Electronic Devices - DC Biasing-BJTs - Discussion
Discussion Forum : DC Biasing-BJTs - General Questions (Q.No. 6)
6.
Calculate the voltage across the 91 k
resistor.
resistor. 
Discussion:
12 comments Page 1 of 2.
Swati singh said:
1 decade ago
For D.C. analysis all capacitors will be open circuit then equations will be,
18 = 3.3(b+1)I1+(91+110)I1+0.7+0.510I1.
I1 = 0.0352mA.
Hence voltage across 91 kohm resistor will be,
91 kohm*0.0352mA = 3.2V.
18 = 3.3(b+1)I1+(91+110)I1+0.7+0.510I1.
I1 = 0.0352mA.
Hence voltage across 91 kohm resistor will be,
91 kohm*0.0352mA = 3.2V.
Kaushal wagh said:
1 decade ago
3.3(b+1)Ib+ 201 Ib + 0.7 +0.510(b+1)Ib = 18.
3.3*76 Ib + 201 Ib + 0.7 +0.510 * 76 Ib = 18.
250.8 Ib + 201Ib+38.76 Ib = 17.3.
Ib= 0.035 mA.
So voltage = 0.035 * 91.
=3.20 v.
3.3*76 Ib + 201 Ib + 0.7 +0.510 * 76 Ib = 18.
250.8 Ib + 201Ib+38.76 Ib = 17.3.
Ib= 0.035 mA.
So voltage = 0.035 * 91.
=3.20 v.
(1)
Vinita said:
1 decade ago
@Kaushal singh.
In your 1st equation its .510*b*Ib not 0.510(b+1)Ib.
As it is Ic=b*ib.
In your 1st equation its .510*b*Ib not 0.510(b+1)Ib.
As it is Ic=b*ib.
Musharique said:
1 decade ago
Why you use (b+1)?
Raymond said:
1 decade ago
Ic = B Ib.
Ie = (b+1)Ib.
Ie = (b+1)Ib.
Sujith said:
10 years ago
Please any one can elaborate the answer because I just start to study this.
Poojadeep gaikwad said:
9 years ago
Ie = Ic + Ib.
Ie = Ib*b + Ib (where b=beta).
Ie = Ib (1 + b).
= 3.3Ie + 201 Ib + 0.7 + 0.510 Ie = 18.
By solving this equation,
Ib = 0.035.
So v/g across 91k resistor will be equal to 0.035 * 91 k = 3.2 v.
Ie = Ib*b + Ib (where b=beta).
Ie = Ib (1 + b).
= 3.3Ie + 201 Ib + 0.7 + 0.510 Ie = 18.
By solving this equation,
Ib = 0.035.
So v/g across 91k resistor will be equal to 0.035 * 91 k = 3.2 v.
(3)
Poojadeep gaikwad said:
9 years ago
Ie = Ic + Ib.
Ie = Ib*b + Ib (where b=beta).
Ie = Ib (1 + b).
= 3.3Ie + 201 Ib + 0.7 + 0.510 Ie = 18.
By solving this equation,
Ib = 0.035.
So v/g across 91k resistor will be equal to 0.035 * 91 k = 3.2 v.
Ie = Ib*b + Ib (where b=beta).
Ie = Ib (1 + b).
= 3.3Ie + 201 Ib + 0.7 + 0.510 Ie = 18.
By solving this equation,
Ib = 0.035.
So v/g across 91k resistor will be equal to 0.035 * 91 k = 3.2 v.
Uapzner said:
9 years ago
Hi, I just want to clarify if which loop did we use in this solution?
(2)
Petrus said:
7 years ago
Ib = Vcc-Vbe/Rb +(bet+1)Re.
Ib =18-.7 / 201000+(76) * 510) =17.3/239760 = 72.155 micro Ohm.
72.155 micro ohm x 91000 = 6.566V.
Ib =18-.7 / 201000+(76) * 510) =17.3/239760 = 72.155 micro Ohm.
72.155 micro ohm x 91000 = 6.566V.
(3)
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