Driver transformer question....

[kc2ifr]

This has been hashed over in the past but Im still cornfused.....so here is the question.
I need a transformer to drive the grids of either a pair 572B's  or 811A's  in a class B  modulator. I have an audio amp I will use to drive the transformer. I hope to use a backward connected Hammond transformer but I get different opinions as to the correct impeadence of the primary (secondary when connected backwards). Also I understand u have to swamp the grids with a resistor to make the transformer happy. What is the value...or how do u determine the value of this resistor? Do I swamp the whole secondary with one resistor or do I use 2 resistors.....one from each grid to ground. The transformer will be Push Pull.
 Thanks in advance for the help.
 Bill

[W1RKW]

Bill, I'm no expert on this but from the configurations I 've seen them go either way as far as having a resistor at the grids or not.  If there's a resistor it's usually a low value on the order of a couple of a hundred ohms. And they're on the grid of each toob. Again I'm no expert. I'm just repeating my observations.

[K1JJ]

Bill,

I would match the transformer's required impedance as shown in the datasheets with a pair of power resistors across the output to center tap.  Assuming the input is 8 ohms driven by an 8 ohm amplifier, then duplicate the transformer's output requirements/specs at the grid.

Though, I have run tone tests while watching the scope very carefully and could never see any difference between a terminated transformer or not in various rigs using that backwards type config.

There IS a very big varying load between zero grid modulator current and full current for AB2. Or in the case of driving the modulators in AB1, then no current at all, thus infinite impedance reflected back all the time.

So, either way, for the arc over safety of the windings and presenting some kind of constant load for the amplifier, I would use whatever 2K-6K? resistors in the secondary at the grids.  Putting 8 ohms in the primary will provide a pure resistive load for the amp too. Maybe do both.

In a few months I plan to build up the solid state modulator driver the guys here designed for my new 813 rig. Once debugged, you might be interested in doing it too, to eliminate the transformer and amplifier.

73,
T

[W2VW]

1. Listen to the Tron. I was not listening too carefully but it seemed like you are asking your driver power amp to play into a very low impedance. Yup it will work but it could work better.
2. Swamp the grids as mentioned. The modulator might try to oscillate if you have increased plate Voltage much from the original value in the JV-500. Been there.

[wavebourn]

Depending on audio amp loading it on non-terminated transformer you may damage it. Also, frequency response curve may be bad, again depending on an amp, if it has high output impedance like tube type. I would load the amp such a way it will see it's nominal load impedancen in case of an amp with high output impedance (tube type with light negative feedback), or ten times of a nominal load in case of transistor power amp. It is better to put the resistor right on the output of the amp (i.e. primary of your transformer) rather than on secondary (desired load impedance multiplied by ratio power 2), because inductive load may cause oscillations of some audio amps.

[WA1GFZ]

Do not use inductive resistors as the tube could use the inductance as a tank and oscillate. I would use 2 resistors and load the winding a little higher than the rated Z of the winding. The higher you go the better the low end response but the higher you go the more phase shift possible for high frequency response.

[WBear2GCR]

Bill,

It seems like the transformer ratio is simply a function of getting the required voltage to drive the tubes into AB2. If you are running an output transformer "backwards" that shouldn't be too difficult to achieve since they tend to have 4, 8, 16 ohm taps... the grids are only likely to draw a few watts - the RCA handbook should specify, so that's almost a non issue.

Making the low Z side look more resistive to the amp is probably not a bad idea - but I think the amp ur using is fairly stable into wierd reactive loads, being a pro audio device... something like a 16-32 ohm resistor seems reasonable across the amp's terminals... of course in actual use it won't take much of an amp to drive the thing... about a 10w P-P amp is quite sufficient iirc...

I'd run it on the test bench first with a big ass  resistive dummy load on the secondary of the actual mod iron sitting on the 811s and look at the waveform going into the grids with a square wave applied at a modest level and look for ringing. If there is ringing then you can try a simple resistor from grid to ground, an RC from grid to ground, and/or series resistor with the grid, etc... (one per grid of course). The leading edge of the square wave will show off any ringing...

But the best way to run the grids of a tube like the 811a is to use a pair of tubes like 6L6 or 6W4, or similar tube as cathode followers with a choke on the cathode feeding the grids... makes a very simple driver, all you need to supply is voltage to those grids equal to the swing required - the current is supplied by the driver tubes...

    _-_-WBear2GCR

[k4kyv]

The idea is to be capable of delivering an undistorted audio waveform to the grids of a class-B modulator stage, right up to the saturation point of the modulator tubes (i.e. the flat-topping or clipping point).  This is despite the fact that the load impedance to the driver stage, as presented by the modulator tube grids, varies irregularly from infinity down to about 500 ohms at each grid, over the course of the audio cycle.  Once the modulator tubes have reached the saturation point, it is of little importance what the waveform at the grid looks like, but you want to avoid driving the tubes all the way up to that point in the first place.

Swamping resistors can be used if there remains an excessive amount of reserve power output available from the driver stage once the modulator reaches saturation.  Just load down the driver transformer until the modulator and driver stages saturate at the same time.  But the same thing can better be accomplished by using more step-down in the driver transfromer turns ratio.  As long as the total plate-to-plate load on the driver stage is still within acceptable impedance range, swamping resistors will help maintain a constant load on the driver stage and reduce distortion.  If the driver stage is designed to just barely drive the modulator to saturation to begin with, as would be the case in a well-designed driver-modulator stage, adding swamping resistors will actually increase distortion.

It is desirable to use as much step down as possible at the driver transformer.  That is more effective than adding swamping resistors to an existing turns ratio.  But if you don't have the option to vary the step-down ratio of the driver transformer, while having plenty of reserve power available from the audio driverr, go ahead and use swamping resistors. 

Increasing the stepdown turns ratio will decrease the peak power output capability of the driver, while lowering the internal impedance of the driving source.  That's exactly what you want.  The internal resistance (impedance) of the driver stage should be as low as possible, and the peak power output of the driver should just barely be sufficient to drive the modulator tubes into saturation.

[W2VW]

For clarification: we are talking about a solid state power amplifier into a backwards connected tube type power output transformer driving triodes in class B. The triodes have no parasitic supression devices such as resistors in each plate connection.

[steve_qix]

This has been hashed over in the past but Im still cornfused.....so here is the question.
I need a transformer to drive the grids of either a pair 572B's  or 811A's  in a class B  modulator. I have an audio amp I will use to drive the transformer. I hope to use a backward connected Hammond transformer but I get different opinions as to the correct impeadence of the primary (secondary when connected backwards). Also I understand u have to swamp the grids with a resistor to make the transformer happy. What is the value...or how do u determine the value of this resistor? Do I swamp the whole secondary with one resistor or do I use 2 resistors.....one from each grid to ground. The transformer will be Push Pull.
 Thanks in advance for the help.
 Bill

Hi Bill,

At the risk of muddying the water    I have a new driver circuit that uses NO transformers (at all), and is perfect for the
application.  Using two source follower MOSFETs, you get an output impedance in the SINGLE DIGITS - and it is highly
unlikely the grid load will have any effect whatsoever on the driver output.  The circuit is derrived from the class H modulator, and is
proven to work very well. 

I'll help you build the low-level circuitry (those darn ICs!!!) if you're interested.  Maybe this deseves a PC board, if enough
people would want them.  With no transformers, the cost is minimal, and the quality is outstanding.  Low phase
shift means negative feedback around the modulator is also practical.  Protection circuitry is provided for the output devices,
in the event of a grid-to-cathode short or arc-back (grid to plate).

Something to think about...

Regards,

Steve

[WBear2GCR]

Steve,

The 811 is a zero bias tube, so it wants to see Zip volts on the grid - how ya gonna get close to ground with a class H style modulator unless you use a negative supply? And then how r U gonna keep it near DC? (servo it?) I'm interested though...

My solution, Bill, r u listening?, is to use the cathode of a pair of relatively tame, cheap tubes and a choke. Choke = cheap almost free. Keeps the 811 grids on ground through a very low DCR, which is good.

looks like this:

   B+
   |
tube
   |
   |---------> to 811 grid
choke
   |
 gnd.

Now you can use a step up xfmr to get the voltage swing to the grids of the driver tubes, which are followers. Would ya rather fork out big $$ for a transformer that needs to handle the output of a big solid state power amp, or use something that might run off something relatively lowly, like a tube preamp??
Lessee... an old tube preamp capable of 50 volts of output swing should be enough... and RCA calls for 175 volts (at only 50 ma) to drive the bejabbers out of the 811? So that's a turns ratio of only 1:4. Not hard to find... should be fairly inexpensive. I'd rather do the voltage step ups at low level than high... fwiw.

The follower provides the current (power) to control the 811 grids, the step up tranny (or other gain stage - see below) provides the voltage gain to swing the grids fullly.

Or you could mod a simple old mono tube amp by trashing the output iron and turning the output stage into a follower, per above! All done in one package. In fact you can simply turn the primary of the output stage into the choke... not much problem there. 

Another advantage of this circuit is that there is no HF limit on the choke driver stage - the LF limit is set by the H of the choke! 

     _-_-WBear2GCR

[steve_qix]

Steve,

The 811 is a zero bias tube, so it wants to see Zip volts on the grid - how ya gonna get close to ground with a class H style modulator unless you use a negative supply? And then how r U gonna keep it near DC? (servo it?) I'm interested though...

Hi Bear,

No problem :-)

The class H modulator employs a floating driver, and the driver moves up and down with the output.  Same thing applies here. 
The output of the "modulator" is, in essence, the grid of the tube instead of a class E amplifier.  There is a negative power supply
and a positive power supply.  The negative supply is there to pull the grid voltage below 0, and very little current is required.  The positive supply
comes into play when the grid is driven positive, and at this point, considerable current could be involved depending on how positive we take
things.

The modulator floating up and down creates an isolation problem whereby the input and output must be isolated.  This is accomplished
by using an op amp (floating) operated as a differential amplifier, and connecting both inputs to ground referenced sources.  As the op amp
(and associated circuitry) floats, the voltage between the floating driver (the output) and ground will appear at BOTH inputs of the op-amp,
and is in common mode.  The differential amplifier rejects the common mode signal.   Only the voltage difference between the two op amp inputs
appears in the output.  Of course, the actual voltage appearing at the inputs of the op amp is reduced by 20 or 30 times using resistors.

This system is very effective in the class H modulator, and that modulator is designed to swing up to 150 volts.  The same concept, and perhaps
the same circuit board should be applicable here.  I'll be trying it soon 

Regards,

Steve

[WBear2GCR]

Interestin idea steve... I'm worrying about it though... 

The differential between the output of the opamps and the ground still can't be permitted to exceed the device maximum... it's good that you've figured out how to scale the CM voltage back down...

I'm presuming a servo to control the "bias point"?

If ur interested I found an article on extending that range & CM range considereably, online in one of the engineering mags... might be very useful here. Email me.

To drive the 811s, of course, we're not talking much power at all, mostly voltage swing... and getting the grids to sit around "0 volts" quiescent. So how is this class "H" much different than a standard P-P solid state amp, maybe without complementary output devices, albiet with the negative rail set up to not run power?


        _-_-WBear2GCR