Another goal was to be able to use a miniature HC-49/U crystal in a tube oscillator. Will that work?  Some say tube oscillators have too much crystal current for these crystals.  My research said that a tritet oscillator has low crystal current, so I settled on figure 8, page 249 of Orr’s The Radio Handbook (1959), substituting a 6CL6 for the 6AG7.  It sort of complicates things - there's a tuned tank output so one more thing to adjust.  That's part of the fun though, right? And there's a parallel tuned circuit in series with the crystal tuned 20% or so above the crystal frequency.  How do you work multi-band operation with that thing?  BTW - find the Bill Orr book and lots of other vintage tube technology books here -

http://www.pmillett.com/tecnical_books_online.htm

Let's show the schematic before further discussion ...

The amplifier stage is a more or less standard circuit with a pi-net output and capacitor coupled input, shunt feed to the plate and cathode keyed.  The 5763 seems to be a favorite tube among glowbug QRPers, after the 6L6 maybe.  The grid leak and plate and screen operating voltages were taken from the data sheet. OK then, a pi-network is designed by specifying Q and input (plate) and output (50 ohm) resistances.  What should the plate resistance be?  I assumed it should be calculated like you do for a class C transistor amplifier, basing it on the desired power out and supply voltage.  My notes on the subject:

Resistance at pi-net input:

There are plenty of calculators for doing a pi-network, but I need to know what resistance to convert the 50 ohm load to.  I think for class-A it’s a matter of a simple load line, but I’m not sure about class C.  Can I use the same formula we use for class C transistor finals, which is Z = Vcc^2 / 2*Po, or supply voltage squared divided by twice power output? 

 Using the above with 250 volts and 6 watts, I get 5,208 ohms.

The 1969 ARRL Handbook has design guides and charts for a pi-network on page 158. You start with the ratio of plate voltage to plate current in mA.  That should be 250/34.3 or 7.3 in my case. Using those charts and a chosen Q of 15, with C1 on the tube side and C2 on the load side,

XC1 = 240 ohms, XL = 265 ohms, XC2 = 36 ohms.  Using these values and a formula for computing Rin from C1 and C2 (QST 1/84 page 48) in my spreadsheet, these values indicate that R at the tube side is 3360 ohms. 

Reg Edwards has a pi-L calculator that also gives plate load resistance for the power and voltage level.  For my case of 6 watts and 250 VDC, he gives about 5,200 ohms as “ideal” and 5,000 as “actual”, whatever that means.  It seems that “ideal” maybe have been calculated using the formula I used above.

After a posting to EMRFD, Randy AD7ZU suggests a formula of 0.6 * Vsupply / I, where I guess I is the DC supply current.  This gives me 4411 ohms so it lands in the middle. Randy also suggests that a pi-net won’t keep me legal for harmonics and I should use a pi-L.  The pi-L is described in Elmer Wingfield’s QST 8/83 article so maybe I’ll give it a try. 

Glen Leinweber VE3DNL goes along with the rough formula V^2 / 2Po and thinks a pi-net would with Q of 15 would just about keep things legal. Dan Tayloe also thinks the same formula would be used.

Screen voltage regulation for 6CL6 oscillator

This was more of a pain than expected.  The data sheet shows 7 mA at 150 volts.  Is it OK just to use a series dropping resistor?  I eventually decided to use four 36 volt zeners, 0.5 watt rated.  With 15k in series, I can get 6.66 mA through the zeners if the tube doesn’t draw anything.  I mounted the zeners on a terminal block with bare lugs crimped to leads cut short in the interest of power dissipation or heat sinking.  Trying out the assembly on the bench with 150 volts, I got 148 VDC immediately which drifted up to 150 VDC.  I’ll leave it on for about an hour to make sure the power level (about 0.25 watt per zener) doesn’t blow up the zeners.

Tank in series with crystal-

This is what makes it a tri-tet, I think.  It’s supposed to be resonant above the crystal frequency by 20% or so, I think.  The coil I calculated to be 2.0 uH for 40 meters (maybe a little high for 30 meters) and I was going to use another real air core inductor but they’re such a pain to mount I cheated and went with a toroid instead.  I used 20 turns #22 on a T50-2 core.

RFC to shunt feed the final-

I needed another RF choke and was low on pi-wound chokes.  Came up with a couple 1 mH small ones and a big 2.5 mH.  I hate to waste one that’s big enough for a 100 watt or bigger rig on this rig, but 1 mH only gives about 22,000 ohms on 80 meters, which is only 4 or 5 times the plate load resistance, a little marginal.  I decided to use both 1 mH chokes in series.  They’re nice in that they have a 6‑32 stud for mounting on a panel. 

Plate current or relative power metering-

I initially wired the B+ through a miniature lamp rated 12 volts at about 60 mA, so I assume I’m only dropping 5 volts or so at 40 mA or less. But I noticed that using the plate current indication lamp on the tube TX wasn’t effective at all in tuning for maximum output power.  There was not a good sharp dip.  So I decided to try measuring the output current at the antenna jack and tuning for maximum antenna current. I initially wanted to have that drive an LED and tune for maximum brightness, but I found that LEDs go from dim too bright to abruptly so I used a meter. The circuit is shown on the schematic and there's some more discussion on my blog entry-

http://wa5bdu.blogspot.com/2010/04/rf-output-current-meter-for-qrp-tx.html

Backwave?

I noticed that the TX was keying my RF-sensed antennas switch when not transmitting. With key up I measure 20 mW out. So there’s only 23 dB of difference.  Not sure how this happens.  Maybe capacitively coupled from grid to plate of the amplifier tube?  I don’t see it as a big problem.  I added a B+ switch so I can put the TX in standby with the filaments still on.  I have to be able to do that anyway when I go to receive since my oscillator runs continuously.

About voltage and power

I was sort of disgusted getting just a little over a watt out on 40 meters with my 200 volt power supply that drops to 190 on key down.  So I hooked up to the bench supply and with key down, cranked up voltage to design value of 250.  Now I’m able to get four, maybe five watts out.  My design calculations were probably OK as far as target power level, but one needs to remember to allow for voltage drop with unregulated B+ supplies.

About building tube rigs-

Wow, it's a major pain in the neck, ain't it?  Mounting just about any component is an adventure in mechanical engineering.  For lots of hams with skills in this area, I know that can be lots of fun and the final product can be really nice to look at.  But for me, if it weren't for Manhattan, Ugly, Wire Wrap and Proto Board methods, I'd probably never get anything built.

How does it work?

Looks pretty good.  I figured if the HC49/U crystal was being over driven it would let me know either by failing completely or by chirping excessively.  But I don't seem to have much if any chirp, so I'm calling it success.  I've only had it on 80 and 40, so despite the schematic, this is a two band rig.

I've had two QSOs so far, matching the TX perfectly with a Hallicrafters S40A receiver.

And just because I love the looks of those pi-wound chokes, here's a picture of the rear of the rig.  Don't they look like futuristic hi-rise apartment buildings, as in The Jetsons? Also shows the RF current metering board with the antenna output lead passing through the toroid.



There you have it -

Nick, WA5BDU

A couple of updates:

1) I was very pleased to have my little transmitter described in  CQ magazine's QRP column, written by Cam, N6GA.  It's in the April, 2011 issue.

 Cam also came up with a neat power supply circuit (should be in the June issue) that uses something I didn't know existed -- a little LR8 3-terminal TO92 regulator (the size of a 78L05) that can handle 450 volts on the input side.  He follows it with a TIP50 pass transistor to get his 250V.

2) I received an interesting email concerning my transmitter from David Newkirk, W9VES.  David has a lot more experience and knowledge in the area of tube transmitter design than I do and offered his suggestions and comments which I reproduce below.  I haven't had time to try them yet, but plan to.  Meanwhile anyone want to follow along the lines of my design might want to consider the ideas below.

From the email by David Newkirk, W9VES:

I've been enjoying your 6CL6-5763 transmitter writeup ( http://pages.suddenlink.net/wa5bdu/2_tube.htm ), and have a few suggestions for how you may be able to get more power output and reduce your backwave. These suggestions have to do with determining (and controlling) the drive to the 5763, and neutralizing the 5763.

Overdriving a power amplifier can actually reduce its output; pentodes and beam power tubes (like the 5763) are particularly sensitive to this. An easy way to determine how much drive your 5763 is getting is to install a 100-ohm resistor between the ground end of the tube's 22-k grid resistor and ground, and then bypass that junction to ground. Now, by probing that junction with a DMM, you can easily determine the 5763's grid drive: 0.1 V across 100 ohms equates to 1 mA. I see from a quick look at the 5763's specs that its grid current should be around 1.6 mA during normal CW operation. If you're driving your 5763 much harder than this--say, at 2.5 mA or higher--you may be reducing its output.

If you discover overdrive to the 5763, the best way to adjust it is to reduce the screen voltage to the 6CL6. Maintaining a regulated screen supply to the oscillator is a good idea, so I'd put an adjustable voltage divider between Zeners and the oscillator screen, with the screen at the divider tap. I recommend not adjusting the 5763's drive by detuning its grid.

A 23-dB-down backwave is not out of line with your 5763 because it's not neutralized--that is, because power can flow through it bidirectionally as a result of feedthrough through its grid-to-plate capacitance. One result is that the oscillator signal is getting through to the output network--at a reduce level--and out to the antenna through the final's output matching network. I recommend installing capacitive bridge neutralization to take care of this. When you've neutralized the 5763, you'll find that the tube is easier to drive *and* that your backwave is much reduced. (The quickest way to try out neutralization would be to replace C3 with a few hundred pF--say, 220 pF--and connect a 1-to-10-pF variable, of suitable voltage rating, between the bottom of the oscillator plate tank circuit and the plate of the 5763. Adjust the neutralization cap for minimal variation of the 5763 grid current as the 5763 plate is tuned through resonance--a good, old-fashioned electromechanical meter shows this better than a digital meter. You can get close by adjusting the neut cap to null the backwave.)

The main reason for neutralizing your 5763, however, is that in its current configuration it's almost certainly operating as a locked oscillator. That is, it's actually oscillating on its own at a frequency finally determined by the crystal-controlled signal from the 6CL6. To prove to yourself that this is likely the case, pull out the 6CL6 and key the 5763 (just dits; now it's getting no drive from the 6CL6 and will draw too much plate current if it doesn't take off on its own). Very likely you'll still have output--possibly more output than with the 6CL6 in play! This happens because the grid-to-plate capacitance of the 5763 is more than enough for it to act as a tuned-plate, tuned-grid oscillator when its grid and plate are tuned to, or close to, the same frequency.

_________________

So that's all pretty interesting.  Especially, I think, that you can measure grid current in that way.  My understand is that RF grid current only occurs on one half cycle, or close to it, so it has a DC level. It's blocked by the coupling capacitor and  so the DC return path is through R3, where David would add the 100 ohm resistor with which to make the measurement.

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