Using mismatched sweep tubes in amplifiers

[loosecannon]

Hi all,

In doing some reading/searching around the interwebs tonight, i came across an old article by Doug Demaw about how to balance the bias current when you can't get a set of matched tubes to use in an old sweep tube amplifier.

I thought this was a cool idea, and might incorporate it into my next project that will be using 6LF6 tubes (well, 6LB6 unless i can find someLF6's).

here is the portion of the article in question:

here is a link to where i found the article, which has 5 parts:
http://www.tetrode.co.uk/sweep.html


if you look at this article about building a sweep tube amp, it looks very similar:
http://www.qro.it/amp/schemi/pdf/72hb176.pdf


notice the part in the QRO article in the schematic where it looks like a negative 25.2 volts is being sent to a remote control.
can anyone tell me what that is being used for?
looks like its for external keying, but why is the voltage negative?

in the article about balancing the bias current, it shows them using negative 25.2 volts on the control grids of the tubes.

i am hoping someone here can clue me in to why a certain voltage level is used for tube biasing?
it seems to me that they went with a 25 volt secondary on the filament transformer because that's four filaments in series, so do we just run whatever we have in the filament transformer, rectified to negative DC?

any advice or insight into this type of design is much appreciated.
LC

 

[nomadradio]

i am hoping someone here can clue me in to why a certain voltage level is used for tube biasing?
it seems to me that they went with a 25 volt secondary on the filament transformer because that's four filaments in series, so do we just run whatever we have in the filament transformer, rectified to negative DC?

The power supply schemo will presumably show the heater voltage being rectified to obtain negative 25 Volts DC.

The tubes will typically need 8 or 10 Volts of negative bias. The 10k pot for each grid is fed from a 10k fixed resistor. Makes the max voltage from the pot half the negative 25-Volt source.

Using the same negative voltage to power the antenna relay just simplifies the thing.

What it really needs is a way to cut off the tubes' anode current while receiving. Typical way to do that would be to break the ground connection to RFC5 through a third normally-open circuit of the antenna relay.

And maybe the power supply diagram shows a relay that shuts off the 900 Volts in receive mode. That's another way to shut the tubes down in receive. You don't want them drawing idle current all the time. Wears out the tubes prematurely, and makes the amplifier efficient at only one thing, heating the room. Keying the "high" side of the power supply puts more stress on the relay contacts as a rule. Keying the 'ground' side of the tube circuit puts a lower voltage across the contact points when they open up, and reduces the size of the arc in a big way.

The input is feeding the cathodes of the four tubes in parallel directly, with no impedance-matching network. This will result in a high SWR between the radio and amplifier when it is keyed and amplifying.

The original article's date of 1968 is a clue. No ham transceivers with a solid-state final yet in 1968. In the era a tube-type radios this was no big deal. A tweak of the radio's Tune and Load controls would take care of this. A solid-state radio won't be so happy about this. A pi-network to match a 50-ohm transmitter to the four tubes is not a big deal. But it's not shown, either. Wasn't needed 52 years ago.

73

 

[loosecannon]

thanks so much for the response.

and thank you for the idea of shutting off the bias current on receive!

i won't be building this amplifier, just using some of the ideas from it.
the four paralleled capacitors feeding the output pi circuit had me wondering also.
im used to seeing a total of .0022 or .0047uF at that point.

my real question though, is where the figure of negative 8 to 10 volts DC bias for each tube comes from?
that's the type of knowledge i seem to be having trouble finding.
LC

 

[nomadradio]

Ahh, 'rule of thumb', pretty much.

D&A puts 8 Volts of negative bias on the control grids of the final tubes in a Maverick or Phantom. We routinely add this to the driver tubes. Factory left it off of them. Go figure.

Can't remember ever attempting to do the math to calculate this parameter.

Never needed to.

73

 

[loosecannon]

thanks for that!

i have been spinning my head around reading through a bunch of stuff, and i think some things are starting to stick.

here are some links to the articles i've been looking through in case anyone else wants to read up on this stuff.

https://www.bellscb.com/cb_radio_hobby/Files/Designing Ampiifiers.pdf

https://pdfs.semanticscholar.org/fa3f/ff89cd885c582cd1530dfe4842adba59d324.pdf

https://www.cpii.com/docs/related/22/C&F3Web.pdf

honestly i find the scope of these articles just a bit too wide and am having to parse them down to certain sections in order to learn what i need to.

switching off the bias in receive seems to be a wide subject in and of itself.
pin diodes, relays, high speed RF switching, etc.
seems a bit too complicated for a CB sweep tube amp.

i think i need to find something else written by Doug Demaw, as he seems to have a lot of practical advice included with the theory in his writings.

anyway, thanks for any interest so far, and please, anyone with tube amp knowledge please feel free to make this a repository for practical knowledge involved in building and modifying tube amps.

oh, another great tool for learning about this stuff is the youtube channel called "uncle doug". he's into guitar amps but his tutorials are great.
LC

 

[loosecannon]

What are the main differences in biasing the cathode or using the grid for biasing?
LC

 

[ExitThirteen]

At least from my experimentation, I've noticed that cathode biasing provides more gain than grid biasing... but at a cost.

You have to be careful of your input drive.

Excessive drive can damage the tube, and can cause potential oscillation issues.

For sweep tubes, this is generally a no-go, as grids and plates on these tubes are quite fragile, and will not hold up very well.



~Cheers~

 

[loosecannon]

thanks Exit13,

i appreciate the advice.
i doubt i would ever try something like that on a sweep tube amp, but when i saw it in a schematic it made me say huh.

maybe you can shed some light on my current conundrum, the grid biasing on the skipper 300 seen here: https://web.archive.org/web/2011011...300/graphics/palomar_skipper_300_sch_6lf6.jpg

what i can't seem to figure out is what voltage potential that control grid is sitting at.
it seems isolated from ground, and low power is determined by the 4.7K resistor to ground or switching in the 100 ohm resistor in parallel with it for high power.

from what i think i understand, for it to be class AB1, the DC voltage should be negative and never reach zero, but i can't find any DC source for a voltage on these pins (5 and 9)
what the heck am i misunderstanding here?
LC

 

[nomadradio]

The Skipper uses an odd setup that 'borrows' negative grid-leak voltage from the final grids and uses it to feed the grid of the driver tube.

When the switch is open for "low" side, the final tubes' collective grid current goes to ground through a 4.7k resistor. This limits the grid current and with it the peak plate current of the final tubes. It also produces a negative DC 'grid leak' voltage that serves to reduce the gain of the driver tube.

Closing the switch for "high" side puts a 100 ohm resistor in parallel with the 4.7k grid-return resistor. The grids of the final tube can now draw more current. The lower resistance reduces the grid-leak voltage and also the negative bias placed on the grid of the driver tube.

It's an odd setup that was used in some other sweep-tube amps, some of them with a pot in this circuit as a "dial-a-Watt" feature. The pot used in these never seemed to have a sufficient power rating and would routinely burn out.

I still prefer fixed bias, especially if you plan to use it for sideband.

73

 

[kopcicle]

 

[loosecannon]

thank you nomad!

looks like my next reading will be on how DC voltage can 'leak' from grid to grid inside a tube.

Kop, looks like the diagram you posted has to do with cathode bias and different ways to do it.
thanks for that.
LC

 

[kopcicle]

NP...



...unless it melts your brain

 

[Tokin]

NP...



...unless it melts your brain

And grid leakage comes with age...

 

[OldTech03]

And grid leakage comes with age...

Most likely it does. I know Brain leakage does?

 

[nomadradio]

Um, okay. The term "grid leak" used to be commonly understood back in the dark ages. But it's a misleading name for what it is.

The grid of the tube is normally placed at a DC voltage either equal to the cathode. So-called "zero bias", *OR* placed at a negative DC voltage, relative to the cathode. The greater the value of this negative voltage the less current flows from cathode to anode (plate). The negative polarity of the voltage on the grid pushes the electrons back toward the cathode, and reduces how many can pass through, on their way to the positive voltage of the tube's plate. The higher the negative voltage, the less electron flow passes between the grid wires.

What the tube cares about is the difference between those two terminals of the tube. The tube doesn't care whether the grid is grounded, or the cathode or whatever. It cares about that difference in voltage between the grid and cathode.

When a large drive signal has a peak-to-peak amplitude greater than the fixed DC voltage difference between grid and cathode, things change.

If you have a drive signal that's 25% larger peak-to-peak than the negative DC bias voltage on the grid, the grid has "pulses" of voltage that is now positive, compared to the cathode. The grid only has that positive polarity during that portion of the waveform. The rest of the time, the grid is more negative than the cathode.

Whenever the grid has a positive voltage, the grid now draws current from your signal source. Think of the tube as a two-element circuit. Making the grid positive now causes the electrons flying off the cathode to flow into the grid. So long as the grid is negative, it repels electrons, and no grid current occurs.

During those short pulses of positive polarity, the grid now acts like the anode of a diode. It rectifies the input signal.

Partly. A circuit that puts a diode across your input signal now produces a negative DC voltage that's greater than any fixed DC bias voltage you set up when no signal is there. The higher the drive level, the higher the negative voltage on the grid, created by this 'partial rectifier' effect.

This "partly rectified" DC voltage has a negative polarity, and is called the "grid leak" voltage.

Terrible name.

The only reason to do this is to get more plate current. Power amplifiers deliver more peak output current when you drive the tube harder. And grid current is what you get when the drive signal exceeds that threshold.

Tubes are not all created equal in this respect. If the grid wires are really thin, they won't tolerate much grid current, since it mostly just heats up the grid. The 4CX1000 tube is a good example. It has incredibly high signal gain, because the grid wires are spaced really close. But to leave a space between them for the electrons to sneak through, the grid wires must be really thin. As a result the 4CX1000 tube has a power rating for its control grid of ZERO Watts. You're really not allowed to push this tube hard enough to cause current to flow in the grid circuit.

Contrast this to the 3-500Z tube. It can't produce full output unless the grid gets driven positive for some part of the input waveform. The grid of that tube is rated for 25 Watts.

Sweep tubes fall between these extremes. The tubes in a D&A Phantom or Maverick only produce full output with some grid current.

Some.

But that's the story of 'grid leak' without graphic aids or animation. Would probably make it clearer.

73