Johnson DKW modulator

[W9BHI]

I would like to fabricate a stand alone speech amp for my Johnson DKW.
The manual says that the modulator tubes (805's) need 15 watts of drive and the impedance is 3500 ohms grid to grid.
Could I use a Hammond 125E push-pull output transformer backwards and drive it using the 8 ohm output of a solid state amp?
The 125E is rated at 15 watts and the impedance across pins
 3 + 6 of the primary is 3400 ohms for a secondary impedance of 8 ohms.
Am I on the right track?
Thanks,
Don W9BHI

[VE3AJM]

http://amfone.net/ECSound/K1JJ10.htm covers what you're asking.

The modulator tubes in the Johnson KW are 810s.

Al VE3AJM

[W9BHI]

OOPS,  I meant 810's.
Thanks for the info.
Don W9BHI

[AB2EZ]

Don

Again:

The specification page of the manual states that the required audio drive power is 15W ... but that is almost certainly the average audio drive power with a sine-wave audio signal.

Modern audio amplifiers, and transformers like the one you are considering are specified at peak power. The peak audio power will be at least twice the average audio power associated with a sine wave. Therefore, you should use an amplifier rated for at least 30W of peak output power, and (for 100% modulation) preferably more than 48W of peak output power (see below)

The 15W transformer will saturate at a low audio frequency that is 40% higher than it is rated for (i.e. you will have less low-end capability)... because the required peak audio output voltage is 40% larger than the rms output voltage.

The specification page of the manual states that the input impedance to the Desk KW is 3.5k ohms. But (see below) this impedance is associated with a pair of (external) 810 modulator tube grid-to-ground resistors that must be added as part of the driver. This is not the intrinsic impedance looking into the grids of the 810 modulator tubes.

From my previous post, in answer to your earlier question:

See pages 19-21 of the Johnson Desk KW manual (available for download from BAMA)

The manual correctly points out that you need a balanced driver for the class B push-pull modulators that can deliver the required grid-grid voltage. According to the RCA 810 data sheet, this is around 380V grid-to-grid.

The manual also states that since the impedance looking into each grid (from a center tapped driver transformer secondary) will vary with the grid input voltage (because each 810 will draw grid current when its grid-to-cathode voltage is positive), one should place 3000 ohms or less of resistance from grid-to-grid (actually 1500 ohms or less between grid and ground on each side) at the secondary of the driver transformer.

It is these added input load resistors (1500 ohms or less from each grid to ground) that will result in the 3000 ohm (or less) grid-grid input impedance of the Desk KW modulator. The schematic of the Desk Kilowatt shows that there are no built-in 810 modulator tube grid-to-ground resistors. These need to be included as part of the audio driver. Also, the modulator bias supply voltage, that is produced inside the Desk KW, must be connected to the center tap of the secondary of the external audio driver transformer.

Since you are only building one driver, and since the peak audio drive power is modest (380V x 380V / 3000 ohms = 48W peak power) it would be a good idea to over-design the driver so that it is capable of delivering 400V peak grid-to-grid voltage into a 1000 ohm load; even though the two added grid-to-ground resistors are 1500 ohms each. That driver would produce much less distortion when the 810's draw grid current.

Stu

[W9BHI]

Stu,
I see in the manual that they use 1000 ohm 20 watt resistors on the secondary of a Viking 1 mod transformer.
I wonder why they have no load resistors when they use the Ranger as a driver?

Don W9BHI

[WD8KDG]

The magic is all in the 9 pin octal plug of the Ranger. Plus I betcha Johnson had a plan. The Ranger came out first in 1954, followed a year later with the KW. On page 9 of the KW manual note the mod xformer of the Ranger now becomes an audio driver output transformer by moving a few jumpers, etc on that 9 pin plug.

They were made for each other & a hard combo to beat.

Craig,

[AB2EZ]

Don

That's a very good question.

I don't know exactly what the EF Johnson design engineers had in mind with regard to the Ranger as a driver for the Desk Kilowatt.

The Ranger's modulator (in stock configuration) will (just barely) deliver around 81W of peak audio power into the Ranger's  (approximately) 650V/125ma = 5200 ohm modulation resistance. I.e. about 125mA of audio output current on 100% modulation peaks. I suspect that the Ranger's stock modulator (limited by the transconductance, etc. of the 6L6 modulator tubes; and the stock modulation transformer's turns ratio) would not deliver enough peak audio current to drive the push-pull 810 grids if the extra grid-to-ground input resistors were included. E.g. 380V/3000 ohms = 127mA of peak resistive load current, plus the extra grid current on modulation peaks (somewhere around 50ma, maybe more).

So, the Desk KW designer(s) probably opted to run the Ranger's audio output directly into the Desk Kilowatt's non-linear load; and to rely on the Ranger's negative feedback to provide some help on positive peaks (i.e. to produce more current gain, when the incremental load impedance drops)

Since you are building the audio driver from scratch, using modern audio components... if you want to get the best audio quality out of the Desk KW, you should use a driver configuration that can accommodate the grid current v. grid voltage characteristics of the 810's.

Note:

If you use a modern audio amplifier having a nominal 8 ohm or 4 ohm output, and a reverse connected audio driver transformer (having a sufficient peak power rating), the effective source impedance of the associated driver can be quite low.

For example, if the audio output transformer has a 2500 ohm output winding, and a 4 ohm input winding, and if the amplifier is specified to be able to drive a 4 ohm load... then the effective source impedance of the amplifier + transformer combination will be much less than 2500 ohms. I.e. the modern amplifier acts almost like an ideal voltage source with close to zero series resistance (because it has lot's of negative feedback inside). It can produce a peak output voltage that is a few volts less than the voltage of its positive and negative power supplies. Its source impedance is much less than its specified minimum output load impedance.  The specified minimum output load impedance (e.g. 4 ohms) is primarily specified because the amplifier will shut down (i.e. go into "protect" mode) if the current into the load is above some maximum allowed value.

For example:

Consider an amplifier with a peak output voltage capability of 16V (corresponding to 18V positive and negative power supplies). Suppose this amplifier was designed to operate into a 4 ohm load. 16V peak audio output voltage / 4 ohms = 4 amperes of peak audio output current. Suppose 4A was also the maximum current that the amplifier could supply without damaging its components. The peak power rating of the amplifier (into a 4 ohm load) would be 16V x 16V / 4 ohms = 64 watts. However, it is really the 4A current limitation in combination with the 16V peak output voltage limitation that determines how much power the amplifier can deliver to any specific load. This same amplifier would deliver a maximum of 32 watts into an 8 ohm load (limited by the 16V maximum output voltage). This same amplifier would deliver a maximum of 32 watts into a 2 ohm load (limited by the 4A maximum current that can be delivered without activating the "protection" shutdown circuit)

This would be a modern version of the low-source-impedance audio driver that Johnson was looking for.

The voltage step-up ratio of the 4 ohm input/2500 ohm output transformer would be the square root of 2500/4 = 25.

To obtain a peak transformer output voltage of 400V (to drive the 810's), one would need a peak transformer input voltage of 400V/25 = 16V.

Since most modern amplifiers are rated to drive a 4 ohm load (even if the actual load is higher than 4 ohms), the peak power rating associated with a 16V peak amplifier output is: 16V x 16V / 4 ohms = 64W

This does not mean that the amplifier will actually produce 64W when driving the Desk KW... because the actual load on the output of the transformer, driving the grids of thee 810's is always more (or much more) than 2500 ohms... What is does mean is that the amplifier is capable of producing (with high linearity) up to 16V peak output voltage into the transformer's input winding, for whatever load the transformer's output winding is connected to (even a non-linear load)... provided that load is at least 2500 ohms (and, therefore, the load on the amplifier is at least 4 ohms).

The step-up ratio of the transformer is the most important parameter in this application, because it determines the peak voltage that the modern amplifier must produce at the input of the transformer in order to obtain the needed peak output voltage (400V)

Bottom line:

I suggest that you use a modern amplifier rated at around 100W per channel, with each channel rated to feed a load having as low as 4 ohms resistance. Such amplifiers are amazingly "cheap" these days. You will only need to use one channel, but you could also use both channels in a "bridged" configuration to get 100% more peak output voltage.

I suggest that you use a push-pull output transformer, rated for 4 ohms input and 2500 ohms output, that is specified as being capable of handling a peak audio power level of 64 watts (again, to be able to produce the needed 400V peak output voltage... not because it will have to deliver 64 watts of audio power)

Stu

[AB2EZ]

Don

For the backward-connected output transformer, you might try a 9V 100W Antek (current production) toroidal ferrite core transformer, which comes with a pair of independent 115V (rms) primary windings. ($21.60 + shipping)

With the two primary windings in series (providing a center tap for applying the 810 grid bias), and the two secondary windings in parallel, you will have a voltage step up ratio of 230/9 = 25.6

These transformers work great as modulation transformers in modified Heising modulators... and they will probably work great in this audio driver application (no Heising reactor needed here). There will be some unbalanced DC in the output winding (since the 810s will draw somewhat different average grid currents), but it is certainly worth trying one of these.

Stu

[W9BHI]

Stu,
Did you mean the 100VA unit?
It is part #AS-1209.
Will I still need to put resistors across the grids and if so, what value.
Also how big of an audio amp will I need?

Thanks,
Don W9BHI

[AB2EZ]

Don

1. Yes... the AS-1209 100VA transformer

2. You should use a modern, solid state 2-channel audio amplifier that is specified to be able to deliver 100W peak power into a 4 ohm load, on each of its outputs. Again, the average power delivered by the amplifier to the audio input of the Desk KW will be much less than that. Example:

Samson Servo 200
http://www.sweetwater.com/store/detail/Servo200/

Minor detail: some amplifiers have a small amount of residual DC offset at their outputs. They do not like looking into a load that is too close to 0 ohms DC resistance. The 9V side of the Antek transformer will have a very low DC resistance. I suggest that you be prepared (if necessary) to add a 0.5 ohm 10 watt (wire wound is okay) resistor in series with the 6V winding of the transformer (i.e. at one of the output terminals of the audio amplifier). You may not need it, but it is good to have one on hand.

3. I suggest that you include a 10,000 ohm resistor between each 810 grid and ground (not 1500 ohms)... to soften the transition seen by the audio amplifier when each 810 starts to draw grid current. Each of these resistors should have a rating of 10W or more (wire wound is okay).

Stu

[W3RSW]

You guys keep adding to your driver amp requirements and you won't even need the 810's. 

Well, maybe not so  , after all if your looking at a nominal 350 watts carrier, then 200 good, clean audio requirements ought to satisfy it.   so why not try a modern 150 to 250 watt solid state amplifier already complete with scads of feedback and freq./gain leveling circuitry, and with distortion less that 0.1%.   Then feed the 4 ohm output into a large tube audio output transformer and then Heising mod that higher Z with the usual inductor and 2 to 4uf cap to the Class C output's plate?  You keep HV off the hi Z winding of the old audio xfor.

Of course if you want to go with a full gallon, transformers are harder to come by.

[WD5JKO]

With an external audio driver, make sure you kill the drive to the 810's when the transmitter is not keyed.

So what about a class D 50+50W audio driver that can be bought for $16.49 shipped:

http://www.aliexpress.com/item/TPA3116D2-amplifier-board-50w-50-w-digital-two-channel-amplifier-board-24V/1629195683.html

Some of the translated English chooses interesting words..makes me want to screem. 

Look at Mono Mode (PBTL):
http://www.ti.com/lit/ds/slos708c/slos708c.pdf

Jim
Wd5JKO

[WQ9E]

Jim,

Interesting feedback on that site    I like the one where the seller canceled the order and then refunded 50% of the amount paid to the buyer.  I guess that is Chinese version of gambling, you pay full amount you may get products ordered for cheap price or you may get nothing and we keep half of what you pay.

[AB2EZ]

Rich

Focusing only on the technical aspects of your post (attached below):

If a plate modulated transmitter is operating at 350 watts of RF output power, at carrier, then the plate input power, at carrier, will be 522 watts (assuming 67% efficiency, which is realistic for a Desk KW, but admittedly less than what is theoretically possible)

The average audio power needed to modulate the transmitter with a sine wave, at 100% modulation, is 50% of the plate input power... if you assume that the modulation process is 100% efficient.

However, the peak audio power needed to modulate the transmitter, at 100% positive and/or negative modulation peaks (regardless of whether the audio signal is voice or a sine wave) is 100% of the plate input power.

Therefore to achieve 100% positive and/or negative modulation peaks, while running 350 watts of RF output power (at carrier), one needs an audio modulator capable of delivering at least 522 watts of peak power.

To achieve 125% positive peaks, one needs an audio amplifier capable of delivering 522 x 1.25 x 1.25 watts of peak power = 816 watts of peak power.

Modern amplifiers are specified in terms of the peak power they can deliver.

Stu

You guys keep adding to your driver amp requirements and you won't even need the 810's.  

Well, maybe not so   , after all if your looking at a nominal 350 watts carrier, then 200 good, clean audio requirements ought to satisfy it.   so why not try a modern 150 to 250 watt solid state amplifier already complete with scads of feedback and freq./gain leveling circuitry, and with distortion less that 0.1%.   Then feed the 4 ohm output into a large tube audio output transformer and then Heising mod that higher Z with the usual inductor and 2 to 4uf cap to the Class C output's plate?  You keep HV off the hi Z winding of the old audio xfor.

Of course if you want to go with a full gallon, transformers are harder to come by.

[W3RSW]

Correct and thanks. " I knew that." In an effort to make a funny, I thought of output power rather than plate input power,  then went on to justify my error with real world miss-application.

Now that's sadly funny. 

[Opcom]

well I don't think I can add anything to this, save to agree with anyone who may have hinted that the driver of a class B modulator ought be able to do, undistorted, 3x the required max. drive power, with an additional caveat is that if the driver is push pull screen grid tubes, go even bigger for reasons of regulation of the rive voltage over the half cycle. That's the old 'book' value from the days when brute force was easier to use to overcome something than trying to explain or add windings to transformers.

Another way to regulate and drop the output Z a whole lot is to use cathode feedback on the screen grid driver amp. NFB from a low voltage secondary winding if you have such. The idea is to use a secondary winding, like a 25VCT winding (or the speaker terminals 0-4-16 Ohm, making the 4 Ohm GND) and bucking the cathodes with the two signals. So if bias is -35V, and you have 70V p-p grid voltage, you need another 36V p-p (12.5V RMS) of grid drive to the AB1 screen-grid tubes to overcome the cathode NFB. Usually the driver can do this, or, you may have to raise its plate volts a bit.

Here's an old article on that: http://www.montagar.com/~patj/aph1050.htm so making a 50W screen grid amp work much better for driving a class b modulator.

My work and article is kinda sloppy and used an existing 25VCT winding, so attached is a 1956 thesis by a Mr. Lockhart on the method and madness as applied to a McIntosh amp using the same kind of additional OPT winding except his has many more turns on the cathode winding of a custom transformer. Custom transformers are not necessary for this work, just the speaker or 25V line winding. Benefit is regulation like an ultralinear or better, but with good efficiency and full power of a straight beam tube amp. It also allows running much closer to class B1 without crossover distortion, in my case it came out to be half the usual idling current.

[W2PFY]

3. I suggest that you include a 10,000 ohm resistor between each 810 grid and ground (not 1500 ohms)... to soften the transition seen by the audio amplifier when each 810 starts to draw grid current. Each of these resistors should have a rating of 10W or more (wire wound is okay

Hello Stu, do you mean chassis ground or the center tap of the audio transformer for the 10k resistors. My goal is to get a different triode modulated transmitter on the air this summer that uses 810's.

Thanks

[AB2EZ]

Terry

The center tap of the audio transformer would be best... since the two resistors would not draw extra current from the grid bias supply.

Stu

[Tim WA1HnyLR]

Hi Don,
 The best audio drive for the Junkston Kilowatt is the cathode follower design used in the Gate BC1T and BC1G transmitters . The component values used would be pretty much the same in both cases. The biassing requirements for the 810 and the 833 are much the same. Using the negative feedback ladder that is used in the Gates transmitters as well would result in a very high fidelity low distortion signal. Another route would be the direct coupled FET audio driver that Steve WA1QIX has recently designed. As with any direct coupled driver be it hollow state or solid state the fidelity is not compromised by a driver transformer. The fidelity and frequency response is almost unlimited.The only limiting factor is the mod transformer.The stock mod transformer the Chicago CMS-3 is a very fine piece of iron but @ full power its' limits are being tested. The mod iron out of a Gates BC1T or G would work very well. I have a Junkston Kilowatt that I am going to shoe horn in a mod transformer from a RCA BTA1-R It will fit! some of the rocks will need to be re-arranged though. The mod transformer from the BC1T is smaller . I going to use a mod reactor from a Gates BC1G . The reactor can be placed down below . Solid stating the HV power supply and eliminating the rectifier filament transformer will free up much needed space .I may change the modulator tubes to a pair of 250 TH s or 3-500 Zs. Who knows when I will start in on the project. Much metal work to be done. Good luck with your project. Tim WA1HnyLR