So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output.

Uwe, The 10 Vrms that your meter indicates may not be accurate at 3.5 or 7
MHZ. It is probably OK at 60 hz. What you can derive from the 10 Vrms is that
the AC-1 is producing power, just not sure how much.
One thing you might try is a method to remove the frequency dependency of
your measurements. Use a 10:1 voltage divider (10k and a 1.11k). Run this
thru a Germanium or Schottky detector diode and a .01 filter capacitor. You
now have a DC voltage that is proportional to power, and relatively frequency
independent.
To calculate the RMS voltage across the 50 ohm load: Read the DC volts out
of the detector-Vdc. Then Vrms=(Vdc*.707)*10.
Example: You read 2Vdc out of the detector. Vrms=14.14 volts. Power into the
50 ohm load is then: 14.14^2/50=4 Watts. The diode drop in the dector will
introduce some error at QRP levels, hopefully not too much for what you are
trying to do.
73 Gary N4AST

Gary, you may be right regarding my power output measurement, but I thought
I knew how you calculate DC power and AC power of sinusoidial waveforms on
my scope, which, by multiplying with 0.707 you have to reduce first to the
aquivalent of a DC voltage.

What confuses me in your suggestion (printed below) is that even after you
rectify an AC signal and send it through a filter capacitor you STILL
suggest to multiply it times 0.707.
I would have thought that would give you the wrong result??

Where do I go wrong, if at all??

Uwe


in article , JGBOYLES at
wrote on 3/12/04 17:19:

So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output.

Uwe, The 10 Vrms that your meter indicates may not be accurate at 3.5 or 7
MHZ. It is probably OK at 60 hz. What you can derive from the 10 Vrms is
that
the AC-1 is producing power, just not sure how much.
One thing you might try is a method to remove the frequency dependency of
your measurements. Use a 10:1 voltage divider (10k and a 1.11k). Run this
thru a Germanium or Schottky detector diode and a .01 filter capacitor. You
now have a DC voltage that is proportional to power, and relatively frequency
independent.
To calculate the RMS voltage across the 50 ohm load: Read the DC volts out
of the detector-Vdc. Then Vrms=(Vdc*.707)*10.
Example: You read 2Vdc out of the detector. Vrms=14.14 volts. Power into
the
50 ohm load is then: 14.14^2/50=4 Watts. The diode drop in the dector will
introduce some error at QRP levels, hopefully not too much for what you are
trying to do.
73 Gary N4AST

What confuses me in your suggestion (printed below) is that even after you
rectify an AC signal and send it through a filter capacitor you STILL
suggest to multiply it times 0.707.
I would have thought that would give you the wrong result??

Uwe, Wouldn't be the first time I was wrong.
If you rectify and filter an AC waveform, and the read the DC voltage, this
is Vpeak of the AC. In the example I gave, you read 2 vdc out of the detector
circuit. Multiply this times the X10 divider, and you get 20vdc which is Vpeak
across the 50 ohm load. To get the rms voltage across the load
Vpeak*.707=20*.707=14.14 Vrms. 14.14Vrms across 50 ohms=4 watts.
BTW, the reason I suggested a voltage divider (attenuator) rather than
reading directly with a VOM or scope is to reduce the likelyhood of the scope
probe or meter impedance from detuning the AC-1 causing further errors.
If you are using a scope, measure the Vpeak across the 50 ohm resistor,
multiply by .707 to get Vrms.
73 Gary N4AST

What confuses me in your suggestion (printed below) is that even after you
rectify an AC signal and send it through a filter capacitor you STILL
suggest to multiply it times 0.707.
I would have thought that would give you the wrong result??

Uwe, Wouldn't be the first time I was wrong.
If you rectify and filter an AC waveform, and the read the DC voltage, this
is Vpeak of the AC. In the example I gave, you read 2 vdc out of the detector
circuit. Multiply this times the X10 divider, and you get 20vdc which is Vpeak
across the 50 ohm load. To get the rms voltage across the load
Vpeak*.707=20*.707=14.14 Vrms. 14.14Vrms across 50 ohms=4 watts.
BTW, the reason I suggested a voltage divider (attenuator) rather than
reading directly with a VOM or scope is to reduce the likelyhood of the scope
probe or meter impedance from detuning the AC-1 causing further errors.
If you are using a scope, measure the Vpeak across the 50 ohm resistor,
multiply by .707 to get Vrms.
73 Gary N4AST

Gary, you may be right regarding my power output measurement, but I thought
I knew how you calculate DC power and AC power of sinusoidial waveforms on
my scope, which, by multiplying with 0.707 you have to reduce first to the
aquivalent of a DC voltage.

What confuses me in your suggestion (printed below) is that even after you
rectify an AC signal and send it through a filter capacitor you STILL
suggest to multiply it times 0.707.
I would have thought that would give you the wrong result??

Where do I go wrong, if at all??

Uwe


in article , JGBOYLES at
wrote on 3/12/04 17:19:

So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output.

Uwe, The 10 Vrms that your meter indicates may not be accurate at 3.5 or 7
MHZ. It is probably OK at 60 hz. What you can derive from the 10 Vrms is
that
the AC-1 is producing power, just not sure how much.
One thing you might try is a method to remove the frequency dependency of
your measurements. Use a 10:1 voltage divider (10k and a 1.11k). Run this
thru a Germanium or Schottky detector diode and a .01 filter capacitor. You
now have a DC voltage that is proportional to power, and relatively frequency
independent.
To calculate the RMS voltage across the 50 ohm load: Read the DC volts out
of the detector-Vdc. Then Vrms=(Vdc*.707)*10.
Example: You read 2Vdc out of the detector. Vrms=14.14 volts. Power into
the
50 ohm load is then: 14.14^2/50=4 Watts. The diode drop in the dector will
introduce some error at QRP levels, hopefully not too much for what you are
trying to do.
73 Gary N4AST

Uwe wrote in message ...
So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output.

E squared over R, yes, two watts is about right.

All this at B+ of 200v and an anode
current of about 35mA (I am running the transmitter off an adjustable bench
tube power supply at this point).

Seven watts input, two watts output is a bit low. How are you
adjusting the controls?

Does all this compare with the numbers you get with your transmitter??
I will need to build a power supply for the final version and would be
interested to know what you use for B+. In order to get 5 Watts I would have
to crank up the B+ quite a bit.

IIRC that design called for 300 or 350 volts B+, which would work out
to 15 watts or so input. The output network is not optimized for 200
volts B+, and in addition the efficieny is better with more plate
voltage.

Actually I don't care at this point since I am still practicing for my code
exam, so I can't even use the transmitter right now.

What study methods are you using?

How do you monitor your transmitting tone with a device like this.

Several ways:

1) Connect an audio oscillator to the key so that both it and the
transmitter are keyed at the same time. This can be a bit tricky
because the two circuits must not interact.

2) Listen to the transmitted signal on your receiver when
transmitting. This requires that you have a way of reducing the
receiver gain while transmitting, but not completely silencing the
receiver. What sort of receiver do you have to go with the
transmitter?

73 de Jim, N2EY

So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output.

Uwe, The 10 Vrms that your meter indicates may not be accurate at 3.5 or 7
MHZ. It is probably OK at 60 hz. What you can derive from the 10 Vrms is that
the AC-1 is producing power, just not sure how much.
One thing you might try is a method to remove the frequency dependency of
your measurements. Use a 10:1 voltage divider (10k and a 1.11k). Run this
thru a Germanium or Schottky detector diode and a .01 filter capacitor. You
now have a DC voltage that is proportional to power, and relatively frequency
independent.
To calculate the RMS voltage across the 50 ohm load: Read the DC volts out
of the detector-Vdc. Then Vrms=(Vdc*.707)*10.
Example: You read 2Vdc out of the detector. Vrms=14.14 volts. Power into the
50 ohm load is then: 14.14^2/50=4 Watts. The diode drop in the dector will
introduce some error at QRP levels, hopefully not too much for what you are
trying to do.
73 Gary N4AST

Doug, thank you soo much for your email, you just fixed my transmitter.
Well, sort of.

As I tried to do what you had suggested I noticed that someone (I could
never have done such a thing) had connected the 18k resistor (which is
connected to grid # 2 on one end) on the wrong end of the choke.
I guess that choked it.

Now we don't just have oscillations, there is even a little bite to it.

I am relatively new to all this and I am just putting together the necessary
equipment. So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output. All this at B+ of 200v and an anode
current of about 35mA (I am running the transmitter off an adjustable bench
tube power supply at this point).

Does all this compare with the numbers you get with your transmitter??
I will need to build a power supply for the final version and would be
interested to know what you use for B+. In order to get 5 Watts I would have
to crank up the B+ quite a bit.

Actually I don't care at this point since I am still practicing for my code
exam, so I can't even use the transmitter right now.

How do you monitor your transmitting tone with a device like this.


Doug, thanks again, I am quite happy now


Uwe

KB1JOW





in article , Troglodite at
wrote on 3/11/04 14:26:

It oscillates nicely but does not have any power.
I know, it is not supposed to have much power but this one is weaker than
weak, the output couldn't drive any antenna.

I was wondering if there are AC-1 owners/builders around here who could
suggest a few things to try, who are knowlegable about this thing.

Did you do any voltage or current measurements? While I haven't played with a
real AC-1, I've built many similar 6V6 single tube transmitters. I have the
circuit for the AC-1, and it looks pretty similar to some I've built. First
thing to do is measure the B+ to the plate and screen. Next, open the plate
feed where it connects to the filter capacitor and insert a milliameter. With
the "load" capacitor at maximum, adjust the "tune" capacitor for minimum
current. Take a set of readings and get back to me.

Doug Moore KB9TMY (Formerly K6HWY)

In article , Uwe
writes:

I recently put together the old AC-1 tube transmitter for the 40 m band.

From a kit, or from your own collection of parts?

It oscillates nicely but does not have any power.
I know, it is not supposed to have much power but this one is weaker than
weak, the output couldn't drive any antenna.

How are you tuning it up and measuring the output power?

I was wondering if there are AC-1 owners/builders around here who could
suggest a few things to try, who are knowlegable about this thing.

First thing is the voltages and currents, as already suggested. Then there's
the size of the coil and your tuneup technique.

What sort of load do you have connected to the output?

73 de Jim, N2EY

It oscillates nicely but does not have any power.
I know, it is not supposed to have much power but this one is weaker than
weak, the output couldn't drive any antenna.

I was wondering if there are AC-1 owners/builders around here who could
suggest a few things to try, who are knowlegable about this thing.

Did you do any voltage or current measurements? While I haven't played with a
real AC-1, I've built many similar 6V6 single tube transmitters. I have the
circuit for the AC-1, and it looks pretty similar to some I've built. First
thing to do is measure the B+ to the plate and screen. Next, open the plate
feed where it connects to the filter capacitor and insert a milliameter. With
the "load" capacitor at maximum, adjust the "tune" capacitor for minimum
current. Take a set of readings and get back to me.

Doug Moore KB9TMY (Formerly K6HWY)