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I've been self-studying electronics and am currently trying to design a high-gain single-stage BJT amplifier. Here is what I have so far:

enter image description here

Q1 serves as a current source "active load" for the common emitter amplifier. R4 and R5 form a biasing voltage divider supplied from Q2's collector. C2 bypasses R6 to increase the gain at signal frequencies.

  1. I've simulated this circuit in MacSpice and found a gain of about 250. This is comparable to a collector resistor of about 12.5 kΩ (assuming total emitter resistance around 50 Ω at signal frequencies). I expected the current source "active load" in Q2's collector to present a higher impedance and result in a higher gain.

  2. On the other hand, since I'm using Q2's collector to supply the voltage for the biasing voltage divider (a form of negative feedback) I might have expected that the gain would roughly be (R4 + R5)/R5 ≈ 10. Specifically, the open loop gain is 250, I'm returning 10% of the output signal to the base through feedback, and the gain equation gives 250/(1+250*0.1) = 9.6 expected gain. What is faulty about this logic?

  3. How would you generally improve this circuit for higher gain, sticking to a single stage using only BJT transistors?

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  • \$\begingroup\$ Forgot to include on the diagram, Vcc is 20V. \$\endgroup\$
    – broken.eggshell
    Jan 22, 2022 at 7:37
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    \$\begingroup\$ What are your source and load impedances? Your design profoundly depends on both. Working frequency? You have reactive elements as well. \$\endgroup\$
    – fraxinus
    Jan 22, 2022 at 8:09
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    \$\begingroup\$ In general, it doesn't make a lot of sense to try for a very high gain with one BJT. There are very few circumstances where that works out without gross, terrible distortion resulting or else the high gain applying in such a way that the output only has a very tiny output signal (meaning the input has to be very very much smaller, still.) It's just not useful. If you want high but constant gain, then you also want NFB. And that means more than one active device (BJT.) \$\endgroup\$
    – jonk
    Jan 22, 2022 at 8:09
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    \$\begingroup\$ @broken.eggshell What are your design goals here? What is the source impedance? What is the peak-to-peak input expected to be, unloaded? What does this circuit drive? \$\endgroup\$
    – jonk
    Jan 22, 2022 at 8:17
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    \$\begingroup\$ You can't seriously ask for improvement with no design specs or on;ly 1 transistor. Pls define Zin, Zout, Av, THD and Pd max. \$\endgroup\$
    – Tony Stewart EE75
    Jan 22, 2022 at 8:53

3 Answers 3

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First, some remarks on the circuit diagram:

  1. The current source is voltage dependent. This can be improved by replacing R1 with 2 silicon diodes connected in series, so that the current will depend mainly on the value of R3.
  2. As already mentioned in the comments, the gain (as well as the bandwidth) depends on Rload and the negative feedback Rsource/R4.

Here are the results (Fig1) with some changed values ​​and the resulting current source of about 1mA, LTSpice simulating : Fig1 At 1kHz frequency, Vin=1mV sine and Rload: 1M - gain 2500; 10k - 350; 1k - 40;

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  • \$\begingroup\$ I put your new values into Spice and am still getting the same gain. It's very possible my Spice circuit is wrong, but supposing it isn't, it makes me wonder a little bit about how accurately Spice can model circuits like this. How much should I actually trust it and rely on it for designs? Also, ignoring questions about whether it is desirable to do so, is a gain of 2500 for a single stage BJT amplifier really possible?! \$\endgroup\$
    – broken.eggshell
    Jan 22, 2022 at 17:31
  • \$\begingroup\$ @broken.eggshell There is no cheating, I have tried similar schemes, everything is exactly as in the simulation. \$\endgroup\$
    – Peter MP
    Jan 22, 2022 at 19:32
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    \$\begingroup\$ Just took another look at this and indeed there were mistakes in my Spice circuit. I'm using a 2N3906 model for the pnp and getting a gain of >2000! \$\endgroup\$
    – broken.eggshell
    Jan 22, 2022 at 22:28
  • \$\begingroup\$ @broken.eggshell Yes, it's my mistake but these transistors are almost identical for this purpose. By the way you can discard R5 and bypass R6 and C2. \$\endgroup\$
    – Peter MP
    Jan 23, 2022 at 9:13
  • \$\begingroup\$ Interesting. The mistakes were mind though... I had just implemented your circuit incorrectly. May I ask how you arrived at the values for R4 and R5? \$\endgroup\$
    – broken.eggshell
    Jan 23, 2022 at 16:37
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Assuming your signal input is a voltage source (zero impedance) and you didn't connected any load at "out", here's what we have:

I expected the current source "active load" in Q2's collector to present a higher impedance and result in a higher gain.

Your "active load" is somewhat negated by R4. Your Q2 collector sees somewhat less than 154k load (in AC).

I might have expected that the gain would roughly be (R4 + R5)/R5 ≈ 10.

Wrong as well because from the feedback viewpoint, your R5 is in parallel with C1 and the source. At high enough frequency, R4 is not a feedback at all.

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  • \$\begingroup\$ I think I get your first point, but in Spice if I run R4 from a 17V voltage source rather than from the collector, the gain drops to 28 instead of going up. Second point I'm less clear on. At high enough frequencies I just imagine C1 as a short and not impacting the circuit. \$\endgroup\$
    – broken.eggshell
    Jan 22, 2022 at 17:18
  • \$\begingroup\$ I am not exactly sure what happens to your Q2 collector DC bias. In R4 to 17V Q2 probably gets saturated. \$\endgroup\$
    – fraxinus
    Jan 22, 2022 at 21:24
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I've simulated this circuit in MacSpice and found a gain of about 250 ?

AC Analysis ... Seems ok. But TRAN and DC Analysis seem "weird".

(Your R3= 1k8 should be 7k5 for good "DC quiet point".)

enter image description here

Zin, Zout ...

enter image description here

After "relooking" ... this seems better. Zout = ~ 100k ! (EE&O)

Test TRAN Analysis:

enter image description here

Bandwidth:

enter image description here

Zin, Zout impedances:

enter image description here

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  • \$\begingroup\$ Your first point is to center Q2's collector voltage by appropriately choosing R3...makes sense. Also, it never occurred to me that I could get input and output impedances directly from Spice, so learning that already made this answer helpful. I think you are proposing to change the circuit by bypassing Q1's collector resistor, but I don't think I understand the benefit of doing that. \$\endgroup\$
    – broken.eggshell
    Jan 22, 2022 at 17:41
  • \$\begingroup\$ I tried adjusting the DC quiet point by changing only (my choice) the resistor of emitter Q3. But certainly, there is another means of doing so. I left the 154 kOhm because it makes negative feedback for polarizing Q4 (all benefits), but I don't have measured the incidence on QP by changing its value. \$\endgroup\$
    – Antonio51
    Jan 22, 2022 at 22:15
  • \$\begingroup\$ Be careful that Zin and Zout have weird values. \$\endgroup\$
    – Antonio51
    Jan 23, 2022 at 10:17
  • \$\begingroup\$ Re "quiet point": Don't you mean "quiescent point"? \$\endgroup\$
    – Peter Mortensen
    May 31, 2022 at 21:20
  • \$\begingroup\$ Yes, it is. My french "mistake" ... \$\endgroup\$
    – Antonio51
    May 31, 2022 at 21:29

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