R G's mosfet driver diagram - power Voltages to mosfet

[pjd3]

Hello.

I was wondering if someone could verify one thing regarding the voltages supplied to the Out put tube mosfet driver in R G's "mosfet follies" diagram.

http://www.geofex.com/Article_Folders/m ... tfolly.htm

The driver mosfet shows +Vs connected to the drain and the -Vs connected to the 10k resistor that is connected between the source and the output tube grid.

Are we to safely assume that the +VS and -Vs are simply representing the Plate voltage and ground that would normally be applied to a tube cathode driver stage?

Thank you. I an receiving a few of the recommended mosfet from Digi key today and want to make sure I am understanding the diagram as R G has laid it out.

Best,

Phil D

[R.G.]

That was written 23 years ago. I think it needs some updating and documenting some of the things you're asking.

First, are you using a MOSFET as a source follower directly connected to the plate of the PI tube and a capacitor to the grid of the output tube? If so, yes, the power supply can be the normal B+ and ground. This will cause some grid blocking issues on the output tube if you overdrive the grid.

If you're wanting to use the directly connected MOSFET source to the tube grid, then V+ and V- need some caveats. If all else lets it, the MOSFET grid will follow the bias adjuster setting for DC, and the cap feeding the grid will drive it with the AC signal. If the power supplies let it, the source will be about 3-8V lower than the grid; the adjuster on the bias setting lets you dial this offset out.

The output tube grid needs something like -38 to -50 volts as a static bias, so the source needs to sit at least that low, and have some swing lower than that. -Vs will then need to be at least -50V, probably better to be -75V to -100V. The more negative -Vs is, the higher the voltage across the source resistor gets, and the more you have to juggle power in that resistor versus resistor value. The drain of the MOSFET needs to be a bit more positive than you want to drive the grid. A MOSFET will "saturate" to a small voltage. For the (now kind of obsolete) IRF820s, this will be something like 10 ohms. I'd need to go refresh my memory from the datasheets, but for driving a grid, 10 ohms is effectively zero. If you're only driving the grids to as high as the cathodes at 0V, then the drain voltage on the MOSFET can be only a few volts positive. If you're planning to go to Class AB2 with the grids going positive, you need to supply the drain with enough voltage to get as high as you're wanting the grids to go. Probably +20 to +50V should cover it.

Of course, +50V and -100V are PITAs that don't exist already in the amp. The easy voltages, B+ at +350 to +500 and ground are not where you want +Vs and -Vs. High Vds MOSFETs can take +500V fine, but this makes for a high power dissipation if the MOSFET conducts much current. Using the amp's bias supply at -50V to -75V is probably not a good idea, as bias supplies are low current and the amp is fragile if anything goes wrong. Making a negative supply from the two-diode B+ by adding two more diodes and another filter cap can get you to +/- 350V to +/- 500V, but then there's a lot of power wasted and a lot of load on the power transformer. Might work OK if you don't insist on more than a few ma of positive grid drive, or if you mess with dropping resistors and zeners to lower the actual voltages on the MOSFET. This last is what I tested with. IF ... I remember correctly from 2000.

A good choice for a power supply might be to use a separate small PT rated at 120Vac to 120Vac CT. Full wave rectifying this ought to get you +/- 80Vdc around the CT. Mouser has a suitable one from Hammond for about $20: 546-162H120. This removes the loading from the main power supply, but is kind of a PITA itself.

[pjd3]

Thank you very much R.G. for stepping in to help clarify for me.

I'm new with implementing mOSFETS in this kind of role so sure can use the guidance. I find them to be a very interesting and worthwhile option going by the descriptions of how they can work in these guitar amp circuits and what their roles and benefits can be. I really want to learn them better.

What I think I'm getting here is that the mOSFET can be implemented with the same B+ node voltage as would be on the plate of a tube cathode follower "if" there is a capacitor between the source and the grid of the subsequent tube stage "But", if you wanted to remove the capacitor to go directly to the grid of the next stage (to eliminate the potential of blocking distortion when grid is driven positive to the cathode) then, you would need to consider applying the more custom/optimal voltages to the drain and source of the mOSFET.

Since this up and coming 2-channel tube amp will only ever be relegated to totally clean, sparkling tones and sounds, which means keeping the signal always well between saturation and cutoff, would you suppose I am OK to go with the capacitor driven mode of the NOSFET follower circuit, using the available B+ and ground already provided by the amp?

Thank you for taking the time for my pondering - I am actually excited and looking forward to experimenting with these mOSFETS both driving the passive EQ and implementing the output tube drivers to hear what they are actually doing to the output and tone of what is otherwise an amp of fairly straight forward topology.

Thanks again R.G. and everyone!
Best

Phil D.

[martin manning]

Some time ago I proposed a bipolar supply running off the 60V bias tap for the SSS cathode follower drivers. It provides +/-160V for a 12AX7. A number of people have built that, and I have not seen any complaints. No reason that wouldn't work for MOSFET drivers.

[Colossal]

Some time ago I proposed a bipolar supply running off the 60V bias tap for the SSS cathode follower drivers. It provides +/-160V for a 12AX7. A number of people have built that, and I have not seen any complaints. No reason that wouldn't work for MOSFET drivers.

This works great for a tube-based cathode follower driver

[R.G.]

If it works and doesn't lose you output tubes, great. I'm just very conservative about not wanting to overload the bias tap, not knowing how the PT maker actually wound it. Careful design of the MOSFET drivers to keep their current loading down would be prudent.

[R.G.]

What I think I'm getting here is that the mOSFET can be implemented with the same B+ node voltage as would be on the plate of a tube cathode follower "if" there is a capacitor between the source and the grid of the subsequent tube stage "But", if you wanted to remove the capacitor to go directly to the grid of the next stage (to eliminate the potential of blocking distortion when grid is driven positive to the cathode) then, you would need to consider applying the more custom/optimal voltages to the drain and source of the mOSFET.

That's pretty much the case. A MOSFET is pretty much a drop-in replacement for a tube cathode follower, working from B+ and ground. The actual source voltage will be a few volts lower than a 12AX7 in the same position, but this is negligible for a single supply follower. The direct coupled case needs a negative supply so it can have the source running at the negative bias voltage.

Since this up and coming 2-channel tube amp will only ever be relegated to totally clean, sparkling tones and sounds, which means keeping the signal always well between saturation and cutoff, would you suppose I am OK to go with the capacitor driven mode of the NOSFET follower circuit, using the available B+ and ground already provided by the amp?

I think so. Being able to drive an output tube grid positive to get a little more current from an output tube is a specialist kind of thing. If you just want clean audio and are not trying to wring a few more watts out, I would stop with the MOSFET follower capacitvely coupled to the output grid.

[pjd3]

Thanks so much all for stopping by for this.

When the Bias tap as the source of the Nosfet voltage was mentioned it did occur to me that bias source may not be expected to source a lot current and has to wonder if activity in the Nosfet could pull around (or pull down) the negative bias voltage but, at this point I don't have a developed intuition on what the current draws would be in the Nosfet for this application.

I think for now it nay be best for ne to keep things within my means and try the capacitor driven source follower with the amp B+ and ground. Of course, I will need to determine what that B+ voltage should be, what portion of the power node I ought to tap from. But I certainly won't go forward compulsively until I know why I'm providing what drain (faux plate) voltage to the Nosfet.

Thank you all and I'm in a much better place now to experiment with this Nosfet application. A Hammond OT (1650N) will be in in a day or three and that will provide the final critical component to spark up this amp.

Best

Phil D

[R.G.]

When the Bias tap as the source of the Nosfet voltage was mentioned it did occur to me that bias source may not be expected to source a lot current and has to wonder if activity in the Nosfet could pull around (or pull down) the negative bias voltage but, at this point I don't have a developed intuition on what the current draws would be in the Nosfet for this application.

The important thing is to decide how hard you'll let the MOSFET drive the output tube grid, whether through a capacitor or not. Pentodes (EL34, etc.) and power beam tubes (6L6, and cousins) have grid impedances that are quite high, even compared to the 220K typical grid leak resistors that are commonly used to connect them to the bias voltage. So call them 2.2M or more. The instant the grid is pushed above the source, this drops down to a few or several K ohms. So you might need 1ma or so. The MOSFET can pull up several amperes, so it has no trouble with this. But the resistor pulling the MOSFET source and the grid down can't pull as hard. Normal old-school design is to make the idle current in this resistor 10X the current it would have to pull down. Contrary to this, the source resistor only has to pull an overdriven grid down maybe 10V or so before the grid stops conducting and the grid current isn't needed any more.

The direct coupled case probably needs 5-10ma flowing in the source resistor. If you go for the capacitor coupled thing, the output tube grids never (intentionally ...) go positive, so you're always running in the high impedance grid mode. For this one, a couple of milliamperes in the source resistor is probably fine. The MOSFET and the source resistor have to eat all the power dissipation of the idle current in the MOSFET. Running from B+ to ground, a MOSFET and source resistor then have to dissipate B+ times the idle current. If you set these at 3ma and have a 450V B+, they have to get rid of 450* 0.003= 1.35W. This is split between the MOSFET and resistor by the idle voltage, but they're both heating by under a watt.

Going the direct coupled route, you need several times more source current in the MOSFET. If you do the obvious thing and make + and - B+, you're having the MOSFET and source resistor cope with 900V * 10ma, or 9W of power. Not a show stopper, but you'll need a 10W rated source resistor and have to heat sink a TO-220 MOSFET to get 5W of heat out. Plus you need two of these, so you're up for 18-20W of waste heat. This is the train of reasoning that says to run the MOSFETs from a lower total power supply. If you can get down to 200V total for the MOSFET plus source resistor, even at 10ma of idle current, you're only seeing 2W total in the MOSFET and source resistor.

I think for now it nay be best for ne to keep things within my means and try the capacitor driven source follower with the amp B+ and ground. Of course, I will need to determine what that B+ voltage should be, what portion of the power node I ought to tap from. But I certainly won't go forward compulsively until I know why I'm providing what drain (faux plate) voltage to the Nosfet.

For the capacitor driven version, just bias the MOSFET so it's conducting 2ma or so at idle. You do this by finding out what voltage is on the gate, assuming the source is 5V lower, and then computing the source resistor as R = Vsource/2ma. The resistor is then eating Vsource * 2ma of power, and the MOSFET is eating (B+ - Vsource) * 2ma of power. If the MOSFET is under 2W, it probably doesn't need a heat sink.

[pjd3]

Hi RG, thank you. You're explanations are clarifying a host of things that bring much more sense to all of this.

So as a sanity check for my understanding of this (or lack of), I'm getting the idea that by virtue of placing a high B+ kind of voltage on the drain, this voltage will sort off distribute itself to the gate and source of the mosfet based on this intrinsic design characteristics of the mosfet.
When you say check the voltage on the gate of the Nosfet, I'n assuming that this voltage is somehow deriving from the B+ at the drain. Is that assumption correct?

Then, where you said that we can assume that the source voltage would be perhaps a few volts lower than the gate, I'm taking that as a characteristic voltage relationship or drop intrinsic to the Nosfet itself. I say that believing that there is a cap going to the gate that would be blocking any DC that exists at the previous gain stage/triode.

But regardless if I'm getting the picture well or not, I would want to ensure that I' choosing a source resistor that is allowing around 2mA of quiescent current to flow from source to ground through the source resistor. I think I can manage that!

Thanks again for your valuable and generous guidance. I'd love to get a handle on the how's and where's of using these mosfets to see and hear how they change the presentation of the audible output!

Best,
Phil D

[pjd3]

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