I'm trying to design a CE BJT amplifier in Multisim with 100mVp 1kHz input, and target gain of around -208.8. Starting from my choice of Vcc = 50V and Ic(DC) = 15mA, here is the design I came up with: enter image description here

The Ic(DC) during simulation looked close enough to my expected value: enter image description here

So I calculated gain using -Rc/(Re'+Rsw) = -1613/(1.725+6) = -208.8, everything was looking reasonable.

Then I simulated the circuit with virtually no load (1Tohm load) as seen in the first screenshot. For 200mVpk-pk I'm expecting 41.76Vpk-pk, however the output is only 42.03 - 8.31 = 33.72Vpk-pk, ~20% drop from expected output. I'm quite certain the output is not clipping, but what is causing this drop, and is it possible to improve the gain accuracy as close to 100% as possible? I've looked around the Internet and found no similar situations so I'm perplexed and don't know where to start looking for design issues. enter image description here

I'm very, very new to circuit design and would appreciate a detailed explanation.

  • 2
    \$\begingroup\$ RL of 1Tohm is a bit excessive: the time constant in connection with Cc is about 3 hours. You want a shorter time span to get rid of 70% of your DC output offset. And that's assuming that this is significantly smaller than the isolation resistance of Cc, at 10µF also an audacious assumption. \$\endgroup\$
    – user107063
    Commented Nov 28, 2023 at 13:18
  • \$\begingroup\$ KYT, I get a close match. But I estimate VA (3%), IKF (2.5%), Ce in quadrature at 1kHz (2.5%), beta (> 2%), and Cb in quadrature at 1kHz (0.5%). Cc won't matter, I figure. But I'm already at 11% or so. Which is getting close enough to stop. \$\endgroup\$ Commented Nov 28, 2023 at 15:58
  • \$\begingroup\$ I think 10u input cap with such a low input impedance is too few. Try 100uF. \$\endgroup\$ Commented Nov 28, 2023 at 16:22
  • \$\begingroup\$ @MichalPodmanický Cb is in quadrature. There will be a small impact. I mentioned it. But almost ignorable. \$\endgroup\$ Commented Nov 28, 2023 at 16:38

2 Answers 2


When you try and make a single stage BJT have a gain that is realistically pushing the limits, you have to factor in the miller capacitance feedback effect due to collector-base capacitance. It may only be a few picofarads but, at unfeasibly large anticipated gains, it will bite you.

BJTs are so cheap that it makes no financial sense to use a single transistor to provide the whole gain. As a rule of thumb, I stick with a voltage gain per stage of no-more than 50.

how to improve simulation accuracy?

The simulation accuracy is far-far better than your hand calculations for gain.

  • \$\begingroup\$ My bad english. I have reworded it to calculation vs simulation accuracy. \$\endgroup\$
    – KYT
    Commented Nov 28, 2023 at 13:37
  • \$\begingroup\$ This is my assignment for an introductory course to BJTs, so I'm unable to get creative with cascading amplifiers or changing the gain. \$\endgroup\$
    – KYT
    Commented Nov 28, 2023 at 13:56
  • 1
    \$\begingroup\$ @KYT you do really need to check on that because I've never seen a BJT that can effectively work at a gain so high. \$\endgroup\$
    – Andy aka
    Commented Nov 28, 2023 at 14:10

You are operating the BJT outside its limits.

Even at "quiescent" the dissipation is ~0.4W which is almost twice the allowed maximum dissipation. If Multisim is able to simulate thermal voltage w.r.t. temperature or dissipation then your \$R'_E\$ assumption of 1.73 Ohms will no longer be valid because thermal voltage, \$V_T\$, will no longer be 26 mV.

Also it's not a good idea to select the emitter resistor that appears in AC gain (\$R_{sw}\$, in your case) not high enough to neglect the intrinsic emitter resistance which is temperature-dependent i.e. \$V_T/I_{CQ}\$. The idea behind adding emitter degeneration is to bring temperature compensation which basically means having almost constant collector current and voltage gain at different temperatures i.e. \$A_V\approx-R_C/R_E\$.

Now your emitter degeneration is 6 Ohms which is not higher enough than the intrinsic emitter resistance, so any small change in temperature will make significant difference in your gain equation. You can see this from your gain equation: Initially, the intrinsic emitter resistance was 1.73 and the calculated gain was 209. Now if it increases to 1.9 Ohms due to temperature which comes from dissipation, the gain will reduce to 200 which is almost 5% change. If it was 600 Ohms, for example, then the gain would be less but at least it'd be more "predictable". That's one of the reasons of making multi-stage (cascade) amplifiers for higher gain instead of a single-stage.

  • \$\begingroup\$ I see, thanks for the insight. I just assumed all is well if the collector current and Vce don't exceed the rated maximum (100mA, 65V). Maybe I'll try again with a smaller current. \$\endgroup\$
    – KYT
    Commented Nov 28, 2023 at 13:53
  • \$\begingroup\$ @KYT see my edited answer. That'll answer your question better. \$\endgroup\$ Commented Nov 28, 2023 at 13:54
  • \$\begingroup\$ I am well aware of the Rsw not being significant, but I still had to reach the target gain with one BJT. I will check with my lecturer if I can cascade amplifiers but I'm not feeling hopeful about it. \$\endgroup\$
    – KYT
    Commented Nov 28, 2023 at 14:05

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