Horizontal SRCFPP
The SRCFPP is the third brother of the SRPP and White cathode follower. This topology confuses even seasoned tube gurus, which I have discovered whenever I bring it up with them. What makes it so hard to understand is that the bottom triode seemingly does nothing, as its grid is grounded. Moreover, the psychological snag occurs when the brain sees that as the top triode pulls up, the cathode resistor must also pull up, which would seem to imply that the bottom triode's conduction has also increased. In fact, it decreases.

I have already shown several horizontal SRCFPP versions in the past, the last time being in post 296, where we see this version.

Okay, I was wrong. I just did a Google search and it revealed that the true last time was in post 337, where we see this variation on the horizontal theme. (Please read post LM337, if you need more help in understanding how it works.)

Dang, I had forgotten that I came up with this circuit. I find this topology compelling. I would love to hear it. It's a push-pull output stage that uses a single-ended output transformer. Isn't that a liability, not a feature? Not necessarily.

One reason that I believe that single-ended tube amplifiers offer an advantage over push-pull tube amplifiers is found in the air-gapped output transformer, which—due to it's being stressed by the heavy DC current flow through its core— sidesteps hysteresis, the stickiness inherent in the core's flipping magnetic polarity; or, as the dictionary informs us, "the lagging of an effect behind its cause, as when the change in magnetism of a body lags behind changes in the magnetic field." In a typical tube push-pull amplifier, the output transformer's core flips magnetic polarity once per sinewave. In contrast, a single-ended amplifier's output transformer's core is constantly magnetized in one direction only, as the current may alter, but it never reverses direction.

Returning to this output stage, which is a buffer, the input and driver stage must deliver a huge output voltage swings, say +/-240Vpk, which does not fit within the 400V B+ voltage. What to do? The easiest solution is to use an inductor as the plate load on the driver stage.

The inductor will displace some voltage due to its DCR against the driver triode's current draw. Note that +390V is a truly hot plate voltage. A 6SN7's plate-voltage limit is 450V and the ECC99's is 400V. The inductive load allows the driver triode's plate to swing far beyond the B+ voltage.

Two constant-current sources can be used to make the horizontal transformation.

To prevent voltage starving the input triode's constant-current source we can use an input coupling capacitor and a two-resistor voltage divider.

With 12 volts, the constant-current source has no problem. The 110Vdc B+ voltage is a nice relief from the more typical 200V to 300V B+ voltages used in tube amplifiers.

The net move is to insert some Aikido Mojo, which explains resistors R1 & R2.

The way it works is easy to explain. The input triode is configured as a cathode follower that is loaded by a constant-current source, so its PSRR equals 1/mu. SO, if a triode with a mu of 20 is used, then 5% of the B+ ripple will appear at the output. But if we force the same 1/20th of ripple to appear at the right triode's cathode, then the headphone driver will not see a differential voltage, so a stellar PSRR obtains.

The above schematic shows a fleshed out version. here is an example of what often happens: a reader sees the above schematic, builds it, then complains to me that the PSRR is not all that stellar. I ask to see what values he used. He then sends me the following schematic. Note the difference.

He added an RC filter tot he first triode. Why? He didn't get the the idea behind the original circuit and he figured that the more RC filters the better. I then tell him to pull out all the Aikido resistors and capacitors. This brings the PSRR back down.

Actually, I quite this version, as it ensures equal cathode-to-plate voltages between triodes. Here is another approach.

The NPN transistor functions as an emitter follower, while the PNP transistor functions as a constant-current source. The LM317 is configured also as a constant-current source. Its resistor must equal 1.25V/A; in this example, we desire 20mA of current flow, so 62.5 ohms is needed. Since they do not make such a value, we can use a 62-ohm resistor. The two diodes are there to protect the transistors at turn-off, as they allow a discharge path for the capacitors.

 

Next Time
No more horizontal transformations.

 

//JRB

For those of you who still have old computers running Windows XP (32-bit) or any other Windows 32-bit OS, I have setup the download availability of my old old standards: Tube CAD, SE Amp CAD, and Audio Gadgets. The downloads are at the GlassWare-Yahoo store and the price is only $9.95 for each program.

http://glass-ware.stores.yahoo.net/adsoffromgla.html

So many have asked that I had to do it.

WARNING: THESE THREE PROGRAMS WILL NOT RUN UNDER VISTA 64-Bit or WINDOWS 7 & 8 or any other 64-bit OS.

I do plan on remaking all of these programs into 64-bit versions, but it will be a huge ordeal, as programming requires vast chunks of noise-free time, something very rare with children running about. Ideally, I would love to come out with versions that run on iPads and Android-OS tablets.