Guys,

Thanks for all of the insight - I appreciate the help.

Bruce Raymond/ND8I



"Roy Lewallen" wrote in message
...
The equation appears in Terman's _Radio Engineering_, where it's derived
with what looks like a bit of hand-waving. However, a paper is
referenced (Terman and Roake, "Calculation and Design of Class C
Amplifiers", Proc. I.R.E., vol. 24, April, 1936) which apparently has
more of an in-depth derivation. However, it appears to be depend
somewhat on characteristics unique to vacuum tubes. I recall having a
homework question in college that looked like it required a derivation
of the formula, and I struggled for several hours before giving up. The
professor told me that it was a somewhat emperical formula which
couldn't be rigorously derived.

Anyway, in Terman there's a graph (fig. 7-29 in the Third Ed.) which
does include the effect of conduction angle, but it's not immediately
obvious how the formula would be affected.

The formula commonly seen isn't generally used to predict power output,
but rather solved for R to give a nominal impedance that a class C
amplifier with output power Po would like to see. It's a good starting
point for designing output networks, but by no means universal.
High-efficiency amplifiers I've designed do require a different
impedance than predicted by the formula. For example, see Fig. 2.97 in
_Experimental Methods in RF Design_. The impedance seen by the collector
of that amplifier is about 18.7 + j8.5 ohms(*), while the formula would
predict about 9 ohms. There are more comments about that in the book.

I'm not aware of any rigorous formula that can be used to precisely
predict the optimum load impedance for a class C amplifier.

(*) I don't recall right off how much C the zener adds, but think it was
around 30 pF. So that's what I used for the calculation. Note that the L
values in the figure caption should be uH, not H and mH as shown.

Roy Lewallen, W7EL


snip

Guys,

Thanks for all of the insight - I appreciate the help.

Bruce Raymond/ND8I



"Roy Lewallen" wrote in message
...
The equation appears in Terman's _Radio Engineering_, where it's derived
with what looks like a bit of hand-waving. However, a paper is
referenced (Terman and Roake, "Calculation and Design of Class C
Amplifiers", Proc. I.R.E., vol. 24, April, 1936) which apparently has
more of an in-depth derivation. However, it appears to be depend
somewhat on characteristics unique to vacuum tubes. I recall having a
homework question in college that looked like it required a derivation
of the formula, and I struggled for several hours before giving up. The
professor told me that it was a somewhat emperical formula which
couldn't be rigorously derived.

Anyway, in Terman there's a graph (fig. 7-29 in the Third Ed.) which
does include the effect of conduction angle, but it's not immediately
obvious how the formula would be affected.

The formula commonly seen isn't generally used to predict power output,
but rather solved for R to give a nominal impedance that a class C
amplifier with output power Po would like to see. It's a good starting
point for designing output networks, but by no means universal.
High-efficiency amplifiers I've designed do require a different
impedance than predicted by the formula. For example, see Fig. 2.97 in
_Experimental Methods in RF Design_. The impedance seen by the collector
of that amplifier is about 18.7 + j8.5 ohms(*), while the formula would
predict about 9 ohms. There are more comments about that in the book.

I'm not aware of any rigorous formula that can be used to precisely
predict the optimum load impedance for a class C amplifier.

(*) I don't recall right off how much C the zener adds, but think it was
around 30 pF. So that's what I used for the calculation. Note that the L
values in the figure caption should be uH, not H and mH as shown.

Roy Lewallen, W7EL


snip

It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?

More current is flowing thru the transistor when conducting than would normally
be used if class-A ? ... this would average out because it's being pulsed rather
than continuous.

As it's a tuned circuit, the tuned circuit would carry on dishing out current
into the load thru the rest of the cycle - stored from the pulse of current when
the transistor conducts.

Clive

In article om,
"Bruce Raymond" wrote:

Thanks for the reply. I'm actually interested in the version
of the formula (Rl = Vcc^2 / 2Pout) listed on page 2.33
of "Experimental Methods in RF Design", by Wes Hayward, ...
(excellent book).

SNIP
It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?


You are missing the effect of the tuned circuit. Class C operation means
that the output transistor(s) are turned on for only part of a cycle and
thus feed energy into the tuned circuit during part of a cycle. However,
the energy stored in the tuned circuit is large enough to maintain the
output voltage through the complete cycle. Thus the formula does not
need to kow about the conduction angle.
Tink about puching a child on a swing. The person pushing is the output
transistor and the child on the swing the tuned circuit. The person
provides small pushes for a very short part of the complete swing cycle
but they do it at just the right times and the total energy in the
system (how high the child swings) builds up and the swing executes
complete cycles even though it is only pushed for a few degrees at the
back of its swing.
Brian.

The equation appears in Terman's _Radio Engineering_, where it's derived
with what looks like a bit of hand-waving. However, a paper is
referenced (Terman and Roake, "Calculation and Design of Class C
Amplifiers", Proc. I.R.E., vol. 24, April, 1936) which apparently has
more of an in-depth derivation. However, it appears to be depend
somewhat on characteristics unique to vacuum tubes. I recall having a
homework question in college that looked like it required a derivation
of the formula, and I struggled for several hours before giving up. The
professor told me that it was a somewhat emperical formula which
couldn't be rigorously derived.

Anyway, in Terman there's a graph (fig. 7-29 in the Third Ed.) which
does include the effect of conduction angle, but it's not immediately
obvious how the formula would be affected.

The formula commonly seen isn't generally used to predict power output,
but rather solved for R to give a nominal impedance that a class C
amplifier with output power Po would like to see. It's a good starting
point for designing output networks, but by no means universal.
High-efficiency amplifiers I've designed do require a different
impedance than predicted by the formula. For example, see Fig. 2.97 in
_Experimental Methods in RF Design_. The impedance seen by the collector
of that amplifier is about 18.7 + j8.5 ohms(*), while the formula would
predict about 9 ohms. There are more comments about that in the book.

I'm not aware of any rigorous formula that can be used to precisely
predict the optimum load impedance for a class C amplifier.

(*) I don't recall right off how much C the zener adds, but think it was
around 30 pF. So that's what I used for the calculation. Note that the L
values in the figure caption should be uH, not H and mH as shown.

Roy Lewallen, W7EL

Bruce Raymond wrote:
John,

Thanks for the reply. I'm actually interested in the version
of the formula (Rl = Vcc^2 / 2Pout) listed on page 2.33
of "Experimental Methods in RF Design", by Wes Hayward, ...
(excellent book).

In the example Vcc is 12 volts and Po is 1.5 watts. Rl
works out to 48 ohms. Wouldn't the peak current be
12 volts/48 ohms = 250 ma? If this were a sine wave then
the RMS power would be 1.5 watts. However, the amplifier
is run as Class C and only produces output less than half
of the time, so the output would then be less than 0.75 watts.

It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?

Thanks,
Bruce Raymond/ND8I


"John Popelish" wrote in message
...

Bruce Raymond wrote:

I've seen the equation Po = V^2 / 2R applied to the design
of Class C amplifiers. This doesn't make sense to me and
I'm looking for corroboration, or somebody to tell me I'm
an idiot ;-). The formula makes sense for a Class A amplifier
which has conduction over 360 degrees, but would seem
to overstate the power output for a Class C amplifier with,
say a 120 degree conduction.

An amplifier with a 120 degree conduction angle would
only produce about 47% as much power as one with a
360 degree conduction angle (if I did the math right).
Therefore, I'm assuming that the formula should be
Po = .47 * V^2 / 2R, or Po = V^2 / 4.2R in this case.
Is this correct?


I think the answer depends on what the letter V stands for in the
equation. If it is the peak voltage of a sine wave that rings out of
a tuned circuit (or any other pretty good sine source), then it
doesn't matter how the sine wave was generated. Somehow, the energy
put into the resonator is driving a resistor with positive and
negative peak swings and that dumps
V^2 / 2R watts into the resistor. The instantaneous peak power put
into the resonance must be higher than that, for the class C amplifier
to pump it up to that voltage in a small fraction of a cycle.

If you want a challenge, figure the power out of a class C DC
amplifier.

--
John Popelish

Bruce Raymond wrote:

John,

Thanks for the reply. I'm actually interested in the version
of the formula (Rl = Vcc^2 / 2Pout) listed on page 2.33
of "Experimental Methods in RF Design", by Wes Hayward, ...
(excellent book).

In the example Vcc is 12 volts and Po is 1.5 watts. Rl
works out to 48 ohms. Wouldn't the peak current be
12 volts/48 ohms = 250 ma? If this were a sine wave then
the RMS power would be 1.5 watts. However, the amplifier
is run as Class C and only produces output less than half
of the time, so the output would then be less than 0.75 watts.

I guess this assumes that with a 12 volt supply, a resonant load could
produce almost a 24 volt peak to peak sine wave. A fair
approximation.

When the class C amplifier comes on, it sees an impedance a lot lower
than the 48 ohms on the tank. The amp loads the tank with all the
energy it will lose in the next cycle, but the tank meters this energy
to the load in sinusoidal form. A piston in an internal combustion
engine puts out more peak torque than the flywheel delivers to the
transmission by the same mechanism. But the average power put out by
the piston is the same as the the average power delivered by the
flywheel to the transmission.

It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?

Energy storage in the tank.

--
John Popelish

Bruce Raymond wrote:

John,

Thanks for the reply. I'm actually interested in the version
of the formula (Rl = Vcc^2 / 2Pout) listed on page 2.33
of "Experimental Methods in RF Design", by Wes Hayward, ...
(excellent book).

In the example Vcc is 12 volts and Po is 1.5 watts. Rl
works out to 48 ohms. Wouldn't the peak current be
12 volts/48 ohms = 250 ma? If this were a sine wave then
the RMS power would be 1.5 watts. However, the amplifier
is run as Class C and only produces output less than half
of the time, so the output would then be less than 0.75 watts.

I guess this assumes that with a 12 volt supply, a resonant load could
produce almost a 24 volt peak to peak sine wave. A fair
approximation.

When the class C amplifier comes on, it sees an impedance a lot lower
than the 48 ohms on the tank. The amp loads the tank with all the
energy it will lose in the next cycle, but the tank meters this energy
to the load in sinusoidal form. A piston in an internal combustion
engine puts out more peak torque than the flywheel delivers to the
transmission by the same mechanism. But the average power put out by
the piston is the same as the the average power delivered by the
flywheel to the transmission.

It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?

Energy storage in the tank.

--
John Popelish

John,

Thanks for the reply. I'm actually interested in the version
of the formula (Rl = Vcc^2 / 2Pout) listed on page 2.33
of "Experimental Methods in RF Design", by Wes Hayward, ...
(excellent book).

In the example Vcc is 12 volts and Po is 1.5 watts. Rl
works out to 48 ohms. Wouldn't the peak current be
12 volts/48 ohms = 250 ma? If this were a sine wave then
the RMS power would be 1.5 watts. However, the amplifier
is run as Class C and only produces output less than half
of the time, so the output would then be less than 0.75 watts.

It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?

Thanks,
Bruce Raymond/ND8I


"John Popelish" wrote in message
...
Bruce Raymond wrote:

I've seen the equation Po = V^2 / 2R applied to the design
of Class C amplifiers. This doesn't make sense to me and
I'm looking for corroboration, or somebody to tell me I'm
an idiot ;-). The formula makes sense for a Class A amplifier
which has conduction over 360 degrees, but would seem
to overstate the power output for a Class C amplifier with,
say a 120 degree conduction.

An amplifier with a 120 degree conduction angle would
only produce about 47% as much power as one with a
360 degree conduction angle (if I did the math right).
Therefore, I'm assuming that the formula should be
Po = .47 * V^2 / 2R, or Po = V^2 / 4.2R in this case.
Is this correct?


I think the answer depends on what the letter V stands for in the
equation. If it is the peak voltage of a sine wave that rings out of
a tuned circuit (or any other pretty good sine source), then it
doesn't matter how the sine wave was generated. Somehow, the energy
put into the resonator is driving a resistor with positive and
negative peak swings and that dumps
V^2 / 2R watts into the resistor. The instantaneous peak power put
into the resonance must be higher than that, for the class C amplifier
to pump it up to that voltage in a small fraction of a cycle.

If you want a challenge, figure the power out of a class C DC
amplifier.

--
John Popelish

Bruce Raymond wrote:

I've seen the equation Po = V^2 / 2R applied to the design
of Class C amplifiers. This doesn't make sense to me and
I'm looking for corroboration, or somebody to tell me I'm
an idiot ;-). The formula makes sense for a Class A amplifier
which has conduction over 360 degrees, but would seem
to overstate the power output for a Class C amplifier with,
say a 120 degree conduction.

An amplifier with a 120 degree conduction angle would
only produce about 47% as much power as one with a
360 degree conduction angle (if I did the math right).
Therefore, I'm assuming that the formula should be
Po = .47 * V^2 / 2R, or Po = V^2 / 4.2R in this case.
Is this correct?


I think the answer depends on what the letter V stands for in the
equation. If it is the peak voltage of a sine wave that rings out of
a tuned circuit (or any other pretty good sine source), then it
doesn't matter how the sine wave was generated. Somehow, the energy
put into the resonator is driving a resistor with positive and
negative peak swings and that dumps
V^2 / 2R watts into the resistor. The instantaneous peak power put
into the resonance must be higher than that, for the class C amplifier
to pump it up to that voltage in a small fraction of a cycle.

If you want a challenge, figure the power out of a class C DC
amplifier.

--
John Popelish

Yes

"Bruce Raymond" wrote in message
gy.com...
I've seen the equation Po = V^2 / 2R applied to the design
of Class C amplifiers. This doesn't make sense to me and
I'm looking for corroboration, or somebody to tell me I'm
an idiot ;-). The formula makes sense for a Class A amplifier
which has conduction over 360 degrees, but would seem
to overstate the power output for a Class C amplifier with,
say a 120 degree conduction.

An amplifier with a 120 degree conduction angle would
only produce about 47% as much power as one with a
360 degree conduction angle (if I did the math right).
Therefore, I'm assuming that the formula should be
Po = .47 * V^2 / 2R, or Po = V^2 / 4.2R in this case.
Is this correct?

Thanks,
Bruce Raymond/ND8I