understanding fet noise curves

[Merlinb]

But it is not irreducible.

When I say 'irriducible noise' I'm talking about the intrinsic noise of the pickups (and other guitar electronics) that you actually have; it's just the accepted jargon, sorry for confusion. Sure, you can design lower noise pickups and guitars, but few of us have that luxury! Your new super designed pickups will still have their intrinsic noise referred to as the 'irriducible (source) noise'.

[Mike Sulzer]

When I say 'irriducible noise' I'm talking about the intrinsic noise of the pickups (and other guitar electronics) that you actually have; it's just the accepted jargon, sorry for confusion. Sure, you can design lower noise pickups and guitars, but few of us have that luxury! Your new super designed pickups will still have their intrinsic noise referred to as the 'irriducible (source) noise'.

But when you say "Reduce it to 10k or even less (with suitable extra shunt capacitance if necessary) and you can approach the irriducible noise of your pickups.", I think that is not quite good enough if you make the "irriducible noise" lower as I described.

[Merlinb]

the lower device has unity gain (gm/gm) and that therefore its noise contributes the same as the top device.

That's true if they're operating in the saturation region as the analysis assumes. But wouldn't our practical circuit be more likely to operate in the triode region? i.e. the impedance seen looking into the top device will be larger than 1/g_m2, so the lower device gain will be higher than unity.

EDIT: I was wrong, the top device is not likely to be operating in triode mode unless your supply voltage is unusually low. So yeah, you need more g_m in the lower device.

[Merlinb]

But when you say "Reduce it to 10k or even less (with suitable extra shunt capacitance if necessary) and you can approach the irriducible noise of your pickups.", I think that is not quite good enough if you make the "irriducible noise" lower as I described.

Well yeah... but "if you make the source even quieter (non standard) then eventually the amp needs to be made even quieter too" seems a little unhelpful and redundant, IMO.

[bob p]

^^^ what publication is that excerpted from? Designing Hi-Fi Tube Preamps? I don't have that one.

Fwiw It looks like Toshiba has obsoleted that FET.

[Gnobuddy]

Fwiw It looks like Toshiba has obsoleted that FET.

I would hazard the statement that most discrete JFETs are endangered species now. Many are already extinct.

-Gnobuddy

[bob p]

357 hits at mouser's parametric search:

http://www.mouser.com/Semiconductors...FET/_/N-ax1rs/

[Enzo]

What actually is going away is the TO92 case. But even in that, MOuser had a whole bunch of stocked types.

[Gnobuddy]

357 hits at mouser's parametric search

That dropped to 96 when I checked the "In stock", "N channel", and "Through Hole" boxes. Still more than I expected, though.

And I found one extremely unusual entry in the list ( http://www.mouser.com/ds/2/827/DS_UJN1208K-1085742.pdf ). A JFET rated for 1200 volts, 80 milliohms channel resistance, 21 amps continuous drain current, and 136 watts power dissipation? It takes both negative and positive gate voltages (wrt the source), too. Sure seems more like a MOSFET than a JFET to me.

-Gnobuddy

[Merlinb]

^^^ what publication is that excerpted from? Designing Hi-Fi Tube Preamps? I don't have that one.

Yes it's from that one.

Fwiw It looks like Toshiba has obsoleted that FET.

Yep, but there are plenty of NOS and you can still get the LSK170 and LSK389 I think.

[Mike Sulzer]

The upper device in a cascode has remarkable properties. If we assume that:
1. There is no grid current (or gate current if we are using a FET).
2. The current from plate to cathode (source to drain) can be modified by the grid cathode (gate source) voltage.
Then the output voltage is just the current times the load resistor (i*Ra), independent of things like transconductance, amplification factor, etc.
This is true because the current that goes in the cathode (source) must come out the plate (drain). You can also prove this for specific cases, such as the common gird circuit, and you see that when you multiply the gain by the input impedance, all the parameters drop out and you just have i*Ra (or more generally it would be called RL).

So, does it matter if we use a triode or a FET? Yes, if the value of the input impedance is important. In order to make the noise of the noisy triode (high voltage MOSFT) irrelevant, we must have enough gain at the point where the two devices meet so that the noise of the upper device is small compared to that of the lower device. If the device is a FET there is only one way too do this: make the transconductance of the upper device lower than that of the lower, so that the input impedance rises. On the other hand, if the upper device is a triode, we also have the effect of RL in raising the input impedance: Ri = (Rp + R_L)/(μ + 1). This can give a factor of two or three over the transconductance alone, and thus the same noise suppression can be achieved with a lower ratio of transconductances. The lower transconductance gives less gain, and thus a wider dynamic range, a higher clipping threshold, which can be important with high output pickups.. (The gain is just gm*RL, that is, the gm of the lower device..)

Anther implication of i*Rl, independent of properties such as gm and Rp, is that the variation of the quantities with various voltages and currents in the circuit does not contribute to the nonlinearity of the circuit. That is, the distortion is almost entirely dependent only on the variation of the gm of the lower device with current. I do not know if this is what you want for a guitar amp, but it is certainly worth listening to. The quote above says that the harmonics are triode like. If true, that means that the variation of the gm of that FET produces trade like distortion, again something that needs to be heard.

[Gnobuddy]

The upper device in a cascode has remarkable properties.

It's essentially a "grounded grid amplifier", which has 100% negative output current feedback. Everything else follows from that.

That is, the distortion is almost entirely dependent only on the variation of the gm of the lower device with current.

A corollary to that statement: if the lower device is a vacuum triode, it will be operating into an almost zero plate resistance, and the resulting almost-vertical load line will maximise nonlinearity of the transfer function. So if you want to wring maximum "valveyness" out of a triode, use it as the lower device in a cascode!

An easy way to visualize why this is true: the anode voltage of the lower triode is held almost constant, due to the low input resistance of the upper stage cathode. Since the (lower) anode voltage is constant, the usual internal negative feedback that occurs inside a triode (because the electric fields from cathode to grid to anode mix) is eliminated. That unleashes the full nonlinearity of the (lower) triode.

I was all set to try this out when I realized that I could buy NOS triode-pentode TV valves for $1 each. The 50-cent pentode inside is the real deal, a triode whose internal negative feedback has been removed by the addition of the screen grid. All the "valveyness" you could ask for, without the headaches of a two-triode cascode!

-Gnobuddy

[Mike Sulzer]

It's essentially a "grounded grid amplifier", which has 100% negative output current feedback. Everything else follows from that.

I do not understand this. The properties of the upper device described in my previous post depend on driving it with a current source. Driving it with a voltage source is very different, and so I do not see how a GGA always has 100% negative output current feedback. Do you mean that the input and output current are the same?

A corollary to that statement: if the lower device is a vacuum triode, it will be operating into an almost zero plate resistance, and the resulting almost-vertical load line will maximise nonlinearity of the transfer function. So if you want to wring maximum "valveyness" out of a triode, use it as the lower device in a cascode!

-Gnobuddy

The opposite happens if you replace the DC supply-plate resistor with a current source and use a very high impedance external load: you get a gain equal to the amplification factor, which is not dependent on the voltages, but only on the internal geometry. Thus, it is has no distortion. Another way to look at this is that Rp and gm change in compensating ways. You can see how this works from the usual plate characteristics of a triode.

[Gnobuddy]

I do not understand this. The properties of the upper device described in my previous post depend on driving it with a current source. Driving it with a voltage source is very different, and so I do not see how a GGA always has 100% negative output current feedback. Do you mean that the input and output current are the same?

I have trouble visualising it too, but I believe that's correct. Consider the properties of a grounded base stage - same input and output current, it's a "current follower", which is what you get when you have 100% negative current feedback - all output current has to also flow through the input. You also get very high output impedance, very low input impedance, extremely linear current transfer function, and so on.

Compare with, say, a common emitter voltage follower, where you have 100% internal negative voltage feedback, and the entire output voltage appears in series with the input voltage. Now you get very low output impedance, very high input impedance, extremely linear voltage transfer function, and so on.

What would happen if you drove a common-emitter stage with a constant current source? Nonlinearities in current gain (hfe) would cause nonlinearities in the output voltage. But if you drive it with a voltage source, the 100% negative voltage feedback makes it extremely linear. The series voltage feedback straightens out the transfer function as far as input voltage goes, but not as far as input current goes - drive it with a current source, and no

What happens if you drive a common-base (or common source, or common gate) stage with a constant voltage source? Nonlinearities in the input I-V transfer function would cause it to be a quite nonlinear amplifier. But what happens if you drive a common-base stage with a constant current source? The 100% current feedback makes it an extremely linear device.

The opposite happens if you replace the DC supply-plate resistor with a current source and use a very high impedance external load: you get a gain equal to the amplification factor, which is not dependent on the voltages, but only on the internal geometry. Thus, it is has no distortion. Another way to look at this is that Rp and gm change in compensating ways. You can see how this works from the usual plate characteristics of a triode.

Yeah, that's what the valve Hi-Fi crowd likes to do.

I have no interest in valve Hi-Fi, because transistors do a vastly better job than valves when it comes to Hi-Fi. But exactly the opposite is true when it comes to electric guitar. So my interest in valves is entirely in coaxing them to misbehave in ways that happen to sound good with guitar signals.

-Gnobuddy

[murrayatuptown]

Ooh, just caught that $1 triode-pentode cascode suggestion...I would probably think about triode-connecting the pentode half

I bought enough to get the $1 down to $0.50, for little PP amps and forgot about the dissimilar cascode idea.

But UN's LND150-J113 is increasingly consuming my thought bandwidth.