Hi, my name is Matt Ozalas, and I'm an RF engineer at Keysight Technologies.
In today's talk, I'll go over the basics of how to design RF power amplifiers.
I think that this material should give you a good foundation
for understanding how these types of circuits work.
If you understand the material that I'm going to cover today,
you'll also be able to build off of it to understand more advanced topics
in the field of power amplifiers.
The objectives of my presentation are first to define power from an AC perspective,
and then demonstrate how power is generated and dissipated in a real power amplifier circuit.
Next, we'll look at power amplifier operation from an intuitive point of view.
I'll build off this to define and discuss efficiency.
From there, we'll look at how to set up a loadline,
and how this is used in PA design.
I'll even share my Workspace with you at the end
so that you can explore the concepts that I'm going to talk about today at your own pace.
Now, part of the Workspace that I'm making available
will be an interactive waveform generator,
which will allow you to design, generate, and analyze your own RF waveforms in real time.
This will help you to understand these key concepts that we're going to talk about today.
All of the waveforms and examples that I'm going to show in this presentation
were generated using this waveform tool.
To understand RF power amplifiers, we need to begin with an understanding of power.
Most RF engineers are familiar with power from a frequency domain perspective.
To be successful as a PA designer,
you'll also need to understand power from a time domain perspective.
Here, I'm showing a set of sinusoidal voltage and current waveforms in the time domain.
Each of these can be described simply by sine wave equations
with an amplitude offset and a phase offset.
We can obtain the average power from this by multiplying these waveforms together
and calculating the area under the resulting curve,
and then dividing by the time.
Now, for the waveforms described by these equations,
the interval works out to be simply a multiplication of one half of the peak values
times the cosine of the phase offset between the voltage and current.
The peak values are, of course, half of the peak-to-peak values,
so we can also express this equation in terms of peak-to-peak waveforms too.
Knowing both expressions for power can be quite useful.
Here, I have a very simple circuit setup in the simulator.
This is just a frequency domain power source.
I'm going to combine this signal with a DC bias through a bias tee.
Then, I'll run a harmonic balance simulation on this.
In the data display, I'll look at both the frequency domain calculation
and a time domain calculation for power.
In the frequency domain, since I have both a DC and an RF component,
I'll get a spectrum at the output for voltage and current.
If I multiply the voltage and current spectra together,
conjugating the current, I'll get a power spectrum.
You can see that both the DC and RF elements are 0.5 W.
Now, it's also easy to isolate the fundamental frequency component, as I'm showing here.
If I want to look at only the RF power,
I can easily do that in harmonic balance by just indexing it,
and it turns out that this is a 0.5 W.
In the time domain, I'll start by multiplying the voltage and current together.
Then, I'll integrate the product and divide by the time to get power.
In this case, I have 1 W of power.
By the way, I get the same result if I just multiply the peak voltage and peak current
and divide by 2.
In other words, this integration produces the total power in the entire spectrum,
which is a combination of both RF and DC power.
What if I only want the RF power?
Well, if I consider the peak-to-peak voltage and current signals instead of the peak signals,
then I'm left with only the RF power.
In the time domain, using this approach can allow you to easily separate RF from DC.
Don't worry if you didn't catch exactly how I calculated all these things.
Again, I'll provide a way to download this Workspace at the end of the presentation,
and you can see exactly how I calculated the power in these different ways.
The phasing between the voltage and current waveforms
is really important for power.
This is a cosine function.
If I sweep the phase angle between the voltage and current waveforms
from 0 to π, or 180 degrees,
turns out that there's a maxima and a minima in the power,
which just arises from the cosine.
Now, at the maximum point, the voltage and current waveforms are in phase.
This means that power is getting dissipated.
In practical transistors, power dissipation acts to convert energy into heat.
Now, on the other side of the graph,
we have the case where the voltage and current waveforms are out of phase.
In this case, we say that the device is generating power.
In the middle of the plot, power is neither dissipated nor generated.
Instead, it's being stored.
Here's an example of a circuit that we're going to use a lot in power amplifier design.
This is an AC current source which is tied to a resistor.
Since these components are connected in parallel,
the current source and the resistor share the same voltage across them.
Since current source pulls current out of the resistor,
the currents are opposite in the two components.
For the resistor, the voltage and current are always in phase
because they're simply related through Ohm's law.
For the current source, since the current is going in the opposite direction as the resistor,
the voltage and current are out of phase.
The current source is the power generator.
The resistor is the power dissipater.
You can determine this simply by looking at the sine of the voltage and current product.
If it's negative, it's a power generator.
If it's positive, it's a power dissipater.
Now, in any physically realizable AC circuit, there must be conservation of energy.
Any power that's generated must either be stored or dissipated somewhere.
For this simple example,
the power generated in the source is equal to the power dissipated in the resistor,
with nothing stored.
Practically speaking, this type of circuit can be realized physically
using a transistor connected to a resistor.
Here, I have a small-signal transistor model, like you might see in a textbook.
It takes the voltage at the input, and then multiplies this by a gain factor,
and converts it to a current at the output.
Like I showed before, the voltage across the load and the current source of the device are the same,
at least for AC signals, while the currents are going in the opposite directions.
There's a little bit of a problem here, it takes energy to get gain.
To provide this energy, we need a DC power source, which will create a bias on the device.
This DC bias will mean that the sinusoidal waveforms
are centered around the DC operating point instead of 0.
We can do a simple nodal analysis at the output
and show that the current that's supplied by the source
is the summation of the current through the device and the current through the load.
Or, to put this another way,
all of the power that's used in the circuit to provide gain and output power
comes entirely from the DC source.
Let me illustrate this.
First, we have to note that the starting point at the current source is the DC offset,
which is provided by the power supply.
Let's see what happens when the input sinusoidal voltage goes negative.
Well, that causes the current through the device to also go negative.
Depending on the bias current, this can go all the way down to 0.
Now, since less current is getting pulled into the current source,
instead, it flows in an AC sense from the power supply into the load.
This causes the current in the load to increase.
Since the load is a resistor, that means that the voltage in the load increases too.
When the voltage in the load increases,
the voltage across the device's current source also increases.
If the input signal goes positive,
then the current source draws more current from the supply.
This is more current than the supply is providing in a DC sense.
Where does that extra current come from?
Well, it gets pulled through the load from ground.
This decreases the voltage at the load, which decreases the voltage at the device.
Now, keep in mind that there's also a DC voltage bias on the device too.
The voltage here might be going from the DC value down to 0.
When I say they're coupled, I really mean they're coupled in an AC sense.
This is fundamentally how a power amplifier works.
There's one more thing that we need to consider.
Since there's a DC offset across the device, power is not just being dissipated in the load,
it's also being dissipated in the current source of the device.
Let's take a closer look at this.
Here, I have voltage and current sinusoids.
These are offset by DC bias components.
Here's the DC voltage, and here's the DC current.
These sinusoids are swinging from a peak of twice the DC value to a minimum of 0.
We can also show the DC power directly on this plot.
Obviously, it's constant over time.
Since these waveforms are always positive, to find the dissipated power,
we simply multiply the waveforms together and integrate.
The result will be positive.
This represents the portion of the DC power
which is dissipated inside of the device at the current source.
Now, if we remove the DC offsets from the RF waveforms,
it's then easy to calculate the power generated
because we can just multiply the signals together, and the result should be negative.
Now, since energy is conserved,
it also shouldn't be surprising that the area of the dissipated power,
combined with the area of the generated power
is exactly equal to the DC power put in by the supply.
What I'm showing here are the waveforms for what's called a class A power amplifier.
I just proved for this type of PA,
at best, half of the DC power will be converted to RF power.
As you might guess, this amplifier will be, at most, 50% efficient.
The concept of efficiency for power amplifiers should now be quite clear.
At the output, the efficiency is defined as the generated power
over the sum of the generated and dissipated power.
Or, in other words, the generated power over the DC power.
This is often expressed as a percentage.
Using the interactive waveform generator, which you can also download,
it's possible to see what different types of waveform configurations will give you,
in terms of generated power, dissipated power, and efficiency.
It even tells you if a waveform is non-physical.
In other words, it tells you if you have a waveform
where the generated power exceeds the DC power.
This is a waveform that you can't build in the lab.
Finally, let's talk about how to pick the load resistor.
To start, I'll take the configuration that I showed earlier,
with current source and a resistor, and I'll sweep the current in a DC sense.
Then, I'll plot the current versus the voltage, current on the Y-axis, voltage on the X-axis.
I get a simple negatively sloped line.
Now, the magnitude of the slope is equal to one over the resistance,
since the slope is Y/X, or current over voltage.
Let's expand this simple analogy to a real device.
For example, here, I have a CMOS transistor,
which is similar to the small-signal model that I showed earlier.
If we take a look at current versus voltage at the drain node of this transistor,
we get a family of curves,
where each curve is the output response for a different input voltage.
To make this circuit operate correctly, we need to bias the transistor.
Let's pick a DC bias point in the middle of the voltage and current range.
Something like 250 mA and 5 V will work for this case.
Now, if I go back to my current source,
and I add this fixed bias current into my current source
and redo the DC sweep, you can see that the scales on the curve change.
Now, the line, instead of being centered around a negative point, is centered at 0.
0, in this case, is the bias point of the transistor.
We can just superimpose the two curves.
This is what is called a loadline plot.
It's very commonly used in PA design.
The load resistor is really setting the ratio between the peak-to-peak voltage
and the peak-to-peak current.
It's important to keep in mind that this is also constrained by the device.
For example, there's a knee voltage in practical devices
which prevents the voltage waveforms from swinging all the way to 0.
This needs to be considered in the design as well.
If you want to learn more about the material that I discussed today,
I'm making this Workspace available for you to download.
It covers everything I talked about today,
from calculating power directly from time domain waveforms,
to drawing a loadline.
There's some more advanced capability here too.
For example, you can set the bias current up to rectify,
which will allow you to explore waveforms for more advanced classes of PA operation,
like class AB, and see how this improves efficiency.
I think it's a great way to master these basic concepts.