1
\$\begingroup\$

I was unable to understand a bunch of stuff related to amplifiers. Specifically my questions are as follows:

  1. What is load modulation? If i have a load of impedance of 10 ohms physically, then does load modulation change the impedance of this load? Does anyone have the theory behind load modulation to share?
  2. What is the difference between "active load modulation" and "active load pull"?. Furthermore what is the difference between active and passive load modulation?
  3. Doherty power amplifier: DPA uses a quarter wave transimission line. Does that mean DPAs are not suitable for low freqencies (less than 100kHz) since the transmission line length will be too long to integrate in chips? Also is it possible to avoid the transmission line?

I have not found any literature which has answered these questions, maybe i missed some. I would really appreciate some help from the experts.

\$\endgroup\$
  • 1
    \$\begingroup\$ Links to terms used would be nice. \$\endgroup\$ – Andy aka Jan 16 '18 at 10:42
  • \$\begingroup\$ @Andyaka : I have no idea why you have never understood my questions! You need the questions to be precise, i get that but i hope you realize, it is not always possible. At the same time i appreciate your efforts to answer! \$\endgroup\$ – RAN Jan 16 '18 at 11:50
  • \$\begingroup\$ @RAN I'm with Andy on this. (1) 'I was unable to understand a bunch of stuff...', what stuff? The stuff may be wrong, it may be assuming some knowledge you don't have, it may need a subtle shift in perspective. We don't know without seeing what particular stuff we're talking about. (2) DPA may be well known to afficianados, but I (and maybe others) regard part of my 'payment' for my free answering service is a widening of my education, through thinking about questions. If you gave me some interesting links to go to, that would help both me and you. Play the game generously, let's all win! \$\endgroup\$ – Neil_UK Jan 16 '18 at 11:56
  • \$\begingroup\$ @Neil_UK The answer to "what stuff?" are asked as questions in the "question" i posted. I agree that DPA may mean anything and I should have clarified, but I think atleast one person understood what I asked(not to mention "Doherty power amplifier: " was mentioned) 3. Link to terms I used? I can use the same line of thought as you and Andy. What links do you want? webpage links to all the literature i read? or links to expand the abbreviation? It is confusing. But like i said, i agree i must be making mistakes, I would really like if experts like you helped me frame my question better. \$\endgroup\$ – RAN Jan 16 '18 at 12:23
  • \$\begingroup\$ @Andy_aka: An apology is in order. I understand that you were only trying to help. \$\endgroup\$ – RAN Jan 16 '18 at 12:42
3
\$\begingroup\$

Consider that in general a load is measured at some point in a network, and can be defined as load Z = V/I at that point in the network.

In an efficient amplifier, we want the load seen by the power devices to be such that the devices are as close to saturation as possible on a moment by moment basis, which if you think about it is saying that the drain load impedance should be inversely proportional to power output.

In a classical power amp the drain impedance is fixed at a level that is low enough to support whatever the maximum power the thing is capable of, which is fine for a constant power mode, but horrible for anything with a significant AM component (Like say just about any efficient modulation scheme) as the input power falls as only the square root of the output power backoff (This is an inherent property of a fixed drain impedance/fixed power rail design). Doherty can be thought of as using the peaking amplifier to lower the drain impedance of the carrier amplifier during modulation peaks this allowing the main amp to run a higher native drain impedance and be more efficient over the vast majority of the time when peak output is not required.

Due to the 1/4 wave phase shift, they are inherently narrow band devices, but IIRC the first ones were indeed built for ~100kHz long wave service by Continental transmitters using valves as the active devices. A 90 degree phase shift does not need to be a transmission line, you can do it using a lumped element hybrid made from a coupled inductor and two capacitors, still narrow band, but workable down well into the audio band if you really wanted to. There are easier ways down that close to DC with modern parts, PWM with a GAN switch will make very efficient power up to at least the medium wave region without the narrow bandwidth of the Doherty designs.

\$\endgroup\$
  • \$\begingroup\$ There are modern solutions that employ this concept without the quarter-wavelength impedance transformer or one built with LC resonance. Instead, they use more classical transformers, and this allows them to operate over wider bands. There have been demonstrated cases of these working with a 3 dB bandwidth from 60 GHz to 81 GHz, which is quite wide band for such an architecture! \$\endgroup\$ – Joren Vaes Jan 16 '18 at 12:34
  • \$\begingroup\$ Half an octave, not bad as there things go! Do you have a reference to that work? \$\endgroup\$ – Dan Mills Jan 16 '18 at 12:49
  • 1
    \$\begingroup\$ ''Transformer-Based Doherty Power Amplifiers for mm-Wave Applications in 40-nm CMOS'' Ercan Kaymaksut, Dixian Zhao and Patrick Reynaert. ieeexplore.ieee.org/document/7059254 and their followup: ''E-band transformer-based Doherty power amplifier in 40 nm CMOS'' ieeexplore.ieee.org/document/6851687 \$\endgroup\$ – Joren Vaes Jan 16 '18 at 12:52
  • \$\begingroup\$ @DanMills: Thank you for the answer. So I guess active load modulation and active load pull are more or less the same technique! \$\endgroup\$ – RAN Jan 18 '18 at 11:51
0
\$\begingroup\$

Since Dan Mills gave a better discription of load modulation than I would be able to, I'll focus on the third part of your question.

  1. Doherty power amplifier: DPA uses a quarter wave transimission line. Does that mean DPAs are not suitable for low freqencies (less than 100kHz) since the transmission line length will be too long to integrate in chips? Also is it possible to avoid the transmission line?

Yes, it is possible to avoid the transmission line! You don't need a quarter-wavelength transmission line for a Doherty power amplifier. This can also be done with LC networks, or with inductive coupling (in other words, a tranformer). The latter makes it suitable for integration into ICs, and this has been done and demonstrated. For example, Ercan Kaymaksut demonstrated a transformer-based Doherty amplifier that has a 3 dB bandwidth from 60 to 81 GHz in 40 nm CMOS. In other words, it is indeed possible to avaoid the transmission line.

As such, they are also possible to operate at "low" frequencies (perhaps not on-chip since the size of the inductors would be impossible to fabricate with a high enough quality factor to be any good). However, in modern integrated circuit processes, the transistors are so fast that it could be easier to build a switching architecture based amplifier, such as a Class D or any of it's alphabet-soup derivatives. They would allow for a much greater increase in efficiency, and in addition if you select the switching frequency high enough, the passives required can be much smaller, saving you area.

A photograph of the die mentioned earlier. Source is the homepage of the promoting professor, link.

I can't link in the picture and I don't want to re-upload it witout permission, so here is the link: link2

\$\endgroup\$
  • \$\begingroup\$ Thank you for the links. The problem remains that at low carrier frequencies, @ 100kHz, it would be very difficult to implement a 90 degrees phase shift component on chip \$\endgroup\$ – RAN Jan 18 '18 at 11:51
  • \$\begingroup\$ Ofcourse (though not impossible). However, as I already said, for operation at 100 kHz the Doherty architecture is likely not a good choice, and some for of switching amplifier is as it can do far better in terms of efficiency. There is a reason why pretty much every audio amplifier that isn't put in a higher-end system uses class-D-derived amplifiers. \$\endgroup\$ – Joren Vaes Jan 18 '18 at 12:06

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.