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!
Edit: My Working
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\$\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 KaulicsCommented May 30, 2016 at 18:10
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\$\begingroup\$ Sorry about that, my working is here, thank you for the advice \$\endgroup\$– ohkneelCommented May 30, 2016 at 18:21
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2 Answers
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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
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1\$\begingroup\$ I also get around 1800 times gain in a quick calculation. \$\endgroup\$ Commented May 30, 2016 at 18:49
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\$\begingroup\$ @FakeMoustache - Yes, and even if there is same mistake in the calculation it should be easy to fix it. \$\endgroup\$– MarioCommented May 30, 2016 at 18:51
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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\$
answered May 30, 2016 at 19:28