From MFJ’s web site listing:
Connects directly to the transmitter with PL-259 connector. No patch cable used, reduces SWR. Finned aluminum, air-cooled heatsink. Handles 100 Watts peak, 15 Watts average. 50 Ohms. Covers DC to 500 MHz with less than 1.15:1 SWR. 1 ⅝” round by 3″ long.
That is pretty stunning for a device with a UHF connector, more on that later.
Bogard wrote a review of the device, making some VSWR measurements using a spectrum analyser with tracking generator and a VSWR accessory (a directional coupler).
The VB1032 VSWR bridge specification directivity is modest at 30dB. That challenges making really low VSWR measurements (as the article does) with low uncertainty.
Above is a calculation of the uncertainty in measurement for indicated VSWR=1.15 using a 30dB directivity coupler, the actual VSWR uncertainty range is 1.08 to 1.23.
Note that the test equipment does not use UHF connectors, so there was some kind of N(M) to UHF(F) adaption used, but no detail.
Above is the reported VSWR response.
In summary, it is pretty good at low frequencies and rises fairly uniformly to just over 1.5 at 500MHz… so on the surface of it, it fails to meet spec in a big way.
UHF connectors do not have a controlled characteristic impedance, and experience is that most UHF connectors are well represented by a transmission line section with Zo in the range 35-40Ω.
Realise that there are at least two cascaded UHF connectors in this test setup, and that the effect of a very short series line section of low Zo is to cause InsertionVSWR.
Let’s look at a model in Simsmith to illustrate the effect.
Above, the Simsmith model that models the UHF connectors as a 37mm length of line with Zo=35Ω and VF=0.66. These values were arrived at by calibrating the model to the published measurements.
The model assumes that the load has little reactance at low frequencies, but the resistance is wrong and causes the low frequency VSWR=1.045. This gives two choices.
Calibration included adjusting the load resistance for low frequency VSWR and the transmission line parameters to achieve the shape, slope and 500MHz VSWR.
It is not a perfect fit, but it gives a very plausible possible explanation the measurements.
So, the question is how much of the non-ideal behavior is due to the test fixture and how much is in the DUT? It is possible that although the DUT has a UHF connector, that the effect of the connector has been compensated inside the DUT and that at some reference plane, it meets the stated specifications. The very slight wavy nature to the VSWR response might hint some level of internal compensation.
So, (Bogard 2021) is a bit unsatisfying. On the surface it shows non-compliant VSWR, but does not address it leaving one wondering about whether a significant part is due to the test setup (coupler directivity and UHF connectors / adapters).
A reader of A common scheme for narrow band match of an end fed high Z antenna commented:
…if the coil is tapped at 1/3, surely then the coil is a 1:3^2 or 1:9 transformer and the capacitor simply ‘tunes out’ the coil reactance, what is the input impedance when it has a 450+j0Ω load?
That is very easy to calculate in the existing Simsmith model.
Above, with load of 450+j0Ω, the input impedance at 50MHz is 8.78+j34.36Ω (VSWR(50)=8.4), nothing like 50+j0Ω.
As for tuning out the reactance…
Somewhat more capacitance does result in a lower VSWR(50), but at 1.6, you could not claim it is matched.
This is not simply a 1:9 transformer by any stretch of the imagination.
As mentioned in the referenced article, solution of these circuits is not intuitive, the amount of flux leakage is critical to behavior, and that depends greatly on the geometry of the coil.
]]>A common variant shows not capacitor… but for most loads, the capacitance is essential to its operation, even if it is incidental to the inductor or as often the case, supplied by the mounting arrangement of a vertical radiator tube to the mast.
A similar arrangement works for vertical radiators around a λ/2, around means from say 3λ/8 to 5λ/8. All of these present an impedance of some fairly high resistance in series with a significant reactance. Because the current at the feed point is relatively low, these are often used as so-called ground independent antennas.
Above is a Simsmith model of such an arrangement. The load L is from an NEC-4.2 model of a 50MHz 5λ/8 vertical over four λ/4 radials in free space.
The tapped coils is a uniform air cored solenoid of radius 10mm, pitch 3mm, 6t tapped at 33% (ie 2t).
C1 represents the equivalent capacitance of self resonance, equivalent mount capacitance, and possibly some additional capacitance.
Though it seems every ham can explain how this circuit works, there is nothing intuitive about the design of the tapped inductor.
The inductor design was assisted by a little python script which found the tap point and c1 for a given inductor, load, etc.
zl: (59.49-175.4j) pitch: 0.00300 m, radius: 0.01000 m, turns: 6.00, length: 0.01800 m, Ql: 300, Qc: 1000 VSWR=1.00009, tap=32.994 (%), c1=4.91618 (pF)
From that starting point, you could play with Simsmith interactively changing the coil parameters and adjusting things for a new match.
#!/usr/bin/env python # coding: utf-8 import math import cmath from scipy.optimize import minimize from scipy import constants f=50e6 zl=59.49-175.4j #network qc=1000 ql=300 #tapped coil p=0.003 r=0.01 n=6 def wcf(n,r,l): return constants.mu_0*n**2*r*(math.log(1+math.pi*r/l)+1/(2.3004+1.622*l/r+0.4409*(l/r)**2)) def vswr(args): tap,c1=args lt=wcf(n,r,l) l1=wcf(n*tap,r,l*tap) l2=wcf(n*(1-tap),r,l*(1-tap)) m=-(lt-l1-l2)/2 rlt=2*math.pi*f*lt/ql y=1/zl y=y+2*math.pi*f*c1*(1/qc+1j) zl2p=rlt*(1-tap)+2*math.pi*f*(l2-m)*1j z=1/y+zl2p y=1/z yl1p=1/(rlt*tap+2*math.pi*f*(l1-m)*1j) y=y+yl1p z=1/y zm=2*math.pi*f*m*1j z=z+zm y=1/z gamma=(z-50)/(z+50) rho=abs(gamma) vswr=(1+rho)/(1-rho) return vswr l=n*p #starting guess tap=0.5 c1=20e-12 x0=[tap,c1] res=minimize(vswr,x0,method='Nelder-Mead') print() print("zl:",zl) print('pitch: {:0.5f} m, radius: {:0.5f} m, turns: {:0.2f}, length: {:0.5f} m, Ql: {:0.0f}, Qc: {:0.0f}'.format(p,r,n,p*n,ql,qc)) print() print('VSWR={:0.5f}, tap={:0.3f} (%), c1={:0.5f} (pF)'.format(res.fun,res.x[0]*100,res.x[1]*1e12))
Above is the code, it elaborates the solution in small steps so that it is easier to understand. The script does not do much error checking, it will produce valid results on sane input.
If you play with the Simsmith model, you may find that there is a solution with c1=0 for some load impedances, but in reality, the incidental capacitances mean that c1 is unlikely to ever be exactly zero.
For the enquiring mind, the relevant files are attached: EFmatch.zip.
]]>The design is for a system power output of about 80W on a 24V supply, it is a combination that should work with practical system components with good efficiency.
Above is a first step, an estimate of an initial load line for the PA. The calculator is written in valve terms, but is quite applicable to this scenario.
Let’s talk in terms of MOSFET devices. So, we expect that the drain to drain load needs to be around 12.6Ω, that is close to 50/4=12.5Ω and would suit a 1:2 turns ratio broadband transformer, so we will base the Smith chart on Zo=12.5Ω.
The article assumes the reader has a good understanding of these type of amplifiers.
This article assumes familiarity with Simsmith, including its right to left signal flow, and concepts of complex numbers, impedance, admittance, circuit analysis, transformer equivalent circuits etc.
The model is a simple approximation of the behavior of the PA and makes some assumptions to make a very complex system into a simpler approximation that leads to an understanding of the main influences on system power out.
The model assumes ideal Class B operation of an ideally linear active device with conduction angle 180°. The conduction angle in Class AB is typically so close to 180° to make little difference to the calculated result, insignificant in terms of other uncertainty.
The model assumes an ideal linear transfer characteristic. The error of this assumption is small in estimating power output at the fundamental, but relevant to IMD / harmonic content… which is outside the scope of this model.
That said, the real test of linearity is an Intermodulation Distortion (IMD) test, statistics like 1dB compression power (P1dB) are pretty weak. The analysis presented hear avoids voltage or current saturation, slightly higher power might be obtainable at acceptable IMD figures.
The nature of the active device, whether transistor or FET is that with a given level of drive, it tends to produce a collector or drain current independent of load impedance. Constraints on output power in linear operation are:
With adjustable drive available (eg using ALC), we can consider that the output device is a nearly ideal voltage source Vmax=Vdd-Vsat able to supply current up to some Imax.
In the Simsmith model, that has been specified in the Zo sub element as
Vdd; Vsat; Imax; absV(Min(Vmax*2^-0.5,Mag(Zin)*Imax*2^-0.5));
where the first three items are declarations of variables so they appear first in the list below. The important part is that it sets the source G to a RMS voltage that is the lesser of \(\frac{V_{max}}{\sqrt2}\) and \(\frac{|Z|I_{max}}{\sqrt2}\). Don’t be confused by absV, it does not set the source absolute voltage or amplitude, but the RMS voltage.
If you have notions of an idealised Thevenin or Norton source, they are not a good fit to this design problem.
The Plt property of G specifies some calcs and plots.
//Plots Vmax=2*(Vdd-Vsat); Idc=2*Mag(I)*2^0.5*2/Pi; efficiency=G.P/(Vdd*Idc)*100; Plot("CoreLoss (Wc)",Core.R1.p,"Wc",y1); Plot("Dissipation (W)",Vdd*Idc-G.P,"PWR"); Plot("Efficiency %",efficiency,"Eff",y2);
Vmax is calculated from Vdd and Vsat, and Idc is calculated from G.I using the factor for ideal push-pull Class B operation. Two plots are specified, one of core loss and one of dissipation of the active devices being the difference between DC input power and RF output power. Active device efficiency is also calculated in %.
Above is an example set of plots for a contrived scenario to show departures from ideal PA behavior. There is a distinct roll-off below about 3MHz and a sharp change around 1MHz, and a broad efficiency roll-off above about 10MHz another sharp change in power out around 26MHz. We will discuss the cause of these later.
C2 models the shunt capacitance at the drain or collector of the active devices. This is an average approximation of what is actually a capacitance that varies instantaneously with voltage.
The Tfmr element models a transformer with a lossless ferrite core, it assumes very high flux coupling factor (which you can tweak). It is used with the Core element which has precedence setting some of the values.
(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)
The core element models the ferrite transformer core, and feeds the Tfmr with key values.
Above is the schematic of the Core element.
The core element models the ferrite core referred to the secondary side as a shunt admittance Ym. The susceptance and turns ratio are used to calculate the primary and secondary inductances of the Tfmr element, and the conductance creates a shunt element to approximate the core loss of the transformer referred to the secondary (LHS). It precedes Tfmr so that it sets Tfmr values prior to execution of Tfmr.
Below is the content of the Core element code.
//Core model //Calculates magnetising admittance from mu data. //Updates Tfmr Hl, Hr. $data=file[]; // core mu aol; Ns; cores; Cse; $u1=$data.R; $u2=$data.X; //u1=$u1; //u2=$u2; Ym=(2*Pi*G.MHz*1e6*(4*Pi*1e-7*$u1*aol*cores*1e9)*Ns^2*1e-9*(1j+$u2/$u1))^-1; Tfmr.Hl=1/(2*Pi*G.MHz*1e6*-Core.Ym.I); Tfmr.Hr=Tfmr.Hl*(1/Core.Ns)^2;
The input values are:
The C1 element models the Tfmr self resonance as an equivalent shunt capacitance.
The L element is the nominal load, for most cases this will be 50+j0Ω. You can change it to observe the sensitivity of the design to variations in load impedance.
The example scenario is contrived to exhibit some departures from ideal for the purpose of discussion.
Above is a Smith chart of the impedance seen by source G normalised to Zo=12.5Ω. At O, freq=5MHz, it is pretty close to the target 12.5Ω. Power out is 84.5W (remember this should be followed by a LPF), efficiency is 75%, close to the theoretical efficiency of an ideal Class B stage.
Core loss is around 15W which is quite high. The design needs lower core loss.
Efficiency rolls off below 3MHz due to increasing core loss, magnetising admittance is too high. The design needs lower core loss, first point would be to address the core selection, cross section, number of cascaded cores, and material characteristics. Change the cores value and observe the result.
The sharp change in power output near 1MHz is due to current saturation, the active device cannot produce the current required to drive the load with the growing magnetising admittance. Increasing MOSFET size and drive can solve this, but that does not solve the attendant efficiency problem.
Efficiency roll-off above 10MHz id due mainly to the increasing effect of C2, but flux leakage in Tfmr contributes also. Fix this by choosing MOSFETs with lower capacitance, and making sure the transformer has low flux leakage (high permeability, structure).
The sharp change in power out near 26MHz is again due to current saturation.
Core loss varies with the magnetising current and depends on core characteristics including the frequency dependent complex permeability. Select material and geometry to improve core loss.
Efficiency is degraded whenever the MOSFETs are not operating at fully voltage swing, so presentation of low load impedance might allow acceptable output power at acceptable IMD, but at degraded efficiency. It is worth measuring efficiency as an indicator of correct setup of the stage.
If you design for rated power into a load impedance with some tolerance, eg 60+j0Ω, that will result in reduced efficiency with a 55+j0Ω load. That is just part of the cost of a tolerant design so that half the units coming off the production line don’t fail to make rated power. In the same vein, the MOSFETs need to be able to drive a 45+j0Ω load. The Simsmith model allows exploring these variations.
Above is an overview of the model applied to a commonly eBay listed nominal 70W Chinese PA. Yes, it falls short of the advertised 70W… but the fine print says that is on 16V… so try it.
But overall, the expected performance is quite good from 1.5-30MHz, and core loss is very low. (Note this core loss is based on VK1EA’s measurement of a single core, there is no assurance the cores used in products sold are of the same type.)
The model will often predict lower power than some products claim. It is likely that they depend on driving the PA into saturation to achieve their stated power, causing degraded IMD performance (rarely stated).
The model provides a framework for exploring the second stage of a design process. At the end of the day, testing some prototypes is necessary to complete a design.
RfBPaSs.zip contains the two Simsmith v18.1g model files and their dependent csv files. The component files are available on github where updates will be posted. They are offered for study without any warranty.
]]>This article offers an explanation of how the the alternative output circuit at Fig 5 of EB104 works.
Let’s look at the schematic diagram of the PA.
Above is the schematic from EB104, of interest for this article is the output circuit comprising T2 and T3 which are intended ideally to provide a drain to drain load of 50/9=5.55Ω.
Let’s run a simple initial load line model of the amplifier (albeit in tube terminology).
Above, we find that if we assume a saturation voltage for the MOSFETs of 0V, a 40.823V supply delivers 600w into exactly 5.555Ω load for ideal Class B. As noted, 5.555Ω is 50/9, so an ideal transformer with 1:3 turns ratio should provide that.
The drain current in an ideal Class B stage is a half wave of sine wave. If we were to perform a Fourier analysis of such a wave of say 1A peak amplitude, we would find that the fundamental RF component is 0.5A. (There are higher order harmonics, but they are not of interest to us in our frequency domain analysis at the fundamental frequency.)
In this case, the calculator results tell us that the drain current amplitude is 29.40A and the amplitude of the fundamental RF current is 14.70A or 10.39Arms. Both FETs are driving such a current but of opposite phase.
Above is a diagram of the ideal Class B current wave shape, fundamental RF component and DC component (albeit in tube terms).
Let’s deal with T2, the magnetising impedance is sufficient that we will ignore the current flowing in its windings for an approximate solution.
Now to the alternative arrangment shown in Fig 5.
Granberg offers a simplification of the alternative to T3 as a cascade of a transformer and 1:1 Guanella balun.
Performing a frequency domain analysis of that simplified circuit, we have two current sources supplying a sinusoidal current to the coupling network.
Let’s annotate the diagram with currents relative to the current into the 50Ω load.
We can simplify the analysis by making some assumptions and depending on coax with well developed skin effect in TEM mode:
Above is from Fig 5. Starting at the right we can annotate current into the upper load terminal (the open arrow) and then observing the rules above, we can annotate currents through the whole network, arriving at the conclusion that the current flowing from the current sources representing the MOSFETs is 3I or three times the 50Ω load current, and therefore the impedance transformation is 1:3^2 from source to load, and the drain to drain load impedance is \(Z_{dd}=\frac{50}{3^2}=5.555 Ω\).
]]>The above pic from an eBay advertisement of the 2020 version of the PA would suggest very strongly that there are three turns on the secondary of the output transformer, and a half turn on each drain. Interestingly the 70W versions also appear to use three turns, alarm bells ring!
Here is a little table that shows the maximum power obtainable with a transformer of this type for various supply voltages and secondary turns.
You would usually design for rated power to be a little less than shown in the power cell for a particular combination. For example, if you wanted 100W out on 16V supply, you might choose a four turn secondary (which is capable of somewhat higher power). As can be seen, three turns is probably ok for up to 75W on 16V. The danger in undersizing (low Nsec) is risk of saturation and resulting non-linearity / intermodulation distortion.
Let’s do some thinking.
Above, an initial load line estimate suggests the required load impedance drain to drain is 3.2Ω, which can be achieved with a Z=16:1, n=4:1 ideal transformer from 50Ω giving nominally 3.125Ω.
Let’s look at a Simsmith model that brings the transformer core to book, and models some other practical aspects of a transformer.
Powered from a 13.8V supply and into a 50Ω load, we just make 100W… using a four turn secondary on T2. Note that this is analysis of an ideal Class B amplifier, though with the stated limits on voltage swing (including an allowance for transistor saturation), and it should not be interpreted to mean that higher power could not be obtained by driving it into non-linearity.
(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)
Note that this module does not include the necessary output filter which will lose 5-10% of the power from this module.
There is more than a little deception in advertising of these things.
It seems sellers do not offer a schematic, manual, or any assurance of the transistor type fitted… prior to or after sale.
One listing of a kit with used transistor and no heatsink states 5.max output power: 100W(13.8V/80W;16V/100W).
Apparently aware of the risk of underachieving. Models suggest that the amplifier and appropriate LPF should deliver towards 120W on 16V supply.
But, they are apparently not short of buyers. A case of caveat emptor, the usual issues of trust and China exist in spades.
More to come…
]]>The PA uses a MRF9180 dual MOSFET operating on 26V supply.
Above is the prototype PA. The text states very clearly that the output transformer uses a secondary of two turns of PTFE insulated wire, and the pic above does not provide evidence to the contrary.
Hmmm, experience suggests that may be too few turns.
Here is a little table that shows the maximum power obtainable with a transformer of this type for various supply voltages and secondary turns.
You would usually design for rated power to be a little less than shown in the power cell for a particular combination. For example, if you wanted 150W out on 26V supply, you might choose a three turn secondary (which is capable of somewhat higher power). The danger in undersizing (low Nsec) is risk of saturation and resulting non-linearity / intermodulation distortion.
Let’s look at an initial load line prediction for M0DGQ’s specification.
Above is a simple prediction of the initial load line of an amplifier of this type with 26V supply and 150W output. Though the calculator is written in terms of valves, the technique is valid in this case, and Rlaa, the anode to anode load resistance is equivalent to the drain to drain load resistance, and we might expect it in the region of 8Ω.
A back of the envelope calculation is that if we have a 26V supply, and we allow for FET saturation voltage drop and other sagging of the supply, we might expect a peak drain to drain RF voltage of 50V. If we want 150W, then \(R_l=\frac{V_p^2}{2P_o}=8.33Ω\). Now if you have a 1t primary and 2t secondary, the secondary load \(R=8.33 n^2=33.33Ω\), quite a long way from 50Ω and the implication is that the PA seems unlikely to develop 150W into 50Ω if the transformer is close to ideal (as needs to be for a broadband PA).
Above is a Simsmith model for the two turn secondary, and it includes a model for the ferrite core, and some allowance for flux leakage in the transformer… so a not quite ideal transformer but probably a good estimate of a good transformer.
(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)
Power output falls way short of 150W. Note that this is analysis of an ideal Class B amplifier, though with the stated limits on voltage swing (including an allowance for transistor saturation), and it should not be interpreted to mean that higher power could not be obtained by driving it into non-linearity.
The problem is that we have delivered a higher load impedance to the drain that needed. The theoretical impedance transformation needed is 50/8.333=6 which implies a turns ratio of 2.45:1… so that would require a 2.45t secondary which is not practical (for balance reasons). We can use 2t which results in lower power, or 3t which will result in higher power.
Let’s examine a three turn option for a drain to drain load around 5.5Ω.
Above is the initial load line prediction.
Above is the Simsmith model adjusted for the three turn secondary.
Reducing supply voltage brings total power down if dissipation is a problem.
]]>One of the first questions to mind is whether it is likely to deliver the rated power, so let’s review the MOSFET output circuit design from that perspective.
Sellers mostly seem to need to obscure the MOSFET type in their pics, so essentially you buy this with no assurance as to what is supplied, no comeback if the supplied MOSFET is not up to the task. Online experts suggest the MOSFET is probably a MRF9120 (or 2x IRF640 in a 70W build). The amplifier claims 100W from 12-16V DC supply.
Note that this module does not include the necessary output filter which will lose 5-10% of the power from this module.
In this case Carlos, VK1EA, connected a sample output transformer (T2) core from a recently purchased MiniPa100 kit to a EU1KY antenna analyser. The fixture is critically important, it is at my specification.
We also need to know the geometry parameter ΣA/l.
Above, from the measured dimensions of a sample core, ΣA/l=0.003415/m.
The saved S parameter file was processed as described at nanoVNA-H – measure ferrite core permeability described a method for characterising an unknown ferrite material and a complex permeability curve produced.
Above, the results are fairly good and fairly much as expected, but let’s remove the noise by digitising the plots.
Above, the points sampled for the digitised output. Though there is a lot less data in the result, when points are obtained by interpolation, noise is greatly reduced.
The above pic from an eBay advertisement of the 2020 version of the PA would suggest very strongly that there are three turns on the secondary of the output transformer, and a half turn on each drain. Interestingly the 70W versions also appear to use three turns, alarm bells ring!
From all this, we can produce an approximation in Simsmith that captures most of the expected behavior of a practical transformer, including core loss.
Above is the RUSE block schematic used for Core which models the frequency dependent magnetising admittance of the transformer and sets the frequency dependent inductances of the Tfmr element.
(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)
More to come…
]]>The radio is an Icom IC-7300. I bypassed the built in tuner, transmitted a tone into my external tuner, adjusted it for SWR=1. I then disconnected the tuner from the radio, and measured the impedance looking into the tuner with a VNA. Surprisingly, (to me anyway) the result was a pretty good 53-j3 Ohms at 14 MHz.
What should we / have expected? It is an interesting case to study.
If:
we should expect that the VSWR observed by the VNA was exactly 1.0.
The VNA measurement translated to impedance is reported as 53-j3Ω which implies VSWR(50)=1.09.
What we can say, subject to the numbered conditions listed above, at the power level tested, and allowing for measurement error is that the VSWR meter in the IC-7300 does not seem perfectly accurate, but it is not very far out. Weakness in any of the conditions above might well lead to a conclusion that the test does not show the IC-7300 VSWR has significant error.
What has this to do with optimal matching and the Jacobi maximum power transfer theorem?
Nothing!
It is wrong to interpret this (as the posted seemed to do) as a means of measuring the Thevenin equivalent source impedance of the transmitter. It is an evaluation of the reference (or null) impedance of the IC-7300 internal VSWR meter. An alternative simple test is to connect a high grade 50Ω load and observe the indicated VSWR (which might be power dependent, one of the weakness of the IC-7300) which you might hope to be less than say 1.1… if only you could read it to that resolution (which plays into the accuracy of the reported test).
The IC-7300 VSWR meter is not too easy to read to high resolution, not that it needs to be, but hair splitting tests fail on the uncertainty of such measurements.
None of this addresses whether such a transmitter is well represented by a Thevenin equivalent circuit.
]]>I was recently asked about FT-8 on the Colling 30-L1 linear amplifier considering my article Collins 30L-1 and AM.
The first thing to note is the Colling 30-L1 manual cautions against AM and FSK:
That said, what are the reasons for such a prohibition?
The RF Power Amplifier Tube Performance Computer gives us a tool for exploring different operating conditions without risking damage to the valves.
Using the previously constructed model for the 30-L1, key parameters were tweaked for the following conditions:
So the amplifier is operating essentially tuned for rated key down CW. That is different to SSB because the power supply sags more for key down CW.
The model extract above shows Idc should be 625mA, and Pout 674W, and anode dissipation is 251W, just within the ICAS limit of 260W. So long as the power supply can sustain that current, it should be possible to safely operate the 30-L1 on FT-8 or similar FM like modes… but Collins warns of the power supply limitation.
Of course it is prudent to monitor plate dissipation in high duty cycle modes, the 811A plates should not colour up.
A question that arises in the context of power supply limitation, is there a lower power level that can be achieved efficiently for a specified Idc? Calculation of an optimal configuration for say Idc=250mA is not too difficult, but it requires a significantly higher resonant load impedance which might not be achievable on all or any bands… because the output circuit is designed for rated power rather than reduced output.
All in all, the 30-L1 is not well suited to ‘high duty cycle’ modes because of the power supply.
A relevant question with FT-8 is whether operation at high power in a shared channel is sociable.
]]>