I have set up the simple common emitter amplifier circuit based on 2N3904. The circuit was first simulated in Multisim and than breadboarded. All works fine as expected. I have tested about 20 different transistor models (also vintage ones and PNP) on breadboard and most of them are working fine with that bias and give about the same voltage gain at the output. But some does not at all (no amplification) like 2N3391A and 2N3417. My question is why the don't? They are "general purpose" so i guess they should work with the same bias as 2N3904 or 2N2222?

Note: In Multisim simulation ALL transistors work also 2N3391A.

enter image description here

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    \$\begingroup\$ Have you checked the data sheets? \$\endgroup\$ Feb 12, 2017 at 14:18
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    \$\begingroup\$ Yes, sorry my mistake - 2N3391a and 2N3417 have different pinout (BCE not CBE as usual) so that's why they did not work! \$\endgroup\$ Feb 12, 2017 at 14:29
  • \$\begingroup\$ And they work as expected as well:) \$\endgroup\$ Feb 12, 2017 at 14:30
  • \$\begingroup\$ 3-wrong pin out \$\endgroup\$ Feb 12, 2017 at 15:15

2 Answers 2


why the don't?

a few possibilities:

1) you could have soldered it in incorretly;

2) you got the wrong pin out;

3) you didn't power it up;

4) the pcb failed on that try;

5) you didn't measure it up;



Just a few thoughts about the schematic you provided. I don't know the exact design criteria you used (or the designer used, if that wasn't you), but I can make some assumptions about it based on common example ideas I've seen on the web and in writings.

I'm guessing here that the designer (you?) used \$V_{C_q}\approx 9\:\textrm{V}\$, implying \$I_{C_q}\approx 4\:\textrm{mA}\$. This suggests \$V_{E_q}\approx 900\:\textrm{mV}\$ and \$V_{B_q}\approx 1.6\:\textrm{V}\$. Also, an expected unloaded gain of about \$\frac{2.2\:\textrm{k}\Omega}{220\:\Omega}\approx 10\$. Given the expected \$\pm\:5\:\textrm{V}\$ swing, this looks okay as \$V_{CE_{min}}\approx 3\:\textrm{V}\$ and this keeps the transistor well out of saturation. So that's good, too. (The high rail voltage makes this possible given your desired gain and the lack of a separate AC emitter gain setting leg.)

The questions I have with the design choices come with the biasing side of things. With \$I_{C_q}\approx 4\:\textrm{mA}\$ and with \$V_{CE_{min}}\approx 3\:\textrm{V}\$, I might estimate \$\beta_{min}\approx 100\$ for a typical small signal BJT. This means that I'd want to be sure I have \$I_{B_q}\ge 40\:\mu\textrm{A}\$. But a rough estimate of the available biasing current in your biasing pair says \$\frac{18\:\textrm{V}}{220\:\textrm{k}\Omega+27\:\textrm{k}\Omega}\approx 70\:\mu\textrm{A}\$. That's probably not enough for the amplifier design to work consistently over a wide variety of BJTs. Sure, there are some common BJTs that will deliver in the \$\beta=300-400\$ range here. Some I have in a box (super beta) will deliver \$\beta=1000\$, even. So the required \$I_{Q_q}\$ may be less than the estimate I gave and this may allow the circuit to still work reasonably well even with such a light biasing current. But if you want to make a design that works consistently, you should plan on a lowish \$\beta_{min}\$. Around the 100 value I mentioned earlier. Regardless, I'd stiffen up that biasing pair.

[All of the above thoughts are true, even if I'm somewhat wrong about \$V_{C_q}= 9\:\textrm{V}\$. Moving \$V_{C_q}\$ up or down a few volts won't change the estimated \$I_{C_q}\$ enough to change my difficulty with the weak biasing you are using.]

For example, assuming \$I_{B_q}\ge 40\:\mu\textrm{A}\$ it is typical to plan on about \$10\times\$ for the biasing pair, or about \$400\:\mu\textrm{A}\$. Given that figure and using the \$V_{B_q}\approx 1.6\:\textrm{V}\$ figure, I'd get \$R_2=\frac{1.6\:\textrm{V}}{400\:\mu\textrm{A}}= 4\:\textrm{k}\Omega \$. Which means I'd use \$R_2= 3.9\:\textrm{k}\Omega \$. This means now that my biasing current is recomputed to be about \$410\:\mu\textrm{A}\$. Now, I need \$R_1=\frac{18\:\textrm{V}-1.6\:\textrm{V}}{410\:\mu\textrm{A}-40\:\mu\textrm{A}}= 44\:\textrm{k}\Omega \$. So I'd use a standard value of perhaps \$R_1=39\:\textrm{k}\Omega \$ or \$R_1=47\:\textrm{k}\Omega \$. The actual choice here will move \$V_{C_q}\$ around a bit and since I don't want it to fall much below the value of \$9\:\textrm{V}\$, I'd rather use \$R_1=47\:\textrm{k}\Omega \$, pulling a little less on the BJT base. It's the closer value, too, which is nice.

So that's the design you should also try out. Use \$R_1=47\:\textrm{k}\Omega \$ and \$R_2=3.9\:\textrm{k}\Omega \$. See how that works for you.

Oh, some final notes. I'd expect to see a gain of less than 10 at the output. \$r_e=\frac{k T}{q I_{C_q}}\approx 6.5\:\Omega\$. This adds to your \$R_E=220\:\Omega\$ value and gives a gain of \$A_V\approx\frac{2.2\:\textrm{k}\Omega}{220\:\Omega+6.5\:\Omega} \approx 9.7\$. And that's still unloaded. Once you place a load on it, it will reduce further, of course.

Also, given your input swing being so much as compared to the biasing point of the BJT base (which is just another reason I'd redesign the whole thing and move the quiescent point of \$V_{E_q}\$ up higher), you can expect quite a swing on \$I_C\$. The low point will be about \$V_{B_{min}}=\frac{3.9\:\textrm{k}\Omega}{39\:\textrm{k}\Omega+3.9\:\textrm{k}\Omega}\cdot 18\:\textrm{V}-500\:\textrm{mV}\approx 900\:\textrm{mV}\$ and therefore \$V_{E_{min}}\approx 200\:\textrm{mV}\$, so that \$I_{C_{min}}\$ falls slightly below \$1\:\textrm{mA}\$. This provides \$r_e\approx 29\:\Omega\$ and the gain drop to about 8.8 or so. This gain variation means that the output will be slightly distorted and won't be as clean as you might want it (though perhaps that is just fine, too.)

To improve that circumstance, it is common to raise \$V_{B_q}\$ to a higher value. Given the large supply rail you have, there should be no problems here. However, this raises \$V_{E_q}\$ and that means a larger \$R_E\$ to keep \$I_{C_q}\$ the same, which then reduces the gain without changing the topology. The answer is to add an AC gain leg to the emitter. This allows you the freedom to set the DC value of \$V_{E_q}\$ in order to reduce gain variation caused by varying \$r_e\$ without sacrificing the final AC gain in the process. It also greatly improves temperature stability, as well.

  • \$\begingroup\$ wow, i'm impressed. yes indeed the circuit is taken from the web. i will go through your explanation and will experiment with the circuit. thanks a lot for your efforts and i really appreciate that a lot. it helps me understand electronics more and more. it's my passion. \$\endgroup\$ Feb 13, 2017 at 21:45
  • \$\begingroup\$ @JerzyPrzezdziecki Try it out and see how well it works with the more difficult BJTs (the 2N3391A and 2N3417 you mentioned.) \$\endgroup\$
    – jonk
    Feb 14, 2017 at 21:13
  • \$\begingroup\$ Dear Jonk, i have changed the resistors as you suggested and the circuit works well, i mean sine wave is nice and clean. Anyway, the values are slightly different than yours: Ibias = 355uA, Ib/Vb= 17uA/ 1.32V, Ic/Vc= 2.85mA/ 11.7V. What do you think? \$\endgroup\$ Feb 20, 2017 at 21:59
  • \$\begingroup\$ @JerzyPrzezdziecki No two transistors are exactly alike. In fact, some parameters vary quite widely. Even the same part number. But you were considering using LOTS of different part numbers, even. So I'd expect quite a variation in base current, etc. But the point of the design I struggled towards wasn't to nail down exact base currents (impossible to do under the circumstances) but instead to make a design that would work, regardless. \$\endgroup\$
    – jonk
    Feb 21, 2017 at 2:10

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