I am facing a problem on how to design an amplitude modulator using a common-emitter amplifier circuit.

I am asked to use a carrier frequency of 50MHz and an input signal of 1kHz to 14kHz. In order to solve this problem, I designed a common-emitter circuit, as in my other question. Then I added a modulator signal at the biasing circuit yet I can't obtain a good amplitude modulated signal.

I have searched various sources but the information is not clear and not providing any calculations. I need to know how the values of the components are calculated and work well in simulation. I have found a sample circuit without calculations for reference.

Transistor AM Modulator

Edit: This is my current configuration right now.

enter image description here enter image description here enter image description here enter image description here

  • \$\begingroup\$ I absolutely love your tenacity!!!! You get another +1! I still think you should read up on the MC1496 and then go out and just buy discretes. You can get them in matched pairs, too!! So they will work pretty well in a similar topology. And, fortunately, the arrangement is even practical. \$\endgroup\$
    – jonk
    Nov 16, 2021 at 6:21
  • \$\begingroup\$ @jonk the problem is my prof wants me to use a transistor based circuit only… \$\endgroup\$
    – Eddy Miner
    Nov 16, 2021 at 6:37
  • \$\begingroup\$ But you can do that. I just said so. Either use some emitter degeneration and/or calibration steps (potentiometers) or else buy yourself some nicely matched BJTs. It's transistor-based that way. And it is practical. (Have you seen the kits you can buy that provides an opamp and the 555 timer, both, entirely with discretes?) Or are you prohibited from certain kinds of part packages (like SOT23-6), too? Do you think the inventor of the MC1496 (Barrie Gilbert) used the IC? Or do you think he built one from discretes, first? \$\endgroup\$
    – jonk
    Nov 16, 2021 at 6:42
  • \$\begingroup\$ @jonk Yup I am only limited to a simple transistor, resistors and some basic components only. \$\endgroup\$
    – Eddy Miner
    Nov 16, 2021 at 6:44
  • \$\begingroup\$ So you have package limitations, specifically? Look up the BCV61, BCV62, BCM61, and BCM62, for example. Those are discrete BJTs. Even if those are disallowed (hard to imagine) you can still do it with random junk box BJTs. You are going to have to work at this, regardless. You should base things upon a proper topology. A bad foundation isn't a good beginning. That said, I guess you know what's best. What do I know? \$\endgroup\$
    – jonk
    Nov 16, 2021 at 6:45

3 Answers 3


I need to know how the values of the components are calculated and worked well in simulation.

Let's work backwards. The voltage that appears on the collector of Q1 should look something like this. (ignore the time-scale, this is taken from another Q/A).

enter image description here

This signal consists of two components. An AM signal plus an amplified version of the modulating signal.

If your collector voltage looks like that, you can get the AM signal by filtering away, or attenuating the component which is the amplified version of the modulating signal. This is (imperfectly) accomplished by the RC high pass filter consisting of RL and the output capacitor. By selecting RL and Cout values such that the cutoff frequency of the RC high pass filter is around that of the carrier frequency.

Use the formula:

$$f = \frac{1}{2\pi RC}$$

where f is the cutoff frequency.

If you pick the cutoff frequency to be equal to the carrier frequency, then you can calculate what Rload * Cout needs to be.

To get the collector voltage to look like that in the diagram, one needs

  • The gain of the transistor to be set properly
  • The biasing of the transistor to be set properly
  • The emitter voltage to reflect the modulating signal, but much less so the carrier signal.

The last point is (imperfectly) accomplished by the low pass filter formed by the combination of Re and Ce.

Choose values for Re and Ce so that the cutoff frequency is the same as the modulating frequency.

Again, use the formula

$$f = \frac{1}{2\pi RC}$$

The emitter voltage should look something like this:

enter image description here

If too much carrier frequency signal appears on the emitter, the voltage on the collector may look something like this:

enter image description here \$\uparrow\$ NOT WHAT YOU WANT AT COLLECTOR! \$\uparrow\$

Again, working backward, one needs to fix the gain so that the transistor neither goes into saturation nor cutoff. Either of these situations will greatly distort your AM signal. Run a simulation of the circuit with the collector being monitored. You need to adjust the gain and bias of the circuit so that the voltage swing is something like from 0.7V to Vcc-0.3V. If you go too close to ground the transistor is in saturation, and if you go too close to cutoff. You can modify the gain in a few different ways. Increasing Re will decrease the gain. Increasing Rc will increase the gain.

As you adjust the gain, you may discover that bias also needs to be adjusted. That is, if the transistor goes near Vcc-0.3V, but doesn't go anywhere near 0.7V. Or alternatively, if the transistor goes near 0.7V, but doesn't go anywhere near Vcc-0.3V. You can adjust the bias by changing the values of R1 or R2.

I can’t understand why the Em is greater than Ec

Both Em (Vmodulation) and Ec (Vcarrier) get amplified. However, from the image above showing the collector voltages, it should be clear that the smallest amplitude of the AM envelope occurs when the top of the amplified carrier gets close to Vcc. The smaller the carrier frequency peak-to-peak voltage at this point, the closer one can get to 100% AM modulation without clipping. Hence the reason for keeping the carrier voltage small. A larger Vcarrier means that the % of AM modulation must be less.

  • \$\begingroup\$ I tried tuning few times using your advice, takes some time to get a very well tuning but for now i managed to get the shape of the graph shown and still the am is weird and cant be seen \$\endgroup\$
    – Eddy Miner
    Nov 16, 2021 at 15:36
  • \$\begingroup\$ could you post a simulation of what appears at the collector? \$\endgroup\$ Nov 16, 2021 at 15:50
  • \$\begingroup\$ updated in the post, since i not so sure how to post in the comments \$\endgroup\$
    – Eddy Miner
    Nov 16, 2021 at 15:59
  • \$\begingroup\$ Is the graph you added the collector voltage or the emitter voltage? \$\endgroup\$ Nov 16, 2021 at 16:08
  • \$\begingroup\$ Oh my fault. My brain already burned up \$\endgroup\$
    – Eddy Miner
    Nov 16, 2021 at 16:09

What is amplitude modulation AM ?

What we know is one of the pictures below. Generally speaking, it is the first (DSB) or the second which are known.

See the formula where the DC component is.

enter image description here

Are this ok ? Design a simple transistor amplitude modulator circuit

Or is this also acceptable ?

enter image description here

Or this with a "real" multiplier (2 transistors used) ?

enter image description here

Note that all suffers of some "distortion", because of the non-linearity of BJT ...

Also use of a tuned LC.


Eddy Miner - did you already consider a simple differential amplifier (long-tailed pair)?

It can be used as a perfect AM-Modulator. The carrier drives one of the main transistors and the modulating signal drives the base of the transistor in the common leg, thereby varying the transconductance (the gain) of the transistors.

  • \$\begingroup\$ I havent consider it, since my knowledge is not enough for me to construct it \$\endgroup\$
    – Eddy Miner
    Nov 17, 2021 at 1:18
  • \$\begingroup\$ Eddy Miner-the diff.amplifier (used as a AM modulator) works better and is easier to understand if compared with the circuit as given by you. The only thing you must know is that the gain of a transistor stage depends on the transconductance (and, hence, on the DC current through the device). And this current can carry the modulating signal, thereby varying the gain of the stage. \$\endgroup\$
    – LvW
    Nov 17, 2021 at 8:24

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