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I am trying to get to the simplified form of the equation of the differential output voltage of a differential pair:

Differential pair

I know the correct equation is:

Correct equations

*Image above from Franco, S. Analog Circuit Design. Please let me know if it is not proper to show this excerpt here.

I am stuck with simple algebra, really. I want to get the equation 4.73. Here is what I did:

Notation: In the book, VoD. In my text, Vodif In the book, ViD. ViD is the complete differential input voltage, vid =vi1 - vi2. In my text, I named it Vdif.

Also I've just considered αf = 1, so it is not explicit in my equations. Considering Ree too large and just ignored.

Derivation

I am stuck at this point for a while now. This is not homework or alike. Just reviewing some basic concepts.

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  • \$\begingroup\$ Hello jonk. I am sorry I did not make it explicit, I'll fix it. No, vid =vi1 - vi2 in the first circuit. How can you format the vid text as you did in your post? Anyway, vid is the complete differential input voltage, and is also vbe1 - vbe2. Summing Ic1 and Ic2 gets to the expected Iee. I am trying to get to the final differential output voltage formula: Vod = Vo1 - Vo2 = alpha * Iee * Rc + tanh(-Vid / 2*VT). I'll also make this clearer at the question. \$\endgroup\$
    – tfm
    Apr 5 '20 at 5:19
  • \$\begingroup\$ For the math usage here, see here. But keep in mind that on EESE you bracket your equations with $$ or with \$. \$\endgroup\$
    – jonk
    Apr 5 '20 at 5:27
  • \$\begingroup\$ Also, are you using resistor \$R_\text{EE}\$? Or current sink \$I_\text{EE}\$? Or both? \$\endgroup\$
    – jonk
    Apr 5 '20 at 5:36
  • \$\begingroup\$ Considering Ree too large and just ignored. \$\endgroup\$
    – tfm
    Apr 5 '20 at 5:49
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You can follow these steps: $$i_{C1} = \frac{I_{EE}}{1+exp(\frac{-v_{id}}{Vt})}$$ $$i_{C2} = \frac{I_{EE}}{1+exp(\frac{v_{id}}{Vt})} = \frac{I_{EE}exp(\frac{-v_{id}}{Vt})}{1+exp(\frac{-v_{id}}{Vt})}$$ $$v_{od} = R_c(i_{C2} - i_{C1}) = \frac{I_{EE}R_c(exp(\frac{-v_{id}}{Vt})-1)}{1+exp(-\frac{v_{id}}{Vt})} = I_{EE}R_ctanh(\frac{-v_{id}}{2Vt})$$

EDIT
There is nothing wrong with your steps, just proceed as follows:

$$v_{od} = IeeR_c\frac{exp(-v_{id}/Vt) - exp(v_{id}/Vt)}{2+exp(-v_{id}/Vt) + exp(v_{id}/Vt)}$$ Multiply numerator and dinominator by \$exp(-v_{id}/Vt)\$: $$v_{od} = IeeR_c\frac{exp(-2v_{id}/Vt) - 1}{2exp(-v_{id}/Vt) +exp(-2v_{id}/Vt) + 1}$$

$$v_{od} = IeeR_c\frac{(exp(-v_{id}/Vt) - 1)(exp(-v_{id}/Vt) + 1)}{(exp(-v_{id}/Vt) + 1)^2}$$ $$v_{od} = IeeR_c\frac{exp(-v_{id}/Vt) - 1}{exp(-v_{id}/Vt) + 1}$$

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  • \$\begingroup\$ sarthak, thank you. Can you tell me what if/what is wrong in my previous steps? \$\endgroup\$
    – tfm
    Apr 5 '20 at 6:54
  • \$\begingroup\$ @HW_SW_Engineer see edits \$\endgroup\$
    – sarthak
    Apr 5 '20 at 7:29
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HW_SW_engineer - I think, the simplest way is to start with the currents:

In general, we have: Ic=Ico[(exp(Vbe/Vt)-1]

Ic1/Ic2=exp[Vbe1-Vbe2)/Vt]=exp[Vd/Vt]

with Iee=Ic1+Ic2 we arrive at

Ic1=Iee/[1+exp(-Vd/Vt)] and Ic2=Iee/[1+exp(+Vd/Vt)]

From mathematics: 2/[1+exp(-x)]=....[1+th(x/2)]

Ic1=0.5*Iee[1+th(Vd/2Vt)] and Ic2=0.5*Iee[1-th(Vd/2Vt)]

Ic1-Ic2=Iee*th(Vd/2Vt)

Vo1=Vdc - RcIc1 and Vo2=Vdc - RcIc2 (Vdc: DC quiescent voltage)

Vd=Vo1-Vo2=-IeeRc*th(Vd/2Vt)

Comment: I have assumed Iee=Ic1+Ic2 (for each transistor: coll=emitter current); hence the factor "alpha" does not appear in the equations)

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  • \$\begingroup\$ Thank you very much LvW. I've followed a lot of posts from you here and in diyaudio.com, I've learned a lot. Same user there as well, right? \$\endgroup\$
    – tfm
    Apr 5 '20 at 17:23
  • \$\begingroup\$ diaudio.com? Never heard about it.... \$\endgroup\$
    – LvW
    Apr 5 '20 at 17:28

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