I am trying to make a 27 MHz carrier wave transmitter from a crystal oscillator and a secondary amplifier. This is the complete circuit:


simulate this circuit – Schematic created using CircuitLab

The first part, left of C6 is a colpitts crystal oscillator. And on the right side of C6 is the common emitter amplifier. Colpitts crystal oscillator that I've built can be found here.

Q1 and Q2 datasheets can be found here.

The problem is the following. If I have CE amplifier disconnected, and measure the voltage with oscilloscope at O1, I get expected 150 mV peak-to-peak. But as soon as I connect the CE amplifier and measure the voltage at O2, I get around 300 mV peak-to-peak (note that antenna at this time is not connected), which is far less than I was expecting.

Values chosen for colpitts oscillator are the same as on that website I posted a link to. For the CE amplifier I calculated the values my self, here's how I did it:

  1. \$\beta=100\$
  2. I chose: \$I_C=I_E=1mA\$
  3. I chose: \$V_E=1V\$, so \$V_B=1.7V\$
  4. \$R_6=\dfrac{1V}{1mA}=1k\Omega\$
  5. \$I_B=\dfrac{Ic}{\beta}=10uA\$, \$I_{R5}=100uA\$, \$I_{R4} = 110uA\$
  6. \$R_5=\dfrac{1.7V}{100uA}=18k\Omega\$, \$R_4=\dfrac{9V-1.7V}{110uA}=66k\Omega\$
  7. \$R_7=\dfrac{9V-4.5V}{1mA}=4.7k\Omega\$
  8. For \$C_4\$ I read somewhere: \$X_{C4}<=\dfrac{1}{10}\times R_6\$, and I get \$C_4 >= 60pF\$

C5 and C6 were chosen arbitrarily, If someone could tell me how to precisely calculate their values I'd really appreciate it.

So shouldn't the gain of the amplifier be:\$r_e=\dfrac{25mV}{I_C}\$ \$A_v=\dfrac{-R_C}{r_e}=-188\$?, while I'm getting gain of only 2.

What could be the problem? I read somewhere that impedance mismatch can affect the power of the signal transmitted, could this be the case here, since the output impedance of colpitts oscillator is relatively low, while the input impedance of the CE amplifier is relatively high?

Also, what if I were to use BS270 MOSFET in common gate mode instead of 2N3904, would the gain increase? I've read somewhere that MOSFETs are faster and seen them used in HF applications. Because I have them at hand and can't buy any components at the moment.

  • \$\begingroup\$ Nice question, (and answer by @Chris Johnson) RE: C5 and C6. C6 is a coupling cap and as long as it's impedance at 27MHz is low compared to what it's driving then it should be fine... (Z (1nF)<~10 ohms) I'm not as sure about C5. Do you really need this cap? What's the impedance of the antenna? Is it mostly capacitive? (A few pF?) \$\endgroup\$ Commented Aug 19, 2014 at 12:30
  • \$\begingroup\$ Ah thanks for the explanation. I haven't actually got a antenna at this moment, but I was thinking of using just a solid copper wire. C5 is there to get rid of DC bias voltage. \$\endgroup\$
    – Golaž
    Commented Aug 19, 2014 at 12:37
  • \$\begingroup\$ Not sure about using a MOSFET in common gate mode - I think you would need a cascode to increase the input impedance of the amplifier. If you are getting a reasonable amount of gain (factor of 5, say) from a 2N3904 and the larger C4 suggested in my edited answer, you could try adding a second identical NPN amplifier stage. \$\endgroup\$ Commented Aug 20, 2014 at 15:54
  • \$\begingroup\$ Larger C4 worked, I will add more stages then, the gains from stages multiplies, right? Thanks again! \$\endgroup\$
    – Golaž
    Commented Aug 20, 2014 at 16:04
  • \$\begingroup\$ Actually the gain is more around 3. \$\endgroup\$
    – Golaž
    Commented Aug 20, 2014 at 16:12

3 Answers 3


One reason is that the transistor gain is degraded at high frequencies. To pick a specific example, the ON semiconductor BC546 has a gain-bandwidth product (GBP) of 100MHz at 1mA collector current (see figure 6 in the linked datasheet). This means that at a frequency of 27MHz, the current gain (beta) is about 100MHz/27MHz = 3.7, not 100.

At 27MHz, stray capacitances in the transistor (amplified by the Miller effect) may well also be playing a role in reducing the gain.

Simply replacing the transistor with one more suited to high frequencies may be sufficient to fix the problem. You may get away with just choosing a different general-purpose transistor: the 2N3904, for example, is a little better with a typical GBP of 300MHz. A better solution is probably to choose one of the many transistors designed for high frequency applications. To pick one at random, the PN5179 from Fairchild has a typical GBP of 2000MHz.

Because of the Miller effect, the common collector amplifier is not especially well suited for high frequency amplification, and topologies such as the common base amplifier are often used for signals at several tens or hundreds of MHz. However, at 27MHz I suspect you will be OK with a common emitter amplifier.

An additional factor limiting the gain is that the impedance of C4 || R6 needs to be added to r_e when calculating the emitter resistance at signal frequencies. Usually C4 is chosen to have negligible impedance at signal frequencies compared to the r_e of the transistor, but at 27MHz the impedance of your R6 || C4 is about 55Ω (dominated by the 59Ω impedance of C4). Switching C4 to a 1nF or 10nF capacitor should increase the gain by more than a factor of two.

  • 1
    \$\begingroup\$ This answer would be improved by suggesting a solution. Perhaps a different arrangement of the transistor? More stages? \$\endgroup\$
    – Phil Frost
    Commented Aug 19, 2014 at 11:10
  • 2
    \$\begingroup\$ I thought that the suggestion to use a transistor with a higher gain-bandwidth product along with the included transistor example WAS an excellent solution. In addition, the suggestion to use the common base arrangement - along with the reason why to - was the icing on the cake. In my opinion, of course. \$\endgroup\$
    – EM Fields
    Commented Aug 19, 2014 at 12:05
  • \$\begingroup\$ Using 2N3904 improved the gain a bit, I'll try your other suggestions also. Thanks for your help! \$\endgroup\$
    – Golaž
    Commented Aug 19, 2014 at 12:22
  • 1
    \$\begingroup\$ EM Fields - The suggested solutions were added after the comment by Phil Frost. \$\endgroup\$ Commented Aug 19, 2014 at 12:43

A couple things to think about - what do the DC biasing resistors do to your signal? If you removed Q2 but left R4/R5, what would the gain at O1 be? Also, you calculate the gain of the second stage as RC/re, but neglect the effect of R6, which is in series with re. With those two things in mind, go back and calculate the gain.


Yes.... the impedance mismatch might be part of the problem. Remember that equals impedances (out and in between stages)allows maximum power transfer. Another extra step you might take is to add a "buffer" with high input impedance, which avoids loading the first stage (the Colpitt oscillator). The suggested stage is a collector amplifier.


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