VVR Design - comments and opinions welcome

[JazzGuitarGimp]

Hi All,

I am on the verge of starting a PCB layout for a fixed bias VVR Board. This came out of a conversation I had with strelok in a separate thread he started a few weeks ago. Strelok and I (mostly strelok) have been working through this design, and we thought it would be a good idea to open it up to the forum and start a discussion that might improve the design's operation and functionality. My goal is to do a layout that will be small enough to mount to the front panel by way of directly soldering the board to a dual-gang panel-mount potentiometer. I realize that heatsinking of the B+ MOSFET will be mandatory, and the intent is to mount the MOSFET close enough to a board edge such that the MOSFET can be bolted to the chassis for heatsinking. There will also be a 0.125" mounting hole in each corner of the board, in the event it is not practical to use a direct-solder pot. In the interest of making the board small enough to comfortably hang off a direct-solder pot, I may have to delete the Bias supply components R7, D5, R8, C1, and C2. If that is the case, I will do a layout for a small Bias Power Supply Board. Once the design is tested, I will make the board(s) available at a reasonable cost - this will be just the pc board, the buyer will need to source their own components and assemble the board. To that end, I will make a BOM available that will include manufacturer's as well as Mouser part numbers.

A cathode bias version will also be made available once the fixed bias version is in production.

If you have any feedback on the design at this point, please don't hold your breath.

Many thanks,
Lou

[roberto]

Hi, may I ask you why R2 and R9 have the same value and Wattage?

[strelok]

They're different wattage, 5w for the B+ and 1w for the bias. I believe the main purpose they serve is to provide some slight decoupling of the MOSFET from whatever is down the line in conjunction with the filter caps. Virtually every VVR circuit that uses this configuration has them, at least from what I've seen.

[JazzGuitarGimp]

I believe it is referred to a "break out" resistor and it's purpose is to keep the MOSFET stable (oscillation free). From what I've read, the breakout resistor should be employed whenever you're taking signal off the source, and especially when driving a highly capacitive load like a shielded cable (think effects loop send) or in this case, the 10uF electrolytic capacitor being driven by the source. This is also reccommended "good engineering practice" for cathode followers, but in that setting, a 100-ohm resistor is more common.

[strelok]

That would have been my next guess. Seen it done for that reason on old tube regulators, 6AS7's and similar types were recommended to have a 5 ohm cathode resistor for stability in regulator service.

[didit]

Liking inclusion of R4 & R11 on the wipers. Perhaps I'm tired & not thinking it through. Why not both direct to ground or both same as R4 to R6 and then ground but R11 via R13?

Thanks .. Ian

[strelok]

They're safety resistors as you probably guessed. There to keep the MOSFET's from oscillating if the wiper opens up. The reason we have it configured this way is so that if the wiper on the B+ opens up it sets the mosfet to the minimum B+ voltage, and if the bias wiper open up it defaults to the maximum bias voltage. This way you avoid damage to the tubes should one or the other become open.

[didit]

Got it. Thanks.

[xtian]

I'm exited about this. Does your circuit have "bias tracking", like L----n P---r?

[JazzGuitarGimp]

I'm exited about this. Does your circuit have "bias tracking", like L----n P---r?

This one tracks by way of a dual-gang pot; one gang for the B+ VVR, and the other for the Bias VVR. But I would like to make a true tracking model available in the future.

[roberto]

They're different wattage, 5w for the B+ and 1w for the bias. I believe the main purpose they serve is to provide some slight decoupling

Why these specific resistance values and wattages?

I believe it is referred to a "break out" resistor and it's purpose is to keep the MOSFET stable (oscillation free).

Usually oscillations are tamed at the gate, can you please provide me more material on the brak out resistors? You can implement current limiters with similar values and a bjt.

[JazzGuitarGimp]

The 5W part in the B+ circuit is probably a bit of overkill, but if the VVR were used in a 100W amp that pulls 400mA from the B+ node at full tilt, then 0.4 x 0.4 x 10 (ohms) = 1.6 watts of dissipation in the resistor, and a 5W part seems reasonable. My guess is most folks using power scaling are probably doing so with amplifiers of 50 watts or less of output power, and in that case, a 5W part is overkill.

And I misspoke; it's referred to a build out resistor, not a break out resistor. Nevertheless, there still is very little infotmation to be found on the internet. Here is a link to Merlin's discussion on build out resistors with regards to cathode followers. Scroll all the way down to the last two sentences.

http://www.valvewizard.co.uk/accf.html

[roberto]

Well, but you are not doing a cathode follower here.

On sloclone we have developed some years ago a VVR, you can find it included in the supply board on the OD3 section.

You can also check this reading: http://www.infineon.com/dgdl/Infineon-A ... 3d1a197d94

This is a kit sold by tube-town, including schematic:
https://www.tube-town.net/cms/?DIY/Amp- ... ge_Control

This is the kit developed by Dana-Hall:
http://searcysw.blogspot.it/2013/04/fan ... 5.html?m=1

There are also others on the market.
They are working on B+ only, but the implementation for the bias (look at the OD3 circuit) is basically a copy-paste concept.

DSCN2200.jpg

[roberto]

Of course there's also the most famous one, but it's way bigger: the london scaling.

[Ten Over]

R2 and R9 are parts of current limiting circuits. They set the maximum currents through the MOSFETs when the filter capacitors are initially charged or when there is a catastrophic failure of some sort.

For the B+ circuit, the Source will be a few volts less than the Gate. If R2 drops another 7 Volts, the 12V zener will break over and the Gate voltage will drop with respect to the Source and current will be reduced.

7 Volts across the 10 Ohm resistor is 700mA for a dissipation of 4.9W. The duration of the initial inrush is very short, so a 5W rating should be sufficient. There is the possibility that R9 should be 5W, also.