On 17/05/2014 08:12, FranK Turner-Smith G3VKI wrote:
"gareth" wrote in message
...
"gareth" wrote in message
...
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop
across the active element divided by the current through it.

Now, clearly as current sources, this would be true, and it could be
measured.,
by changing the loads.

But, how could you calculate such a value up front?

Interesting how the great blusterers display their ignorance, either
by not entering the discussion by the device of side-stepping with
abusive
remarks*****, or trumpeting complete nonsense ....

Consider the case of your homebrewed 1.5kW linear being set up with
QRP for safety.

Let's say 6W output on a 12V supply. Some claim that this means an
impedance of 24 ohms that has to be matched to the 50 ohm antenna.

So, having done your matching, you now crank up the output to 1.44kW
by increasing the drive, and those some claimers will now say that the
ouput impedance is now 0.1 ohm.

Hang on there! By their argument, you will now be in danger of severely
damaging your pride and joy by driving into what you say is a complete
mismatch, set, not for 0.1 ohm but for 24 ohms.

You would need to increase the supply volts to at least 200V to give
headroom, unless you want a square wave output.

Matthew 7:6

On 17/05/14 09:29, Kafkaesque wrote:
On 17/05/2014 07:07, gareth wrote:

So, having done your matching, you now crank up the output to 1.44kW
by increasing the drive, and those some claimers will now say that the
ouput impedance is now 0.1 ohm.

Hang on there! By their argument, you will now be in danger of severely
damaging your pride and joy by driving into what you say is a complete
mismatch, set, not for 0.1 ohm but for 24 ohms.

And this is why valved PAs in older radios (FT-101Es, for example) had
variable impedance matching (the Pi Tank circuit), and why it's good
practice to tweak that matching after increasing the drive to full power
(say 250mA anode current) having first matched at a lower power (100mA
anode current).

Or use a tone-[pulser.


--
Spike

On 17/05/2014 09:35, Spike wrote:
On 17/05/14 09:29, Kafkaesque wrote:
On 17/05/2014 07:07, gareth wrote:

So, having done your matching, you now crank up the output to 1.44kW
by increasing the drive, and those some claimers will now say that the
ouput impedance is now 0.1 ohm.

Hang on there! By their argument, you will now be in danger of severely
damaging your pride and joy by driving into what you say is a complete
mismatch, set, not for 0.1 ohm but for 24 ohms.

And this is why valved PAs in older radios (FT-101Es, for example) had
variable impedance matching (the Pi Tank circuit), and why it's good
practice to tweak that matching after increasing the drive to full power
(say 250mA anode current) having first matched at a lower power (100mA
anode current).

Or use a tone-[pulser.

Or a bug key sending a train of dits. 12wpm would NOT be fast enough.

In message , Rob
writes
FranK Turner-Smith G3VKI wrote:
"gareth" wrote in message
...
"gareth" wrote in message
...
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop
across the active element divided by the current through it.

Now, clearly as current sources, this would be true, and it could be
measured.,
by changing the loads.

But, how could you calculate such a value up front?

Interesting how the great blusterers display their ignorance, either
by not entering the discussion by the device of side-stepping with abusive
remarks*****, or trumpeting complete nonsense ....

Consider the case of your homebrewed 1.5kW linear being set up with QRP
for safety.

Let's say 6W output on a 12V supply. Some claim that this means an
impedance of 24 ohms that has to be matched to the 50 ohm antenna.

So, having done your matching, you now crank up the output to 1.44kW
by increasing the drive, and those some claimers will now say that the
ouput impedance is now 0.1 ohm.

Hang on there! By their argument, you will now be in danger of severely
damaging your pride and joy by driving into what you say is a complete
mismatch, set, not for 0.1 ohm but for 24 ohms.

You would need to increase the supply volts to at least 200V to give
headroom, unless you want a square wave output.

It is a widely distributed misunderstanding that a linear that has
been designed to delever its output in a 50 ohm load always has a
50 ohm output impedance. In fact it almost never has.

My understanding of RF design doesn't include how to calculate the
output impedance of PA stages. However, it's obvious that of the output
impedance was 50 ohms, the PA efficiency would be only 50% at best - and
clearly this is not so (certainly for class-C operation, which is
notionally 66%-ish). I've always understood that, in practice, it's
usually resistively low-ish (around 25 ohms?) and quite capacitive. An
expert opinion is needed!
--
Ian

"Kafkaesque" wrote in message
...
On 17/05/2014 08:12, FranK Turner-Smith G3VKI wrote:
"gareth" wrote in message
...
"gareth" wrote in message
...
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop
across the active element divided by the current through it.

Now, clearly as current sources, this would be true, and it could be
measured.,
by changing the loads.

But, how could you calculate such a value up front?

Interesting how the great blusterers display their ignorance, either
by not entering the discussion by the device of side-stepping with
abusive
remarks*****, or trumpeting complete nonsense ....

Consider the case of your homebrewed 1.5kW linear being set up with
QRP for safety.

Let's say 6W output on a 12V supply. Some claim that this means an
impedance of 24 ohms that has to be matched to the 50 ohm antenna.

So, having done your matching, you now crank up the output to 1.44kW
by increasing the drive, and those some claimers will now say that the
ouput impedance is now 0.1 ohm.

Hang on there! By their argument, you will now be in danger of severely
damaging your pride and joy by driving into what you say is a complete
mismatch, set, not for 0.1 ohm but for 24 ohms.

You would need to increase the supply volts to at least 200V to give
headroom, unless you want a square wave output.

Matthew 7:6

Poor old Beanie, he still hasn't grasped what happens when you over-drive an
amplifier.
Not so long ago he proposed a spectrum analyser with a linear input and
refused to see the problem there.
The worrying thing is his offer to teach others.
--
;-)
..
73 de Frank Turner-Smith G3VKI - mine's a pint.
..
http://turner-smith.co.uk

On 16/05/2014 20:44, gareth wrote:
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop
across the active element divided by the current through it.

Now, clearly as current sources, this would be true, and it could be
measured.,
by changing the loads.

But, how could you calculate such a value up front?

One is a static value the other is dynamic, you seem to be wondering why
a real calculation should be different to a complex one.
Once the signal passes through any phasing element the vector will shift
off the real axis and the V=IR. That's why it is measured in impedance
and not resistance.

As for calculation - for simple stuff it is a matter of number crunching
with complex numbers - for more complicated circuits then you need to
look at some network analysis over the whole frequency range you will be
using.

Andy

"Generic" wrote in message
news
why do I keep posting this crap?

Because, Duncam, despite your years, you display the emotions
of a child.

On 5/17/2014 3:12 AM, gareth wrote:
"gareth" wrote in message
...
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop
across the active element divided by the current through it.

Now, clearly as current sources, this would be true, and it could be
measured.,
by changing the loads.

But, how could you calculate such a value up front?

Consider for a moment the equivalent circuit of a current generator
with a shunt impedance feeding a load.

The voltage across the load is the same as that across the current
generator,
and measuring the current coming out of the generator at any one time will
tell you
about the load impedance, but not the shunt impedance.

However, by changing the load impedance and again measuring the voltage
and the current, then you will ahve sufficient information to calcualte the
source impedance.

Yes. What is your point?

The output impedance of an amplifier depends on the circuit topology and
the component values. I have designed amplifiers with "synthetic"
impedance where a small series resistor was used with positive feedback
to create an output impedance of a larger value. I think the actual
numbers were 12.5 Ohm resistor and 50 Ohm output.

--

Rick

On 5/17/2014 5:14 AM, Ian Jackson wrote:
In message , Rob
writes

It is a widely distributed misunderstanding that a linear that has
been designed to delever its output in a 50 ohm load always has a
50 ohm output impedance. In fact it almost never has.

My understanding of RF design doesn't include how to calculate the
output impedance of PA stages. However, it's obvious that of the output
impedance was 50 ohms, the PA efficiency would be only 50% at best - and
clearly this is not so (certainly for class-C operation, which is
notionally 66%-ish). I've always understood that, in practice, it's
usually resistively low-ish (around 25 ohms?) and quite capacitive. An
expert opinion is needed!

I may be jumping in where I don't understand all the issues as I know
little about RF... but...

The output impedance of an amplifier does not dictate the power
dissipation of that amplifier. In fact the one amp design I have done
used synthetic output impedance for the sole purpose of reducing the
power consumption of the amp circuit. As I mentioned in another post,
it used a small series resistor and positive feedback to synthesize a
larger output impedance. I only had 12 volts power supply available and
needed to swing over 8 Vpp at the output. This circuit worked well for
that.

Is this never done in an RF amp?

--

Rick