# Vacuum Tube Amps

**How the PA Tube**

Converts DC anode

voltage to Radio

Frequency

Power

**How the PA Tube**

Converts DC anode

voltage to Radio

Frequency

PowerConverts DC anode

voltage to Radio

Frequency

Power

A

typical vacuum tube

radio-frequency

amplifier has a high

voltage power

source. This power

source supplies the

energy for the RF

output power. The

vacuum tube acts

like a non-linear

time-varying resistance

in series with a

diode. The

electrical

equivalent of the

anode and output

system is:

R1

represents the

time-varying anode-to-cathode

resistance of the

output device. As

grid-cathode and

anode-cathode

voltage changes, R1s

value changes.

D1

represents the

directional

characteristics of

the tube

anode-to-cathode

current path. D1 is always

on when the

voltage at point A

is more positive

than ground. D1

would also behave as

a current limiter,

the current always

having to be

something less than

the saturated

emission current of

the cathode. The

only exception to

this is if the tube

shorts or arcs

internally.

Lets

examine how this

circuit converts DC

to RF power. We will

assume +HV is 3000

volts.

In

the initial state,

C1 has been charged

through the

following DC path:

Hopefully

the designer was

smart enough to

include RFC2, to

prevent charging or

discharging C1 into

the load, which

might be a sensitive

piece of gear (or the

operator)!

The

voltage across C1 is

the anode-to-ground

voltage at A, in this case 3000 volts while resting without RF drive power.

Everything

past A, at B and VL,

is zero volts in the

resting state. This

is

because of the

low-dc resistance of

RFC2 and the nearly

infinite resistance

(and dc impedance) of

capacitor C1. RFC1,

the plate choke,

establishes a fixed

or static magnetic field,

the intensity of

which

is set by the

quiescent or

steady-state tube

idle current flowing through

anode-to-cathode

resistance R1.

As

a positive-going

grid-to-cathode

RF voltage decreases

anode-to-cathode

resistance R1,

the magnetic flux

stored in

the inductance of RFC1 tries to hold

supply current

steady. RFC1 reacts

to the current

change by increasing

voltage across

its terminals.

Energy stored in the

electric field of C1

tries to hold the

voltage steady, and

C1 sources any

additional current

the tube (R1)

requires as it tries to force C2 and the tank input negative.

The

anode, in effect,

tugs or pulls

point A towards

ground. The initial

peak current can be

quite high, limited

only by

plate-to-cathode

resistance, because

C1 has a

low-reactance (at the operating frequency) path

to ground through

C2. The tank

inductor L1 also builds

up a magnetic field,

as plate tuning

capacitor C2 develops

a charge and

gradually increasing voltage.

This voltage

increase comes from the

current flowing

through C2 and C1

back to the tube. On

the initial RF

cycle, the voltages

hardly change before

the grid is moving

back negative.

At

a time interval

equal to 1/(4F)

(where F is the RF

frequency), the

positive grid

voltage has peaked.

The grid begins to

swing negative (or

less positive) with

respect to the

cathode.

Anode-cathode

resistance (R1)

decreases in step

with decreasing

grid-cathode

voltage.

There is no power-supply-forced

upward swing (the

tube cannot *source*

anything). The

collapsing field in

tank coil L1 (and to a much

lesser extent the collapsing field in plate choke RFC1)

tries to hold

changing anode current at the same

value. This is accomplished by the inductances of RFC1 and L1 increasing voltage as tube

anode-to-cathode resistance increases. L1

(and to a lesser

extent RFC1)

pull

anode voltage above B+ rail

by an amount virtually equal to the amount

anode voltage

was pulled downwards by the tube, *with the anode even more than twice B+
rail voltage under some tank loading
conditions! *

** The**This

actual amount the

tank pulls this

voltage higher than

+HV depends on the

amount of energy

transferred to the

load compared to

energy stored in the

tank system!

is VERY important in

understanding where

arcs come from! Some

switching power

supplies, many

ignition coils used

in automobiles, and

the flyback systems

in television

receivers use this

type of “flyback

effect” voltage to

step up voltages to

very high levels

without use of

transformer action.

This

cycle repeats over

and over. Assuming the tank

system at C2, L1,

and C3 is resonant,

voltage at point

A increases in alternating swing

while the peak

current through R1

(the tube) is

gradually reduced.

Eventually*, in
a fully loaded
amplifier,*

equilibrium is

reached. In a

*properly*

tuned amplifiertuned amplifier

running at maximum

available power,

equilibrium occurs

when point A

swings up and down

an amount just under

twice the anode DC

voltage.

Consider

a class AB 3-500Z

using a 3000-volt

supply. When the

amplifier is tuned

into the load

properly at full

rated power, and

driven to full

power, anode voltage

will swing between

5,500 volts maximum

and 500 volts

minimum at point A.

This would be a ** total**.

anode voltage swing

of 5,000 volts

*anode current*

**Peak instantaneous**in continuous

carrier operation

would typically be

1.12 amperes, with a

minimum R1 value of

446 ohms. This is

typical of

continuous CW power

out of 750 watts

with a single 3-500Z

tube with a

190-degree

conduction angle

running at 400mA

average anode current from the HV supply.

Lets

assume (we actually

CAN have this

condition, if the

amplifier and

exciter relays are

not sequenced

correctly) we have

an envelope that

rises instantly. At

the peak of first

positive

grid-cathode cycle,

the anode resistance

would drop to

someplace well under

400 ohms (anode

resistance is

non-linear with

anode voltage, and

is lower with higher

anode voltages). The

saturated anode

current could reach 7.5 amperes

in the 3-500Z,

if tube grid-cathode

bias

permits.

(Emission

in a directly heated

thoriated-tungsten

tube is typically in

the range of

50-100mA per watt of

heater power, large

transmitting tubes

being at the upper

end of that range. A

full-emission 3-500Z

has a saturated

emission current of

about 7.5 amperes.)

If

we have infinitely

fast envelope rise

and fall times from

the exciter,

amplifier

RF-envelope rise and

fall times are

determined by the

operating Q of the

tank system.

C2/L1/C3 dominate

the high frequency

energy storage. RFC1

and C1, being larger

values, dominate

lower-frequency

energy storage,

while stored energy

in the power supply

dominates long-term

energy demands. The

voltage across C2,

upon initial

application of RF

drive, is a rapidly

expanding sine wave.

It reaches maximum

steady-state swing

many dozens or

hundreds of RF

cycles later than

the initial tube

excitation (and

decaying in a

similar fashion).

The

peak voltage across

C2 is equal to the

peak anode voltage

swing, and the

peak-to-peak voltage is

slightly less than

twice the HV supply

voltage in NORMAL

operation. The

voltage across C3 is

a function of the

load resistance, and

power delivered to

the load.

SWR or Reflected Power Myth:
We often hear people claim reflected power burns up as heat in the power The only effect of reflected power is it changes the loadline of the In one case heat increases, in the other heat decreases. An SWR |

**Incorrect**

Loading or Load

**Incorrect**

Loading or LoadLoading or Load

We

know the tank

circuit stores

energy. We now

understand the

conversion process

where DC is

converted to AC (or

RF) power. We also

must understand *energy
must be transferred
out of the tank at a
rate equal to or
exceeding the rate
at which it is
supplied by the
downward pull
of the tube.*

*If we do not*

remove energy at a

sufficient rate,

voltages and

currents increase

until a new point of

equilibrium is

reached.remove energy at a

sufficient rate,

voltages and

currents increase

until a new point of

equilibrium is

reached.

Voltage at

point A can

actually swing well

beyond twice +HV on

upward excursions,

and below zero volts

(becoming negative)

on downward

excursions.

The

maximum voltage with

a load or drive

fault can be

tremendously higher

under the fault

condition than typical

properly operating

tank working voltages

(when

energy is being

removed at the

proper rate). If we

do not remove energy

from the tank at the

same rate the power

supply is supplying

energy, the voltage

in the tank will

increase until

something absorbs

that energy!

We

have the same basic

tank system as

discussed earlier,

but *with light
loading compared to
drive level* the

tank is pulled down

very hard by the

tube. Minimum Rp is

reached early in the

cycle, before the

tank voltage reaches

its minimum

swing.

This is easy

to do, since the

tube only pulls the

system down and the

tank stores the

energy of that

downward tug.

As

the plate voltage

swings below zero

(negative) from the

tank energy, the

tube is already

cutting off. Nothing

clamps or prevents

point A from going

negative. D1

effectively takes

the tube out of the

circuit. By the time

the tank reaches its

minimum, the tubes

grid-cathode voltage

is already on its

way positive. The

tank free wheels

positive, and can

overshoot the +HV

supply by several

times the supply

voltage. If loading

is light enough and

Q is high enough,

this continues until

the energy stored in

the tank reaches

equilibrium with

energy transferred

to the load, or a

component fails and

the arc dampens the

tanks gyrations.

See

the practical

demonstration

page for a real

working model with

waveforms!

## Proper Plate Load

Impedance or

Operating Impedance

of an RF Amplifier

Since the

anode-to-cathode

resistance of the

tube varies over the

RF cycle, how do we

establish the plate

resistance of the

tube? It all goes

back to energy

transfer. We must

extract the same

energy available

from the power

supply power drawn

by the time-varying

tube resistance,

less any dissipation

in the tube or loss

in the tank

components.

Otherwise the anode

voltage swing will

not be in

equilibrium, voltage

swing will continue

to build until

something eventually

extracts the energy

at the same rate as

it is applied.

If

we try to extract

too much power to

the load compared to

the optimum value,

the output device

current will

increase. The tube

path looks like

this:

The

tank circuit

transforms the load

impedance of the

antenna system or

dummy load up to a

new value. This

resistance appears

at C1, and is

coupled through C1

to the tube and

power supply

junction. This is

the resistance value

we are actually

calculating when we

plan the plate

resistance, Rp.

This

resistance has to

match the optimum

E/I of the desired

RF frequency

appearing at the

anode of the tube.

Let’s consider the

fundamental

frequency only in a

typical class AB

amplifier operating

near the class B end

of the operating

point. Let’s assume

we have 3000 volts

of high voltage, and

the tube performs

best swinging down

from 3000 volts to

500 volts. This is

2500 volts of peak

anode voltage swing,

and the flyback

effect causes the

anode to reach 3,000

(B+ value) + 2,500

(downward swing) =

5,500 volts on peak.

Our vacuum tube has

a peak dc anode

voltage of 5,500

volts in

NORMAL

linear operation. If

you develop an

antenna system

problem or have a

bad lightning

arrestor or relay,

the peak voltage

could be several

times that amount.

I’ve seen test data from Chinese 572B’s where the tubes were tested for arcing and failure at 1,500 volts dc anode voltage. If we think about this, all tubes should be tested to withstand some significant amount more than thetwicemaximum dc anode voltage. This is a noteworthy point because test data from various Chinese manufacturers has indicated they test tubes at . Manufacturers shouldless than therated dc supply voltage actually be testing and gettering tubes to withstand significantly more than twice the highest expected dc supply voltage. Even an 811A should not be tested that low. The old Eimac 3-500Z’s, rated at 4,000 volts supply voltage, used to test to over 12,000 volts of anode-to-grid breakdown voltage. They virtually never arced when new. To be |

With

the 3-500Z example,

we might have a

useful RF output

power of 800 watts

with the anode

swinging 2500 volts

peak (5000 volts

p-p), or about 1700

volts RMS at the

fundamental. There

are harmonics

involved, so to do

this right we would

have to use a

Fourier (harmonic

waveform) analysis

of the waveform, and

there is just such

an analysis that was

developed for vacuum

tubes called a

Chaffee analysis,

but for casual use

we can use

approximations. The

typical rule of

thumb for tubes with

a 15% minimum plate

voltage is we

multiply indicated

anode current

by 1.6, and use the

supply voltage.

Ep/(Ip*1.6) = Rp.

Using this example

we have

3,000/(.4*1.6) =

4,688 ohms.

Now

let’s look at

another rough

method. We know the

swing is about 2500

volts, so RMS is

about .7 times that.

1750^2 / 750 = 4,083

ohms.

Using a Chaffee

analysis the optimum

load resistance

comes out to 4500

ohms. The error in

approximations is in

not allowing for

harmonic content.

## Tank Q

We

often get far too

worried about tank

circuit Q. There

really is very

little change in

efficiency as Q

moves from a minimum

of (SQRT Rp/Rl)+1.

If we had a 4500 ohm

optimum anode load

resistance (Rp) and

a minimum load

resistance of 25

ohms, we would want

a tank Q of at least

SQRT 4500/25= 13.4 +

1 = 14.4. A

typical HF tank

circuit, even with

modest sized

components and such

a large impedance

ratio, only has

approximately 4 %

power loss. If we

doubled the Q, tank

loss would only be 8

%. Lower impedance

ratio tanks have

even less loss and

are less worrisome.

Most

of the power loss is

in the conduction

angle and waveform shape of the tube,

not tank Q. It is

certainly not in a

blocking capacitor,

would would overheat

and disintegrate if

it had more than a

tiny fraction of

just one percent

loss.

Don’t get overly

concerned about Q.

Most of the time you

will be off 20-30%

anyway. Handbook formulas

for Q and Rp are

just rough

approximations, and loaded-Q generally has a small effect on overall efficiency

(unless the tank components have a great deal of loss).

## Efficiency

The efficiency of a power amplifier can be well over 50% even when the

amplifier’s output port is conjugate matched to the load. This is because the

output device can switch on and off hard, and not be in a state where it looks

lie a dissipative resistance very long. The tank circuit filters the sharp

transitions out, and rounds the waveform into a sine wave. At this point we have

an optimum across and through vector, voltage across the tank and current

through the tank, and that is the “impedance” conjugate matched to the load.

The only dissipative part of the system, other than small tank and circuit

losses, is the tube or output device. Since that area is non-linear (not in

transfer function, but in fractional cycle linearity) power conversion

efficiency can be quite high.

As an example of this I measured the output impedance of a pair of class-C

6146’s two ways, by using a traditional load pull and by using a reverse power

generator and a tapped line. When tuned for maximum efficiency at full power,

the output port looked very close to being a 50-ohm source. Despite being a

50-ohm source, efficiency was around 80%! As a matter of fact I measured over a

dozen PA systems and all of them produced near-maximum efficiency and an

efficiency of over 50% while they behaved like a true 50-ohm source.

An amplifier does not need to be conjugate matched to work, and indeed many or

most are not conjugate matched, but they are when tuned for maximum

efficiency!

To have high efficiency conduction angle must be as short as possible, and when

the output device turns off (goes open circuit) the output device must

transition into open circuit as rapidly as possible. This is why a

low-pass tuned input is so

necessary for maximum efficiency in a grounded grid amplifier.

## Reflected Power

Contrary to somewhat popular myth, reflected power does not heat the power amplifier. As

a matter of fact reflected power can actually make the PA stage run cooler! The

only effect of load mismatch is the operating load line presented to the output

device by the tank circuit shifts to a different value. Normally, an adjustable

tank has enough range to correct the load line change. Even if the SWR is 10:1,

if the tank adjustment range allows adjustment to normal grid and anode currents

and proper RF power levels, efficiency and heating will not change.

If tank range is insufficient for the impedance presented to the PA by the

feed line, or if the PA stage is “fix tuned” and lacks any ability to readjust to

a new load impedance value, the load line will shift. To the tube or output

device, this is not any different than mistuning the tank. This is why

non-adjustable systems, like typical solid state PA’s, require a low “load SWR”.

Because the tank cannot readjust, they must see the design impedance. This is

typically 50 j0 ohms.

If loadline impedance decreases, the output device’s current

will increase. The current increases while peak output device voltage is reduced,

and the anode current conduction angle increases. The longer time period of

increased current result is more anode heat. This can result in an anode

thermal failure, if the time and dissipation product exceeds safe

limits.If the loadline changes in a way that reduces anode current, less

anode heating occurs. The output device runs cooler,

but peak voltage increases. If

peak voltage exceeds safe limits, an instantaneous failure will occur.

Tube-type amateur transmitters were often rated for a very wide

load impedance range. This shows the fallacy of thinking feed line reflected

power causes a PA to overheat.