Having some issues calculating the gain in circuit below, it is apparently 1802, but I can't seem to arrive at this answer. If anyone could help explain how to get this answer I'd be very appreciative!

enter image description here

Edit: My Working

enter image description here

  • \$\begingroup\$ Show your work and others will help you spot the mistake. But currently your question is just going to be closed because no initial effort is shown. \$\endgroup\$ – Bence Kaulics May 30 '16 at 18:10
  • \$\begingroup\$ Sorry about that, my working is here, thank you for the advice \$\endgroup\$ – ohkneel May 30 '16 at 18:21

The gain A of a single transistor amplifier stage is given by $$ A = -g_m r_{out} $$ For two stages in cascade it's the product of the gains.

The gm can be found using \$gm = I_C/V_T\$.

The output resistance is \$R_C\$ in parallel to everything else that loads the output node.

OK, here my calculation:

gm1 = 100e-6 *  q/(k*T)
gm2 = 1e-3 *  q/(k*T)
rpi2 = 100 /gm2
RC1 = 47e3;
RC2 = 4.7e3;
A1 = -gm1 * 1/(1/RC1 + 1/rpi2)
A2 = -gm2 * RC2
gain = A1 * A2

which gives the following results

gm1 =  0.0039571
gm2 =  0.039571
rpi2 =  2527.1
A1 = -9.4897
A2 = -185.98
gain =  1764.9
  • 1
    \$\begingroup\$ I also get around 1800 times gain in a quick calculation. \$\endgroup\$ – Bimpelrekkie May 30 '16 at 18:49
  • \$\begingroup\$ @FakeMoustache - Yes, and even if there is same mistake in the calculation it should be easy to fix it. \$\endgroup\$ – Mario May 30 '16 at 18:51

To confirm Mario's answer, I get gain:

\$G = +\frac{I_{C1}I_{C2}R_{C2}}{V_T^2(\frac{1}{R_{C1}}+\frac{I_{C2}}{\beta V_T})}= +1766\$

where \$V_T = kT/q = 25.26\text mV\$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.