.

There is yet another way to do it--determine the voltages across R1 & R2--very simple. Calculate the resulting currents E = I/R. Base current = Ir2 - Ir1

9. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
I did this...

From calculating the base voltage i get:
Vb = 10V * (r1 / (r1 + r2) = 8 V.

From the thevenins circuit - R1 || R2 = 80kΩ

and then doing a kvl in this circuit i get:

Vb = (80k)(Ib) + Vbe + (0k)(Ie)

8V = (80k)(Ib) + 0.7 + 0

7.3V = 80k*Ib

So, Ib = 91.25uA! (which is less than the current flowing through R2)

And then from here I did:

Ic = Ib*beta = (91.25uA) * (50) = 4.56mA!

:[

I don't know what i'm doing wrong!

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Be more careful with your calculations....

Vb=10*(R1/(R1+R2))=10*(100k/500k)=2V.

aly34 likes this.
11. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
Omg.

I feel so dumb!

Vb = 2V
Ib = 1.625 x 10^-5 Amps = 16.25 uA!
Ic = 8.125 x 10-4 Amps = 0.8125 mA!

So, Ie = 16.25uA + 0.8125mA = 0.82875mA!

And then calculating Vce:

10v = (rc)(ic) + vce + (re)(ie)

10v = (2000)(8.125 x 10^4) + vce + 0

so, vce = 8.375

And then vcb:

vce = vbe + vcb

8.375 = 0.7 + vcb

so, vcb = 7.675v

but he asked for vbc and veb...

is that just the negative of these values?

?????????

Last edited: Feb 3, 2012
12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Your results above looks good to me.

What current did you calculate in R2?

hgmjr

aly34 likes this.
13. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
These were incorrect?

I had Vb = 8V - which is wrong!

Can you check my latest post please?

Last edited: Feb 4, 2012
14. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Actually you computed Vth equal to 8V not Vb = 8V. You computed Vth accurately. Then you used the fact that Vb = 0.7V to calculate the total voltage drop across Rth to be 7.3V. Then you calculated Ib = 91.25 microamps. All of these calculations were correct.

hgmjr

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
So much for the 1minute quick answer......

It might be worth referring back to the schematic attachment in post #1.

If I'm not mistaken R1 is from the base to ground and R2 is from Vcc to base.

This would make

Vth=Vcc*(R1/(R1+R2))

and

Rth=R1||R2=80kΩ

With R1=100k, R2=400k and Vcc=10V this would make Vth=2V.

So IB=(Vth-0.7)/Rth=1.3/80k=16.25uA.

aly34 likes this.
16. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0

Yess - Thanks for yalls help though!

So, 16.25uA for iB is correct...

And from there i can get:

ic = (16.25uA) * (beta = 50) = 0.8125mA

ANd then Ie = Ib + ic = 0.8287mA

And then vce = 8.375V.

:]