6.3 In a particular technology, a small BJT operating at conducts a collector current of 100 μA. What vBE = 28VT is the correspo
nding saturation current? For a transistor in the same technology but with an emitter junction that is 32 times larger, what is the saturation current? What current will this transistor conduct at base–emitter voltage of the latter transistor at iC AE1 =400 μm × 400 μm and AE2 vBE vBE = 28VT ? What is the =1mA? Assume active-mode operation in all cases.
1 answer:
Answer:
Saturation current
Isi = 6.91 * 10^-17 A
The saturation current Is2 of transistor Q2 is 2.2 * 10-15 A
Collector current Ic2 at Vbe2 is 3.2mA
Base emitter voltage Vbe2 is 26.84 VT
Explanation:
Complete question is as follows;
Problem 6.3: In a particular technology, a small BJT operating at vBE = 28VT conducts a collector current of 100
µA. What is the corresponding saturation current?
For a transistor in the same technology but with an emitter junction that is 32 times larger, what is the saturation
current?
What current will this (second) transistor conduct at vBE = 28VT ?
What is the base-emitter voltage of the latter transistor at iC = 1 mA? Assume active-mode operation in all cases.
solution;
Please check attachment for complete solution and step by step explanation
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Answer:
r=0.31
Ф=18.03°
Explanation:
Given that
Diameter of bar before cutting = 75 mm
Diameter of bar after cutting = 73 mm
Mean diameter of bar d= (75+73)/2=74 mm
Mean length of uncut chip = πd
Mean length of uncut chip = π x 74 =232.45 mm
So cutting ratio r
r=0.31
So the cutting ratio is 0.31.
As we know that shear angle given as
Now by putting the values
\
Ф=18.03°
So the shear angle is 18.03°.