The following schematic is a class C AM modulator without use of any transformers I've just changed the parameters in spice to get a correct modulation. however I want to find the theoretical value for modulation index.

In class C modulators using transformers modulation index is proportional to the ratio of Vm to Vcc.


simulate this circuit – Schematic created using CircuitLab

My Try:

To find this I tried to think of the transistor as a resistor Rout in the saturation region. Therefore, the Vcc(mod) is

$$\text{Vcc(mod)}=\frac{\text{Vcc}\text{Rout}}{\text{Rout}+\text{Ra}}+ \frac{\text{Vm}\text{Rout}}{\text{Rout}+\text{Rb}}$$

If this analysis is correct, how should the value of Rout be found?


simulate this circuit

Here is the result of the simulation with Vm=Vcc=9v, Vosc=3.5v:

enter image description here


From "Solid state radio engineering", Krauss 1980. page:223

enter image description here

  • \$\begingroup\$ Since Q1 is never in saturation region, you'll have some problems with that. Might want to start with what a class C amplifier is such as this link: en.wikipedia.org/wiki/Amplifier#Class_C and then proceed on to understand what "conduction angle" is such as this link: electronics-tutorials.ws/amplifier/amplifier-classes.html Then you might want to look into small signal modeling and a bunch of other things because you can't solve this in "DC"; it's an RF amplifier. \$\endgroup\$ – Dave Mar 4 '16 at 10:53
  • \$\begingroup\$ @Dave In AM class C modulators, Q1 is in saturation at the peaks of the output (the collector voltage increases up to Vcc(mod) in the above circuit and the transistor will go to saturation). In this way I thought that it might be possible to find Vcc(mod) by considering transistor as a resistor. \$\endgroup\$ – SMA.D Mar 4 '16 at 12:01
  • \$\begingroup\$ I can actually gaurentee you that your transistor is never in saturation. It's either in forward active or cut-off. That's how BJTs are operated in most amplifiers. Plus saturation has high recovery time so it would be too slow to use if you needed it. I'm glad Olin Lathrop had the patience to write you a full guide. \$\endgroup\$ – Dave Mar 4 '16 at 14:52
  • \$\begingroup\$ @Dave I see this in the book Solid state radio engineering. I've attached the related page to the question. What does it mean by using the word "saturation" there? \$\endgroup\$ – SMA.D Mar 4 '16 at 19:24
  • 1
    \$\begingroup\$ The writer doesn't have a negative bias resistor "Rc" unlike yours. That diagram and description are what we like to call "basic at best." They would be right if everything were perfect: perfect voltage sources, perfect transistors, perfect inductors, perfect capacitors, etc. But in the real world we don't put BJT's into the saturation region if you want them to change state quickly. 1.5 MHz is pretty quick. You're more than welcome to simulate the DC bias voltage on the base of the npn transistor to see that it's negative and that the negative bias keeps it from hard saturating. Or look it up \$\endgroup\$ – Dave Mar 4 '16 at 19:34

Start by understanding how this circuit works.

C3 and L2 are a tank circuit. This is a high-Q circuit that resonates at the carrier frequency. Q1 slams the bottom of this tank to ground for a short time once per cycle to keep it ringing. The resulting ringing is then coupled to the antenna thru C9 and L4, which are presumably to provide a impedance match between the tank and the antenna.

Ideally, with the top of the tank held at a fixed supply voltage, with the bottom swinging to ground at its low point and twice the supply voltage at its high point.

To modulate the amplitude of this ringing, this circuit changes the "supply" voltage at the top of the tank. You can consider it fixed over the period of a few cycles, since the modulation frequency is much lower (about 100x lower in commercial AM) than the carrier frequency the tank is ringing at. C1 is meant to be a short at the carrier frequency. Put another way, the impedance at the top of the tank is very low at the ringing frequency.

V2 is the actual supply voltage. If you didn't need to modulate the carrier (change its amplitude), then V2 would be connected directly to the top of the tank. All the other components from the top of the tank to V2 and Vm are just there to allow Vm to change the apparent supply voltage, while letting V2 set the average. C2 AC-couples the modulation signal to guarantee it has no DC affect.

Ideally, the voltage at the top of the tank can be varied from 0 to twice V2. Due to all the passive coupling parts Rs, C2, Rb, RFC, C1, and Ra, there is attenuation from Vm to the top of the tank. Vm therefore needs to be more than ±9 Vpp to achieve maximum modulation. Exactly how much is difficult to predict. You can start by computing the impedances of all the component mentioned above and use resistor-divider math to compute the minimum attenuation. The impedance presented to the modulation signal by the tank itself is harder to predict. I'd probably take a rough stab at it, then get the rest by experimentation.

By the way, I used this concept (although my circuit was different in a few key places) to make a AM transmitter in college once. It worked very well, and we ran the radio station Sunday nights in the dorm by dumping the carrier onto the AC line. As a "audience participation" event, we had everyone flush the toilets at the same time. It took 20 minutes for the water pressure to come back. It must have shaken 100 years of crud from the inside of the Troy NY water pipes, because the water was rust colored for another few hours after that.

  • \$\begingroup\$ Thank you for your complete explanation and that wonderful memory of "audience participation" event :) . \$\endgroup\$ – SMA.D Mar 4 '16 at 13:37
  • \$\begingroup\$ So I think to take a rough stab, I can assume L2 and RFC are short circuit and C9 is open at the message frequency therefore I should only compute the resistance seen from the collector of Q1 which is r_o of transistor (assuming Vosc is at its maximum). Am I true? \$\endgroup\$ – SMA.D Mar 4 '16 at 13:37
  • 1
    \$\begingroup\$ @SMA: That might get you to a rough value. You can compute the impedances of the Ls and Cs easily enough at the modulation frequency, then do the resistor-divider math. \$\endgroup\$ – Olin Lathrop Mar 4 '16 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.