BJT single stage amplifier we write gm=Ic/VT

[pforpashya]

For BJT single stage amplifier we write gm=Ic/VT where VT is Thermal voltage whose value is 26mv. Now if we have a Darlington pair i.e. two transistor connected in CC-CC mode what will be overall gm.

Will the collector current Ic be divided by 52mv or 26mv. (Because in gain equation gm comes)

as their are two forward active PN junctions ...the cumulative change in VT is 52mV?

is this correct?

can someone through light on "why we have to include Boltzmann constant in calculation of VT"
in simple words?

Thanks in advance

 

[cabraham]

Each device has g_m value equal to its own I_c divided by V_T, where V_T is 25.7 mV for each device. For a Darlington, if the input device is no. 1, and output device is no. 2, then I_e1 = I_b2, I_c1 = α₁*I_e1, I_c2 = β₂*I_b2, or I_c2 = α₂*I_e2, .

This should get you started.

Claude

 

[pforpashya]

ok. thank you for your reply. to find overall gm (i.e. gm2 as gm1 of first pair in darlington is very small) of the device should i divide Ic2 by VT1+VT2 i.e. 52mV?

 

[Jony130]

Yes you should use 2*Vt if you want to find voltage gain.

The small signal gain for CE amplifier is equal to

\Large \frac{Vout}{Vin} =-\frac{Rc (\beta 1+\beta 2+\beta 1 \beta 2)}{(1+\beta 1) (re1+re2+re2 \beta 2)}

Where

re1 ≈ 1/gm1
re2 ≈ 1/gm2

 

[pforpashya]

Thanks alot...I have to solve one unsolved problem on darlington pair before saturday. I hope i will get right answer as i have to just multiply gm by output resistance of darlington pair.

 

[pforpashya]

can someone through light on "why we have to include Boltzmann constant in calculation of VT"
in simple words?