Whoever is the Anonymous person, please do not answer any of my questions anymor
ID: 2085032 • Letter: W
Question
Whoever is the Anonymous person, please do not answer any of my questions anymore. If you don't have access to the textbook, you are useless to me.
This is a follow up on a previous question/answer: my new question is at the bottom.
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For this problem:
Sedra-Smith Microelectronic Circuits 7th-edition-chapter-9-problem-61P
Step 3: Ad = alpha*Rc / 2*re where alpha = 1.
why isn't is 2*Rc / 2*re? where did the other Rc go?
In Sedra-Smith 7th ed. Chap 9 p. 623-624 clearly shows that it should be 2*Rc in the numerator.
Thank you
Expert Answer
The differential pair is biased with a constant-current source I and the current I can be steered to either Q1 or Q2 under the control of the input signal
when Vb1 is greater than Vb2 by about 4VT (aprrox 100 mV) all the current I is conducted by Q1,thus for
Alpha =1 and Vb1 = VCC – IRC.the current through Q2 will be nearly zero thus VC2=VCC
when Vb1 is less than Vb2 by about 4VT (aprrox 100 mV) all the current I is conducted by Q2,thus for
Alpha =1 and Vb2 = VCC – IRC.the current through Q1 will be nearly zero thus VC2=VCC
therefore when Q1 is ON Q2 is OFF or when Q2 is ON Q1 is OFF so we will get the output VC1 or Vc2across either Rc1 or Rc2
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New question:
I understand. Now please correct me if I'm wrong, but in the case of Prob 61, we have |vb1-vb2| = vd = I*2*re = 0.5mA*2*50 Ohm = 50 mV < 100 mV, so should we consider both Rcs?
Explanation / Answer
Yes you have to consider both. In this case current I is equally divided into both Rcs.
Even though I am anonynus I understood your question so answering this.
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