- 20th February 2008, 16:08 #1

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## voltage divider bias bjt

Will voltage divider biased circuit give me the same ICq and VCEq as with Emitter biased circuit?

are ICq = Vcc/ (Rc+Re) and

VCEq= VCC ??

Thanks in advance :)[/img]

- 20th February 2008, 16:21 #2

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## bjt voltage divider

yes... but the difference would be in the stability of ICq i.e. variation in Vceq with change in Vi

- 20th February 2008, 16:21

- 20th February 2008, 20:34 #3

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## voltage divider bjt

Hi,

Vb = [ R2÷(R1+R2)] * Vcc

Ie = (Vb-0.6V) ÷ Re

Ie ≈ Ic = Icq

Vceq = Vcc - Icq * (Re + Rc)

Regards,

Laktronics

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- 21st February 2008, 03:33 #4

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## bjt q point

@Laktronics

How come Vceq = Vcc - Icq * (Re + Rc) with an emitter biased circuit yet when it comes to obtaining Q point Vceq = Vcc ?

And yet with voltage divider bias is it Vceq = Vcc - Icq * (Re + Rc) ?

i'm confused :(

Added after 5 minutes:

@ A.Anand Srinivasan

You and laktronics gave me different answer :)

Does this mean that if i have a voltage divider and an emmiter bias circuit with the same Vcc, R1, Rc and Re I will get the same Q point? Yet voltage divider is more stable since Icq is not too Beta dependent? I am right... Im still a bit lost at this

Thanks

- 21st February 2008, 03:33

- 21st February 2008, 16:46 #5

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## q-point bjt

if Vceq=Vcc then no current will be flowing through the transistor

emitter bias just gives a lil negative feedback to make the bias stable but voltage divider bias is more stable...

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- 22nd February 2008, 07:16 #6

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## q point bjt

I see... so Vceq=VCC only when I am looking for Q point... since Ic= 0

Thanks :)

- 22nd February 2008, 18:20 #7

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## bjt voltage divider bias

dude you are confusing things.... Vceq can be equal to Vcc only when Ic=0... at Q point you calculate Vceq by using the Qpoint current Icq and hence the equation Vceq=Vcc-Icq(Rc+Re)....

- 25th February 2008, 14:49 #8

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## voltage divider bias

@A.Anand Srinivasan

hey i'm really really sorry for my TOO elementary questions. I only have a book and this site (not enrolled;no teachers/classmates to ask help)for my usually careless mistake related questions.

I get it now, i got the previous graph and this one mixed up....

Thanks again

- 25th February 2008, 19:48 #9

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## voltage divider biasing

dont feel sorry for these things.... after all we're all here to help each other... all the best... carry on....

- 6th March 2008, 14:37 #10

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## bjt q-point

In voltage divider basing, the basing resistors are replaced by thevenin equavilent circuit.

So equavilent resister Rb= R1R2/R1+R2

and equavilent voltage source VB = Vcc*R2/R1+R2

Therefore from base side you can write

VB=IbRb + Vbe+IeRe that is same as written but the values of Rb is now different and equivalent voltage VB is different is voltage divider biasing.

This is so because we require the value of stability factor S as small as possible. This we can achive by reducing Base voltage supply and hence through R1 and R2 only.

For voltage divider biasing, ICq equation is given by Vcc= IcRc + Vce + IeRe therefore

Ic= Vcc-Vce-IeRe/Rc.

If you only consider Vcc/Rc+Re then it will give you saturation current and Vce= Vcc is cuttoff voltage only.

Added after 1 minutes:

Please read 5th line that is same as written in Emitter biasing method but the----