Trying to understand MMIC PA usage - Page 1
Forum section: electronics / rf-microwave

author: msat
author_rank: Regular Contributor
timestamp: October 15, 2021, 07:18:22 pm
content: I've been trying to learn RF electronics lately for a project. I'm at a point where my current focus is on power amps, and in an attempt to not get bogged down with designing a discrete unit, I've been looking at PA ICs. Actually, now I'm siding more towards complete (and cheap) eval boards. That said, there are a few specs and parameters that I'm having trouble wrapping my head around.First off, I've been seeing output power specs that don't seem to make sense for the specified load impedance and supply voltage. For example, the CMX90A004 (https://surf.cmlmicro.com/wp-content/uploads/2021/03/CMX90A004_ds.pdf) states "Output power 32.5 dBm @ 3.6 V" with an output matched to 50 Ohms. If 32.5dBm is ~1.8W, that requires a Vpp of 26.6V. So what am I missing here? Even if we say the 32.5dBm figure is with the amp operating in saturation, it would still take ~9.2V DC to produce 1.8W of power through a 50 Ohm load. So how do they figure the "32.5 dBm @ 3.6 V"??What are some other common snags I should be made aware of?Thanks!

author: MartinL
author_rank: Regular Contributor
timestamp: October 15, 2021, 09:14:37 pm
content: The AC output voltage of an RF circuit will often exceed its DC supply voltage. Current flowing into the output does not have to come directly from the DC supply. Nor does current flowing back, during the other half of the cycle, have to disappear straight to ground. Charge can be stored in capacitors, and energy can be stored in inductors.Consider what happens when you drive DC current from a 3.6V source into an inductor connected to ground, then disconnect the ground end and connect it to a capacitor instead. The stored energy in the inductor can force current into the capacitor and charge it to far above the input DC voltage. This is the principle used by DC-DC boost converters. They take a low voltage, high current input, and convert it to a higher voltage, lower current output. The power in and out is the same, less any inefficiency.The role of the output matching network in an RF amplifier is actually quite similar. The voltage at the drain of a common-source FET amplifier is constrained by the supply voltage, but the transistor can sink a very high current. So the transistor itself has a very low source impedance, e.g. a few ohms. The output matching network then steps this up to higher voltage, at lower current, to match a 50 ohm load.

author: KE5FX
author_rank: Super Contributor
timestamp: October 15, 2021, 09:27:37 pm
content: I think what's confusing is that the data sheet presents it as a broadband part with 50-ohm output impedance (presumably pure resistive), no matching network needed. If the output impedance is really 50 ohms, it is indeed impossible to turn 3.6V into almost 2 watts at a 50-ohm antenna.

author: MartinL
author_rank: Regular Contributor
timestamp: October 15, 2021, 10:07:02 pm
content: The matching network will be internal to the part. It's an MMIC, not a discrete transistor.Quote

author: KE5FX
author_rank: Super Contributor
timestamp: October 15, 2021, 10:10:38 pm
content: True, but I'm still surprised the math works out. The Q can't be more than 10, so getting 32 dBm out of 3.6 volts is interesting. I'd expect them to require an external boost regulator.

author: MartinL
author_rank: Regular Contributor
timestamp: October 15, 2021, 11:01:39 pm
content: I'll take your word on the max Q - I'm not familiar with what's practical in GaAs passive structures - but I don't follow why that Q constrains the output power relative to the supply voltage? The source impedance is presumably pretty low, and I'd guess there are multiple matching stages.The math seems reasonable to me? The datasheet gives a typical current consumption of 1A, meaning there's 3.6W going into the part at 3.6V. So 50% efficiency is needed to achieve 1.8W output with perfect output matching, or somewhat more than that to allow for matching losses.An efficiency of well over 50% is perfectly feasible for class C operation, which is surely what this PA will be operating in at full output. And that's fine for the applications it's targeted for, which use constant-envelope modulations.

author: radiolistener
author_rank: Super Contributor
timestamp: October 15, 2021, 11:12:37 pm
content: Quote from: msat on October 15, 2021, 07:18:22 pmSinceP = U * Iyou can increase power by increase voltage or current.There is no needs to design amplifier to 50 Ω load, because you can add matching circuit which will transform low impedance output into 50 Ω. This matching circuit will transform high current into high amplitude. The simple example of such matching circuit is a step-up transformer, it get high current on it's low impedance input and produce high voltage on it's high impedance output.So, the actual power supply voltage almost doesn't matters. The power consumption from power supply matters.

author: KE5FX
author_rank: Super Contributor
timestamp: October 15, 2021, 11:44:47 pm
content: Quote from: MartinL on October 15, 2021, 11:01:39 pmJust going by the usual center frequency/bandwidth relation. 100 MHz of bandwidth at 900 MHz simply suggests to me that there's not much energy storage in the tank circuit.Pretty cool/impressive that they are getting that kind of power out of it. I'm used to working with the sorts of broadband 50-ohm MMICs that run class A from DC to X band. With those, you get about the same RF voltage out that you put in.

author: rfclown
author_rank: Frequent Contributor
timestamp: October 16, 2021, 01:41:13 am
content: What MartinL said is correct. There are two things going on here. First is the magic of the RF choke in an RF amplifier (L1 in datasheet of your part). That makes the final transistor of the amplifier act like a boot converter, so you can get 2x the supply voltage swing across the output of the final transistor in the amplifier. So neglecting any losses, if we have 32.5 dBm out, the impedance at the final transistor output (using P=V^2/2R) would be 3.6 ohms (where V is peak voltage). So we then have to make a matching circuit to transform 3.6 ohms to 50 ohms. When you do that, and you measure the voltage across the 50 ohm load (assuming no losses) it would be (using P=V^2/2R) 13.3 peak or 26.6 peak-to-peak. Yes, it's way more that the DC supply voltage.So.. 2x the voltage comes from the RFC, the rest comes from the reactances in the impedance matching circuit.A must read paper for understanding RF amplifiers is from Steve Cripps: A theory for the prediction of GaAs FET load-pull power contours, in the 1983 IEEE MTT Symposium Digest.

author: MartinL
author_rank: Regular Contributor
timestamp: October 16, 2021, 05:59:43 pm
content: Returning to the OP's original question, I realised that there's a much simpler way to explain to a beginner how the AC output voltage of an RF circuit can exceed the DC supply:An RF output is an AC signal, which means you can put it through a transformer. A transformer lets you scale an AC signal to any voltage you like.Consider a simple, low frequency amplifier powered by 3.6V DC, such that the AC output has a peak-to-peak range of 3.6V at most.If you want it to put a higher AC voltage into your load, you can just attach a transformer at the output with a suitable ratio. Two turns on the primary side and 15 turns on the secondary will give you a voltage increase of 7.5x, so an output of 27V.And that's really all a matching network is - it's a transformer. We just often build them from inductors and capacitors, rather than pairs of coils. A transformer formed from inductors and capacitors can have a complex, frequency-dependent transfer function, rather than just a simple voltage ratio, so it can match two arbitrary complex impedances at a target frequency whilst helping to reject other frequencies.Coiled transformers of the traditional form are also used in RF work - especially when converting between two real impedances. They have the key advantage that they can have a very wide frequency range compared to a tuned circuit.In fact, on-chip planar transformers are possible to create in MMIC processes, and it's not impossible that something like that is used in this part. It would be one way to achieve such a wide bandwidth.

author: msat
author_rank: Regular Contributor
timestamp: October 21, 2021, 03:28:52 am
content: Thanks for the replies. The point made about converting the power transistor's output impedance to the final output impedance using the matching LC network (whether internal or external to the PA IC) which inherently necessitates increasing the output voltage makes sense to me, at least in a superficial sort of way. I glanced at the Cripps paper along with some other load-pull information, but it's way over my head. That said, I'm willing to accept the accuracy of the datasheet, after all, who am I to argue with it? I suppose any further interest I might have in this would be well spent trying to better understand matching networks. Up to this point, perhaps due to it's name, I assumed the RFC in the output transistor bias tee of various amp ICs and power transistors was merely to limit injection of upstream noise into the power supply. So it would seem that's not strictly true.