# what is the highest efficiency a tube amplifier can run at

Discussion in 'Amateur Radio Amplifiers' started by KB1SEL, Feb 27, 2011.

Not open for further replies.
1. ### KA4DPOPlatinum SubscriberPlatinum SubscriberQRZ Page

The efficiency of the tank circuit is determined by the loaded Q of the tank. This takes into account the Q of the components and the effect of circuit loading on the tank which lowers the Q.

2. ### W8JIHam MemberQRZ Page

The theoretical efficiency of a class B amplifier is in the high 70% range, around 78%. This is based on conduction angle of the output device. To get into this range, the output device on resistance has to swing from very high values when off (near zero dissipation) the very low values when conducting (again near zero dissipation because resistance is low). Another way to look at it would be the ratio of peak voltage to minimum voltage on the output device.

I don't know who contributed to Wikipedia, but 60% is really too low for plate efficiency as a rule.

Class C goes nearly 100%, and I have seen actual practical PA stages make in the high 90% range for overall energy conversion efficiency. Efficiency is improved by a sharp off to on transition in current, and very low on resistance in the output device.

Class A is 50% or less.

Naturally a class AB would be between A and B someplace, so less than 78% and more than 50%.

Frequency has little effect on output device efficiency, although it does play a role when the frequency limit of the device is approached. This is because waveforms are rounded at transistions and because the output device might have fairly high distributed loses and it becomes an increasing part of the tank circuit at higher frequencies.

Tank efficiency is a function of loss resistances and currents through those losses. While there are very simplistic rough tank efficiency calculations like the ratio of loaded Q to unloaded Q, they only work when Q is defined correctly. Most formulas for Q are actually incomplete shortcuts, just like most formulas we use for operating impedance are. It is always strange to see people become overly concerned about getting Q to some magical number, when there is no magical number and the formulas are just rough approximations.

I've seen efficiencies in the 90% range on class C, and the best I've achieved is around 75% with a modified DX100 Heathkit and around 85% with an entirely homebrew FET amp. I've built class AB cavity amps that have made around 70% efficiency on 300 MHz. Typically 55-65% is very common for common tubes at normal voltages in class AB in amateur products.

3. ### VK1ODHam MemberQRZ Page

Yes. An ideal device (meaning perfectly linear Ia/Vgk, zero saturation voltage) driven to 120° conduction angle has efficiency of 90% at the anode.

A smaller conduction angle will give higher efficiency, but driven to the same peak anode current, power output is reduced.

Real tubes don't have zero saturation voltage, and aren't perfectly linear (though non linearity of the transfer characteristic can help efficiency), and output networks lose some power.

Efficiency in AB2 will depend on how adventurous one is is driving the device to poorer linearity, but operating tetrodes at 750V supply (as in some desktop transceivers) isn't indicative of what can reasonably be achieved.

The tool at RF Power Amplifier Tube Performance Computer gives some insight.

Owen

4. ### WA9SVDHam MemberQRZ Page

You overly complicate and confuscate the issue. Whether you specify tank losses or not, and while they they DO enter the equation, by FCC standards and specifications, efficiency is the RF Power output divided by the DC power input. Pure ansd simple, and THAT is the final answer.

5. ### W8JIHam MemberQRZ Page

What section of FCC rules define efficiency?

6. ### WA9SVDHam MemberQRZ Page

You are correct; the FCC does not specify a definition of "efficiency." It's a concept in standard engineering practice. The actual design of a tank circuit is immaterial to the overall efficiency of an amplifier.

7. ### W5JOHam MemberQRZ Page

Why not use components that meet the FCC standards (measuring devices, meters, etc. on the input) then measure the output and calculate what percentage of efficiency you have. If I recall my upbringing, there was a question on the tests many years ago about this subject. You can apply theory all you want but the final answer is always dependent on the quality of the components and ability of the person who constructed the equipment. It is also, as stated, dependent on the class of operation.

8. ### W8JIHam MemberQRZ Page

How something works is how it works. It is good to know how it works, because then we can make things as good as possible.

Overall efficiency includes all losses. It would be dc power input to the PA final stage compared to overall RF output power. The problem is people often wrong assume that is the PA device efficiency. It is not. Other people assume tank operating Q has a large bearing on efficiency, and struggle to maintain some magical number like 12 for tank Q.

Another common mistake is to assume driver power in a GG amp feeds through as unaccounted "free" power in the output. Driver power actually shows on the metering as additional PA stage current, so only the additional voltage swing is not measured. This means only a small portion of exciter power, perhaps 50 volts average voltage (out of 3000 volts) on a 3-500Z. There might be 20 watts plate input power unaccounted for in a grounded grid amp using a 3-500Z tube, of which 50-70% winds up being RF output (10-14 watts). Since most meters have more tolerance than that, driver feedthrough is "in the measurement noise".

The FCC got that wrong years ago but did so intentionally. They assume all driver stage plate input power contributed as hidden PA stage plate input power. This was to prevent clever devious people from cheating and running low or negative gain PA's with intentionally high levels of exciter power, not for the sake of accuracy.

It isn't so much how to measure overall efficiency that is the subject here, that much is very clear and without debate. It is how the system works in detail.

73 Tom

9. ### WA4ILHSubscriberQRZ Page

As previously stated, Class B is most efficient but, of couse, this requires at least 2 tubes in what is called "push-pull" operation. If I remember correctly from Navy ET school 45 years ago, another benefit of a push pull amplifier is that they also tends to cancel some distortion which may be present. I haven't seen a push pull amplifier in RF service in many many years. Class B amplifiers were very popular in high end audio amplifiers but require tapped heavy and expensive transformers for coupling between stages. In RF service, it would also require a tapped transformer. Seems to me that Motorola had an early push pull ampifier in one of their early power amplifies I think this used an output coupling technique called "swing loop" coupling. Maybe Glen can correct me on this...
Tom WA4ILH