Electronic Devices - DC Biasing-BJTs - Discussion
Discussion Forum : DC Biasing-BJTs - General Questions (Q.No. 23)
23.
Determine the values of VCB and IB for this circuit.
1.4 V, 59.7 
A
A–1.4 V, 59.7 A
A–9.3 V, 3.58 A
A9.3 V, 3.58 A
ADiscussion:
3 comments Page 1 of 1.
Severina said:
2 years ago
1. By KVL, we can find the value of Ie which is = Ic.
-5 + 0.7 + 1.2k ie = 0.
ie = 3.58 mA = ic.
2. we know that ic = Bib. so,
ib = ic/60.
ib = 59.67 micro A =>answer.
3. by KVL on the right side, we can solve for the value of Vcb.
-10 + 2.4k(Ic) + Vcb = 0.
Vcb = 1.408 v => answer.
-5 + 0.7 + 1.2k ie = 0.
ie = 3.58 mA = ic.
2. we know that ic = Bib. so,
ib = ic/60.
ib = 59.67 micro A =>answer.
3. by KVL on the right side, we can solve for the value of Vcb.
-10 + 2.4k(Ic) + Vcb = 0.
Vcb = 1.408 v => answer.
(1)
Weiwe said:
7 years ago
IE = (VEE-VBE)/RE.
IE = IC,
VCB = VCC - ICRC,
IE = (5V-0.7V)/1,200) = 3.58mA.
VCB = 10 - 3.58mA * 2,400 = " 1.4 V ".
IB = IC/Beta = 3.58mA/60 = "59.7 uA".
IE = IC,
VCB = VCC - ICRC,
IE = (5V-0.7V)/1,200) = 3.58mA.
VCB = 10 - 3.58mA * 2,400 = " 1.4 V ".
IB = IC/Beta = 3.58mA/60 = "59.7 uA".
(1)
MRINAL said:
1 decade ago
First we calculate IE:
IE = (VEE-VBE)/RE = (5-0.7) V/1.2K = 3.58 mA.
Now,
IB = IE/(BETA+1).
= 3.58 mA/(60+1) = 58.68 microampere.
IC = BETA*IB = 60*58.68 microampere = 3.52 mA.
VCB = 10-IC*RC = 10- 2.4K* 3.52 mA = 1.55V.
So value near to this answer is option (A).
IE = (VEE-VBE)/RE = (5-0.7) V/1.2K = 3.58 mA.
Now,
IB = IE/(BETA+1).
= 3.58 mA/(60+1) = 58.68 microampere.
IC = BETA*IB = 60*58.68 microampere = 3.52 mA.
VCB = 10-IC*RC = 10- 2.4K* 3.52 mA = 1.55V.
So value near to this answer is option (A).
(1)
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