DC and low signal Analysis of Common Source Amplifier: wrong drain current and amplification?

Question

For a Coursera test I need to analyze a common source amplifier:

I need to calculate the

• DC gate voltage - easy: -2.3077
• drain current Id - should be easy but the results are wrong. I use this formula from the course or from here: $$ I_d=(\frac{\sqrt{1+4*K*V_1*R_{SS}}-1}{2*\sqrt{K}*R_{SS}})^2 \\ V_1=V_G-V_{TO}-V_{SS} $$
• the drain voltage Vd=Vdd-Id*Rd which gives me -3.7719 V with the "wrong" current ??
• the amplification which gives me -24.8174 (which is wrong!) $$ \frac {V_{out}}{V_{in}}=-\frac {R_D ||R_L}{r_s+R_S||R_3}\\ r_s=\frac{1}{g_m}\\ g_m=2*\sqrt{K*I_D} $$

OK.

Here are my calculations from Matlab:

Vdd=15; Vss=-15;K=1.1*1e-3;Vto=2;R1=150000;R2=110000;Rd=10000;Rs=5000;
R3=200; Rl=10000;

Vg=(Vdd*R2+Vss*R1)/(R1+R2)
Vg = -2.3077

V1=Vg-Vss-Vto
Id=((sqrt(1+4*K*V1*Rs)-1)/(2*Rs*sqrt(K)))^2
Id=0.0019

Vd=Vdd-Id*Rd
Vd = -3.7719

gm=2*sqrt(K+Id)
rs=1/gm
gac=-R_parallel(Rd, Rl)/(rs+R_parallel(Rs, R3))
gac = -24.8174

No I wonder am I too dumb to be true or is there a flaw in the automatic grader?

In case I am too dumb, please, tell, where my error is. Thanks.

Answer 1

If I use your equations

$$ I_d=(\frac{\sqrt{1+4*K*V_1*R_{SS}}-1}{2*\sqrt{K}*R_{SS}})^2 \\ V_1= (V_G -V_{SS})-V_{TO} $$

And if I substitute the resistors in kiloohms and all the voltage in volts and all current in milliamps I get this:

$$ I_d=( \frac{\sqrt{1+4*1.1*((-2.3077) - (-15) - 2)*5}-1}{2*\sqrt{1.1}*5} )^2 = 1.87719 $$

http://www.wolframalpha.com/input/?i=((Sqrt%5B1+%2B+4*1.1*((-2.3077)+-+(-15)+-+2)*5%5D+-+1)%2F(2+Sqrt%5B1.1%5D*5))%5E2

And this result is also in milliamps.

\$I_D = 1.87719\textrm{mA}\$

The drain voltage is \$V_D = -3.77V \$

and \$g_m =2*\sqrt{K*I_D} = 2.87396\:\textrm{mS}\$

\$r_s = \frac{1}{g_m} = 0.347952 \textrm{k}\Omega = 347.952 \Omega \$

And the voltage gain is

$$ \frac {V_{out}}{V_{in}}=-\frac {R_D ||R_L}{r_s+R_S||R_3} = \frac{5\textrm{k}\Omega}{0.347952\textrm{k}\Omega + \frac{1}{\frac{1}{0.2\textrm{k}\Omega}+\frac{1}{5\textrm{k}\Omega}}}=-9.25481 V/V $$

Do you have any additional questions? And next time you should track your units with more care.