Bias voltage

[Barrie]

I've been gathering parts for a nostalgia-rig.  A buddy of mine, in junior high school, had a rig using a pair of VT-127As, 40M CW only, built by his ham father.  When I saw that I almost keeled over.  And then, I went home and looked at my HB pair of 6L6s!

Anyway, 50+ years later, I have several good, I think, VT-127As that I'd like to put on 75M phone.

I've got most of the parts I'll need for the RF deck.  My big concern is the bias voltage for the tubes.  The VT-127As are (so I've been told) similar to the 100TL, which requires, at my 1500 volt plate voltage, -300 volts bias.

I've built several transmitters over the years, but they have all been "cookbook" designs.  Well, this is not quite a cookbook design.

Can anyone help with a gridleak value and bias supply voltage?

73, Barrie, W7ALW

[AB2EZ]

Barrie

You could measure this experimentally... pretty easily... and at the same time determine whether the tubes are still good. [I am assuming you have a rough idea of what the maximum plate dissipation (watts) of one of these tubes is]

Set one tube up so that you can apply filament voltage (I am assuming that you know what the correct filament voltage is)... bring it up slowly using a "variac". Let the tube run with just filament voltage for about 8 hours. [No plate voltage yet!]

Place a 5,000 ohm 25 watt resistor between the cathode and ground; ground the grid through a 100 ohm 1/2 watt resistor (which will act like a fuse, in case of a plate-grid arc within the tube); and apply 500 volts to the plate (bringing it up slowly using a "variac").

If (for example) 30 ma of cathode current flows, then this will result in 150 volts (30 ma x 5,000 ohms) of bias from cathode to grid. In any event, the tube will develop enough cathode-to-grid voltage to limit the current to a few tens of ma... and this will give you an idea as to what the proper bias point is.

Then, you can (leaving the 5,000 ohm cathode resistor in place) apply various voltages between grid and ground to determine the change in cathode current ... and plot out a set of tube characteristics with 500 volts on the plate.

Repeat the above with higher and lower plate voltages

Stu

[Carl WA1KPD]

Barrie

When you get that set up could you send me a picture? Or do you have a pic of your buddies rig?
I have several of those kicking around that I want to do something with. And they just beg to be showed off.

I have tried to build a Hartley with one but no joy. However I was running very low B+ on it which may have been part of the problem

They are a cool tube

[Barrie]

Stu:

That sounds like an interesting experiment.

Yes, I do know the correct filament voltage.  I have burned the filaments on all six of the tubes for a period of time.  I hipotted the tubes, with very good results, as well.

The tube is a cross between the 100TL and the 250TL.  The filament and grid are 250TL, and the plate is 100TL.

When you say "place a 5000 ohm, 25 watt resistor (wirewound okay?) between cathode and ground", would that be from one of the filament pins, or from the center tap on the filament transformer?

I take it that I will be metering the cathode or grid current?

I'm still not exactly sure how this is going to work, buit I'm sure willing to try it.

73, Barrie, W7ALW

[AB2EZ]

Barrie

Super!

Since I'm proposing only DC tests... a wire-wound resistor will be fine (actually preferred, because of its ability to handle surges). For these tests, it doesn't matter whether you attach it between one side of the filament and ground or from the filament transformer CT to ground. Since you have a filament transformer with a CT, you might as well keep things symmetrical and run the resistor from CT to ground. Its purpose is to give you a safe starting point for turning on the plate current. Since the grid will be biased negative with respect to the cathode... there will be no grid current.

Once you determine an initial safe biasing point [the voltage across the resistor is the bias voltage; and the voltage across the resistor / resistor value is the corresponding cathode current (which is also the same as the plate current... because there is no grid current)]... you can apply some voltage between the grid and ground. I suggest that you leave the 100 ohm "fuse" resistor in place... just in case there is an arc-over from plate to grid. Again, there will be no grid current as long as the grid is negative with respect to the cathode.

As you change the grid voltage, the plate current will change... and the voltage across the 5000 ohm cathode resistor will, therefore, also change. As you increase the plate current, you will probably want to change to a smaller value of cathode resistor. For example, when the plate current is more than 50mA, you might want to change to a 2500 ohm cathode resistor (50mA x 2500 ohms = 125 volts). When the plate current is more than 100mA, you might want to change to a 1000 ohm cathode resistor (100mA x 1000 ohms = 100 volts)

Just remember: every time you reduce the cathode resistor value (as above) you will also reduce the contribution of the voltage across that resistor to the biasing... and you will want to reduce the grid-to-ground voltage to compensate.

What you need to "plot" is the plate current (i.e., the voltage across the cathode resistor / value of the resistor) vs the grid-to-cathode voltage (i.e. the grid voltage minus the cathode voltage) for fixed plate-to-cathode voltage. In order to keep the plate-to-cathode voltage fixed (e.g., 500 volts or 750 volts, or whatever) you will have to tweak the "variac" on the plate supply.... so that the difference between the plate voltage and the cathode voltage remains fixed as you vary the voltage on the grid. Once you get the hang of it... it will be easy to plot these curves of plate current vs grid-to-cathode voltage... for a fixed plate-to-cathode voltage.


Stu

[W3RSW]

p. v48 , 1952 handbook
vt127a
Pd = 100 watts; fil. 5v @ 10.4amps; mu=15; base N; socket conn. T-4B,
Class C Ep=2000; Eg= -340; Ip=210ma; Ig=67ma ;; Drive pwr=25watts; Pwr out=315
Class B audio, 2 tupes PushPull; Ep=1500; Eg= -150;Ip=242; Ig=44ma;Drive pwr=7.5 watts; Pwr out=200.

Cap, grid to fil=15.5 pf ; C gp = 2.7; C plate to fil. = 0.35; good to vhf; max freq. full ratings = 150 MHz.

Yes it certainly is a buzzardly looking pube; way kuhl.  Looks like it was designed for some sort of strip line inductance mounting on both grid and plate.

Only worry would be mounting without stressing pins under thermnal change and, of course, gas due to age.  Think I'd run them several hours on fil. alone, then several hours with plate voltage; don't know if it has a getter.

Let's see if I'm doing this right....   if your running all grid leak in class C (without any other bias) the resistor would be 340/.067 or 5k ohms; Dissapation would be 22 watts so you'd need at least a 50 watt resistor.
Not a super efficient tube all the way around but then great in those days for 144 mhz.

Hope that helps.

[AB2EZ]

Given the information from Rick, perhaps it might be better/easier to put these tubes on the air using a somewhat different approach... which will result in a little less output efficiency, but a lot less drive power.

Note: The peak grid current is going to be a lot larger than the average grid current (67mA). I would guess that the peak grid current will be at least 10x the average grid current. I.e. Grid current will only flow when the grid is positive with respect to the cathode... which will happen only during a small portion (20% or so) of the positive-going half of the drive voltage cycle. Thus the drive circuit will probably have to deliver more than 0.67 amp of peak current at the same instants when the peak drive voltage is somewhat more than 340 volts. That's a peak (instantaneous) drive power (for a small portion of the rf cycle) of more than 230 watts. If you wanted to deliver that using a swamping resistor at the input (not a good idea) you would need more than 115 watts of average drive power (i.e., the average power produced by a sinusoidal rf drive signal, driving a swamping resistor that presents a fixed load, is 0.5 x the instantaneous peak power). You'll do much better with a properly designed tuned circuit on the input... but you will still need to deliver a theoretical minimum of the 23 watts that Rick calculated if you bias the tube using the flow of average grid current through a resistor. [Note: The tuned circuit stores energy during each rf cycle... so, ideally, it can deliver more than 230 watts of instantaneous peak drive power, while being "refilled" with energy at the average rate of 23 watts. To accomplish this feat... the input tuned circuit must have a Q of at least 10. This implies that you will be tweaking the input tuned circuit when you change frequencies within a given band.]

I would suggest that you consider using fixed negative bias (from a grid bias power supply*) on the grid, so that you don't have to drive the tube to the point where the grid is positive with respect to the cathode. That way, no grid current will flow (no peak grid current, no average grid current), but the tube will still be operating in Class C (plate current flows only for a portion of one-half of the rf cycle) or (if you prefer) Class B (plate current flows only during one-half the the RF cycle).

*Since the grid current will be zero, the grid bias power supply doesn't have to deliver any power.

The tubes will be running at somewhat lower overall efficiency (maybe 60%-70% rf output power v. plate input power, instead of 75% if you drive the grids to the point where grid current flows, as per the numbers in the specification that Rick posted for Class C operation).

However, without grid current flowing it will be a lot easier to drive these tubes.

Example, if you were to run them in Class C with the grid bias set to around -250 volts (somewhere between -150 and -340 volts) and a peak drive voltage of around 245 volts... then you would not draw any grid current... but you would still be able to adjust the plate tank circuit to get a good portion of the maximum rf output that the tubes can produce at whatever B+ value you decide to use.

Just make sure that the plate input power [B+ value x average plate current] minus the rf output power is kept lower than the rated plate dissipation. Looking at the numbers that Rick posted for typical Class C operation... if the B+ is 2000 volts, and the average plate current is 210 mA, then the plate input power will be 420 watts. If the corresponding RF output power is (when the average plate current is 210 mA) 315 watts... then the power being dissipated by the plates (as heat) is 420 watts - 315 watts = 105 watts. Thus, based on the numbers that Rick posted... this tube has a recommended plate dissipation rating of 105 watts (or less). Note, also, that using these numbers... the tube would be operating at an efficiency of 315 watts rf output / 420 watts plate power input = 75%

Stu

[Barrie]

Stu & Rick:

Thanks for the information!

Now I'm wondering if I should continue with the previous experiment, where I conect the HV to the plate and meter the cathod current?

If this experiment would allow me to match up the tubes, then it might be worthwhile.  Plus I drug out my high voltage, variac controlled, 0 to 3000 volt bench power supply last evening.  It still works!

Also, I've gathered all the parts for the project.

Stu, let me know.

Also, with regard to the bias supply:  Would I be totally with a grid-leak?  And, while I have a heap of transformers that would provide the required bias voltage, all of them are big and heavy.  I take it that I only need a very small transformer?

Will the bias supply require regulation?

73, Barrie, W7ALW

[w1vtp]

...Stu & Rick:
Also, with regard to the bias supply:  Would I be totally with a grid-leak?  And, while I have a heap of transformers that would provide the required bias voltage, all of them are big and heavy.  I take it that I only need a very small transformer?

Will the bias supply require regulation?

A small fil XFMR hooked backwards to an existing fil supply (6.3 fil to 6.3 XFMR)  What used to be the primary can now be used as a secondary in either a bridge or voltage double circuit to get you either ~-150V, in the case of the bridge or ~-300V in the case of a voltage doubling circuit.  Might want to be mindful of any hi-potting issues regarding the XFMR.

Oh yeah, it seems to me that you might want to prevent the bias from climbing from what your supply puts out as you provide RF drive I seem to remember having to put some VR tubes in my old HB rig to prevent that from happening.
73, Al VTP

[AB2EZ]

"Now I'm wondering if I should continue with the previous experiment, where I conect the HV to the plate and meter the cathode current?"

I would say that now that Rick has provided you with a source for tube data, and some of the data as well... its probably not worth the trouble to perform the experiment to measure the tube characteristics.

"Also, with regard to the bias supply:  Would I be totally with a grid-leak?  And, while I have a heap of transformers that would provide the required bias voltage, all of them are big and heavy.  I take it that I only need a very small transformer?

Will the bias supply require regulation?"

The bias would be applied as follows: run the 5k ohm resistor (as calculated by Rick for the case of pure self-biasing) to the output of the negative bias supply. That way, the grid will be biased at a negative voltage which is the sum of the bias supply voltage + (5k ohms x the average grid current). To get the bias you need, the average grid current will be much lower (even zero) than in the case of pure self-bias... because a large part (maybe all) of the bias is provided by the bias supply.

As an interesting (to me) side comment: If any grid average current does flow when the rf drive is applied... it will push power into your bias supply (somewhat non-intuitive) rather than requiring your bias supply to deliver power! Thus, if it wasn't for the large bias voltage required, you could use rechargable batteries for the bias supply. Whenever you applied RF that is large enough to cause grid current to flow, that would recharge the batteries. But anyway...

You refer to the resistor as a "grid leak" resistor. While it is attached in the same place as one would attach a grid leak resistor... it is really not a grid leak resistor. The purpose of a grid leak resistor is to remove any charge that accumulates on the grid of a tube due to electrons landing on the grid and sticking to it. The purpose of the self-biasing resistor is to discharge the coupling capacitor between the drive signal source and the grid of the tube. If the grid draws current on positive peaks of the drive signal (which will happen whenever the drive signal makes the grid go positive with respect to the cathode), then this coupling capacitor charges up. The self-biasing resistor provides a (roughly) constant amount of average current to discharge the capacitor. At the same time (actually as a part of the process), the current flowing through this resistor places a negative bias on the grid. The whole process is self adjusting. Its a cute approach for biasing the grid... but it has a number of drawbacks. For example, if the rf drive signal goes away, the bias on the grid goes away... and the tube will typically draw excessive current.

With respect to regulation... it is a good idea to use a couple of 150 volt, 5 Watt Zener diodes to regulate the bias supply. Remember, the supply won't be delivering any power to the tube. If the rf drive is high enough to cause average grid current to flow, then that average grid current will flow into the supply. I would suggest that the Zener diodes float at around 10 ma of average current (10 ma x 150 volts = 1.5 watts of dissipation for each Zener). If the grids draw an average current of 20ma (not the 67mA associated with pure self-biased operation), then the Zener diodes will sink a total of 30mA. 150 volts x 30mA = 4.5 watts per Zener diode.

Make sure to use a large enough capacitor to keep the peak of the residual AC ripple on the bias supply below 1% of the bias supply voltage. Otherwise, the AC ripple on the bias supply voltage will result in an audible modulation of the rf output of the amplifier (audible hum).

Stu

[w3jn]

A REAL man's rig needs 872As for the biass rectifier!