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I try to build a RF amplifier using the CGH40010F HEMT. The amplifier should work between 2.4 and 2.5 GHz and I would like to have at least 5 Watts of output power. While I was searching for information on designing such an amplifier, I found this article

http://www.highfrequencyelectronics.com/May11/HFE0511_Grebennikov.pdf

which I found very interesting and helpful. As far as I understood, the trick with class F is to connect a network to the drain of the transistor, and the network's impedance shall be infinity for odd harmonics, and zero for even harmonics, such that the voltage becomes a square wave. These two IEEE papers give nice information how the output power can be calculated:

F. Raab, “Class-F power amplifiers with maximally flat waveforms,” IEEE Transactions on Microwave Theory and Techniques, vol. 45, no. 11, pp. 2007–2012, 1997

and

F. Raab, “Maximum efficiency and output of class-f power amplifiers,” IEEE Transactions on Microwave Theory and Techniques, vol. 49, no. 6, pp. 1162–1166, 2001.

So, I designed my load network as follows

enter image description here

where the two radial stubs and the 90° lines present an open circuit at the operating frequency. The impedances Z1 and Z2 are chosen such that the 50 Ohms load is transformed to the impedance

$$ Z = \frac{4\,V^2}{6\,P} $$

where P is the desired output power. This equation can be derived using Raab's 2nd paper. I then simulated the network consisting of the 90° lines, Z1 and Z2 in Microwave Office. The transistor was neglected for this first simulation. The simulation shows that the network works as desired: a low impedance is presented to even harmonics, and a high impedance is presented to odd harmonics. At the fundamental, the impedance seen is as calculated above.

However, as soon as the FET is connected to the network, this can no longer work correctly because the FET has a parasitic capacitance from drain to source, which detunes the loading network. What can I do to compensate for the CDS? The article I linked above gives some hints, but I do not understand them.

ADDED INFO:

(from comments done by the OP)

This is what I simulated in MWO:

load network

The input impedance at port 1 is important. I need 38 Ohms at 2450 MHz, 0 Ohms at twice the frequency, and infinity impedance at the triple frequency, as shown here:

Zin at port 1

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From the article you've linked (p.70):

[...] the efficiency of the power amplifier can be limited by the transistor output capacitance [...] \$C_{ds}\$, [...] if this capacitance is not absorbed into multiharmonic load network without compromising the ability to properly terminate the second and third harmonic components.

So, first hint: you have to pre-detune the load network in such a way that, when connected to the HEMT, the output capacitance \$C_{ds}\$ tunes it in.

Let's see what does the author suggest for accomplishing this:

In Class F mode with a load network shown in Fig. 9(b), the electrical length of an open-circuit stub \$TL_3\$ is chosen to have a quarter wavelength at the third harmonic to realize short-circuit condition at the right-hand side of the series transmission line \$TL_2\$, whose electrical length \$\theta\$ should provide an inductive reactance to resonate with the device output capacitance [\$C_{ds}\$] at the third harmonic.

FIGURE 9:

load network

So the author suggested method is:

  1. Select a length for stub \$TL_3\$ so that it "grounds" (short-circuits) the right side of line \$TL_2\$ at \$3\omega_0\$. The author selects a "small" electrical length \$\theta=30º\$ at \$\omega_0\$ that becomes \$\theta=3 \cdot 30º=90º\$ at \$3\omega_0\$, that is a quarter wavelength that effectively transforms the open circuit at the end of the stub into the short-circuit desired at the right side of line \$TL_2\$.

  2. Thanks to this, now we have two reactances in parallel: the HEMT output capacitance \$C_{ds}\$ and a \$TL_2\$ equivalent lumped reactance (let's call it \$X_{TL_2}\$).

  3. Next thing to do is choosing a length for \$TL_2\$ so that its reactance \$X_{TL_2}\$ is inductive at \$3\omega_0\$ and resonates with \$C_{ds}\$, effectively cancelling it out (=open circuit).

  4. Now that \$C_{ds}\$ has been taken care of, what's left behind? We can ignore \$R_L\$ because it was short-circuited by stub \$TL_3\$: one problem less. Only \$TL_1\$ remains...

  5. But \$TL_1\$ happens to have an electrical length of \$\theta=3 \cdot 90º=270º\$ at \$3\omega_0\$ and is "grounded" i.e., terminated with a short-circuit due to the supply decoupling capacitor. So we have a three quarter wavelength line transforming a short-circuit into an open circuit.

  6. At the end everything comes together, and the HEMT "sees" an open-circuit at \$3\omega_0\$, as desired.

The author then does his math and comes up with a formula for the electrical length (at \$\omega_0\$) of \$TL_2\$, which is the element that compensates \$C_{ds}\$ and tunes in the load network:

$$ \theta = \frac{1}{3} \tan^{-1}{(2 Z_0 \omega_0 C_{ds})} $$

BEWARE!!!:

The formula above doesn't match my observations using the Smith chart, in fact the formula seems to be plain wrong! So I've derived my own formula (and then checked it against the Smith chart) as follows.

The shunt reactance of capacitor \$C_{ds}\$ at \$3\omega_0\$ is:

$$ X_{C_{ds}} = \frac{-j}{3 \omega_0 C_{ds}} $$

The shunt reactance synthesised by \$TL_2\$ at \$3\omega_0\$ is:

$$ X_{TL_2} = j Z_0 \tan \beta l $$

The condition for resonance (the parallel of both reactances going to infinity thus becoming an open circuit and canceling each other out) at \$3\omega_0\$ is \$ X_{C_{ds}} + X_{TL_2} = 0 \$, so:

$$ \frac{-j}{3 \omega_0 C_{ds}} + j Z_0 \tan \beta l = 0 \\ Z_0 \tan \beta l = \frac{1}{3 \omega_0 C_{ds}} \\ \beta l = \tan^{-1} \left( \frac{1}{Z_0 3 \omega_0 C_{ds}} \right) $$

And we finally rescale the electrical length back to \$\omega_0\$ ending up with:

$$ \theta = \frac{1}{3} \tan^{-1} \left( \frac{1}{Z_0 3 \omega_0 C_{ds}} \right) $$

Thus, just use the formula above for adjusting the length of \$TL_2\$ and everything should be fine. We'll check it later on in the example below.

WHAT'S NEXT?

Once you have tuned the harmonic behaviour at \$3\omega_0\$ you're free to turn your attention to the fundamental again. You'll find that after having adjusted \$TL_2\$ and \$TL_3\$ at \$3\omega_0\$, the impedance seen at the transistor output may be very different from \$R_L\$ and could require matching.

But this can be easily corrected by using a fundamental load matching network between the right side of \$TL_2\$ and the load. And what's best: we can adjust this new matching network without detuning the harmonic behaviour at either \$2\omega_0\$ or \$3\omega_0\$. Why? Because \$TL_2\$ is short-circuited at \$2\omega_0\$, effectively "isolating" the part of the circuit located at its right. And the same happens with \$TL_3\$ at \$3\omega_0\$.

So we can do fundamental matching however suits our needs better. See my example below, for which I've used a simple line + stub so that the impedance seen by the transistor output is \$R_L=50\Omega\$ again. You can match it to whatever you need.

EXAMPLE

We start by observing that \$C_{ds}\$ does indeed detune the load network at \$3\omega_0\$:

3rd harmonic: detuned

And it also degrades matching at \$\omega_0\$, moving \$Z_{in}\$ away from its desired \$38 \Omega\$ value:

fundamental: mismatching

So the first thing to do is adjust the electrical length \$\theta\$ of \$TL_2\$ in order get the open circuit at \$3\omega_0\$ back.

We use the correct formula (not the one from the article linked by the OP) and find that we need \$\theta=5.43º\$. This electrical length does get the open circuit at \$3\omega_0\$ back!

However, \$\theta=5.43º\$ is way too short, so we add an extra 60º (equivalent to a half wavelength section at \$3\omega_0\$, a full turn of the Smith chart) and just stick with \$\theta=65.43º\$:

3rd harmonic: open circuit

But this also changes matching at \$\omega_0\$, doing nothing to improve the mismatching situation already caused by \$C_{ds}\$:

fundamental: mismatching 2

So the next thing to do is adding a fundamental matching network. A simple line + stub will be enough to get back to \$Z_{in} \approx 38 \Omega\$:

fundamental: matched

Last thing to do is check that the open circuit at \$3\omega_0\$ has not been affected by our fundamental matching network:

3rd harmonic: open circuit 2

So the harmonic behaviour remains unaffected indeed! Mission accomplished.

NOTES BELOW

N.B. 1: I'm no Class-F expert.

N.B. 2: As your main concern seems to be the detuning effect due to \$C_{ds}\$, I'm not looking into the effect of the transistor package lead inductance, \$L_{out}\$.

package lead inductance

N.B. 2: To make things trickier, be aware that \$C_{ds}\$ behaves often as a non-linear capacitor (which doesn't seem to be a bad thing itself if you read the Class-F literature: quite the opposite).

non-linear capacitor

Sources:

A simplified broadband design methodology for linearized high efficiency continuous Class F power amplifiers.

Behaviors of Class-F and Class-F\$^{-1}\$ amplifiers.

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  • \$\begingroup\$ This is what I tried, I also saw that equation in the paper, but unfortunately, the MWO simulation showed that the pre-detuning did not work properly, the open-circuit at the 3rd harmonic was too low in frequency. N.B.: I am no class F expert as well, but I wonder whether a class F expert could verify whether my assumptions in my first post (the calculation of Z) are correct. \$\endgroup\$ – T. Pluess Apr 6 '17 at 7:06
  • \$\begingroup\$ Mmmh... Have you tried to sweep the electrical length \$\theta\$ of \$TL_2\$ in your simulation to get some insight about its behaviour? Could it be that the actual problem here isn't \$C_{ds}\$? \$\endgroup\$ – Enric Blanco Apr 6 '17 at 8:14
  • \$\begingroup\$ This is what I simulated in MWO: imgur.com/a/Ds6oD the input impedance at port 1 is important. I nee 38 Ohms at 2450 MHz, 0 Ohms at twice the frequency, and infinity impedance at the triple frequency, as shown here: imgur.com/a/301UX However, as soon as I add Cds, you are right, I can tune the length of TL2 such that the point of infinite impedance is again correct, but then my 38 ohms are way too low, Zin decreases to about 20 Ohms, which is not what I want. \$\endgroup\$ – T. Pluess Apr 6 '17 at 10:59
  • \$\begingroup\$ I've updated my answer. You just need to add a new matching network at \$\omega_0\$ so you can get your \$Z_{in}\$ back to \$38\Omega\$. \$\endgroup\$ – Enric Blanco Apr 6 '17 at 15:43
  • \$\begingroup\$ IMPORTANT NOTICE: the formula for the electrical angle \$\theta\$ of \$TL_2\$ in the article you linked seems to be WRONG!! I've updated my answer with a correct (and checked) formula. \$\endgroup\$ – Enric Blanco Apr 7 '17 at 0:25

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