0

Trying to understand what is the maximum and minimum for the input signal of a differential pair by solving this example below.

Vi max & min

My solution: To let the circuit operate as a differential pair, it must be in an active region where $$ VCE> VCE(sat)$$

by considering $$ VCE(sat)=0.5V$$ and $$ IC_1=IC_2=1mA$$, and solving in DC bias $$VC_1=20-(15k*(1mA))=5V$$ $$VB_1=\frac{20*5k}{15k+5k}=5V$$ So, $$VE_1=VB_1- VBE_1=5V-0.7V=4.3V$$. As $$VCE_1 (0.7V)> VCE(sat) 0.5V$$ and it satisfies the condition, I will consider what would jeopardize this equation! It will not be in the active region if $$ Vo=VE_1+VCE_1(sat)=4.3V+0.5V <4.8V $$ By considering $$ Vo=Ad_1 *Vd+Acm_1*V_cm$$ Then, $$Vo=\frac{R_{c1}}{2r_e} * vi +\frac{R_{c1}}{r_{e1}} * \frac{vi+0}{2} $$
By placing the values $$4.8V=\frac{15Kohm}{2*25 ohm} * vi +\frac{15k Ohm}{25 Ohm} * \frac{vi+0}{2} $$ I expect to get the maximum of vi=8mV.

I know that there must be a minimum value that stops the differential pair Q1 from acting as a switch and being turned off, but not sure how to approach that.

"Solution" The correct solution to Vi(max) is as follows but I cannot understand and am not sure why my solution is wrong:

Solution

chami
  • 358
  • 2
  • 11
  • This looks like a solution searching for a problem. Is there a compelling reason for this question when most diff amps are operated closed-loop or, are operated as comparators? Or, put differently, what is the potential application for an open-loop diff amp as per the above? – Andy aka Sep 02 '22 at 17:14
  • @Andy aka Not sure about how to answer your inquiry. My goal is to understand what limits vi and this example is one that I could find. I only could find a few examples, but they only mentioned vi must be less than 4Vt to not let it operate as a switch – chami Sep 02 '22 at 17:24
  • Are you sure Vce (sat) is 0.5V and not 0.2V ? If it's the latter (the answer implies it) then vo_max is 0.5V and thus vi_max is 1.667mV – Rahmany Sep 02 '22 at 17:43
  • @Rahmany Thank you. The problem says VCE(sat) is 0.5 V, and I cannot understand Vo_max=0.5 V. I am getting vo_max as 4.8 V (ve1+vce(sat)). Also, common mode gain has not been considered in the equation, why not? – chami Sep 02 '22 at 17:52
  • 1
    Common mode is negligible because RE (not re) is infinite ; ideal current source output impedance. – Rahmany Sep 02 '22 at 18:09
  • @ Rahmany Understood. Thank you. – chami Sep 02 '22 at 18:12

0 Answers0