Common source mosfet amp Rd value

[LeroyP]

So i'm looking at running a super hard on (SHO) and different voltages to see if i can get some more headroom out of it. I have Dc voltages available of 9,18 & 27. My understanding is that the SHO produces about 40db of gain (100x voltage gain) so it's never going to be cleanish at 9 volts with a typical guitar input level.

I've read all i can about these common source amplifies that i can. Aside form the astronomically hard math, none of the articles include how the value for Rd should be calculated. The all seem just to pick an Rd value from somewhere and then use it to calculate other stuff.
So do i need to adjust Rd if i'm changing in Vdd to 18v or 27v? I know that i'm looking for 1/2 Vdd and the drain. Will the existing 5.1k drain resistor suffice for all voltages? Or do i need to install a pot in the drain and bias it that way?

Thanks
Leroy

[R.G.]

So i'm looking at running a super hard on (SHO) and different voltages to see if i can get some more headroom out of it. I have Dc voltages available of 9,18 & 27. My understanding is that the SHO produces about 40db of gain (100x voltage gain) so it's never going to be cleanish at 9 volts with a typical guitar input level.

I'm a little confused at the objectives. By headroom, do you mean the biggest output signal you can get without distortion?

I wonder what you will drive with a 20V+ cleanish signal. Any amps I can think of only have an input clipping range up to units of volts.
If all you're after is a cleanish signal, you could attenuate the input signal so that 100x gain will only produce a 9V swing.
Both of those sound a little nonsensical, so I'm back at - what is is you're trying to do?

But with that aside, a simple common-source amplifier does produce the largest undistorted signal if the drain is biased at about half the supply voltage. Like so many other things in electronics, you pick Rd's value by Ohm's Law. Your choice of currents in the MOSFET are not unlimited. At some point the power dissipated in the MOSFET, which is going to be 1/2 of the supply voltage times the current you pick, will burn up the MOSFET. There is a thermal limit.

The common TO-92 plastic package can dissipate about 200mW through the plastic and by conduction down its leads. If you're using a TO-92 and 18V, then the maximum DC current it can tolerate is  0.2W/(18V/2) = 22ma. At that point it will be burn-your-finger hot. At 27V, that current is 14.8ma. These correspond to Rd of 9V/0.022 = 409 ohms and 13.5/0.0148 = 912 ohms.  You can pick any larger resistance value

As you recognize, these are very low resistor values, so you would probably never have thought of using them. On the high-resistance side, you're softly limited by noise and output impedance issues. Very high resistances have high thermal noise. The output impedance of a common-something amp is usually approximated by the drain/collector/plate resistor itself, so choosing this resistor simultaneously sets the output impedance and the power dissipation of the device (and the resistor...) by setting the current for the desired bias voltage.

Within the range of greater than 1K or less and about 1M, you're free to choose any value. It's likely that this is why you didn't find any articles on how to pick Rd.  Just pulling a value out of the air is probably OK, as long as you don't consider the systems implications - how much current you can afford to spend on the stage, how hot you're willing to let it get, and what output impedance you need for driving the next stage.

[Frank_NH]

I've gone through the math with MOSFET gain stages like the SHO.  Assuming a BS170 MOSFET and 9V source, you will get about a 50x voltage gain (34 dB) from the SHO at the max pot setting.  Drain current calculates to 0.31 mA.  BTW, this low drain current requires that you have a high impedance load resistor (the 100K in the schematic) to work properly (I found this out messing around with a BS170 on my breadboard). 

If you up the source voltage to 18V, the max voltage gain goes to 86x (39 dB) with Id = 0.9 mA, and at 27V, the gain is 109x (41 dB) with Id = 1.5 mA.   I think 18V should be sufficient for a ~40 dB boost.

As for Rd, basically gain scales with Rd (times the transconductance, gm, if the source resistor is fully bypassed).  So you can increase the drain resistor to get more gain.  The price you pay is that the drain current goes down, and may be too small for the load your are driving.  So, decide on the current level you want and determine Rd accordingly.  Smaller Rd = larger current but lower gain.  I guess there's no free lunch.   

[Transmogrifox]

Rd on the large end will limit your bandwidth.  In these situations you trade bandwidth for gain. 

The best way to pick it is more or less how RG explained.  Find your point of maximum power dissipation in the FET and pick something out of the air that is larger.

You will have to optimize the trade-off between gain and bandwidth if you're going for something really high gain.  Best bet is to breadboard it and plug stuff in until you like what you hear.

[LightSoundGeometry]

I believe the Fets are current devices controlled by gate voltage - so you will need to look at the voltage divider on the gate and the source resistor if I am not mistaken.

[LightSoundGeometry]

I believe the Fets are current devices controlled by gate voltage - so you will need to look at the voltage divider on the gate and the source resistor if I am not mistaken.

actually here is a lab we did on this, lowering the R4 will decrease gain. in this lab froma 2k to a 1.5 we lost 1.8 volts going from Av of 4.8 to 3 volts

[LeroyP]

Hi

Thank you all so much for your responses.

So if I've got this right...

The current SHO at 9v with a 5.1k drain resistor is drawing

(9/2)/5100 = 0.0009A or .9ma

So if I want the same current draw at 18v then

(18/2)/.0009 = 10000 or 10k
Power dissipated is 9*.0009 or 0.0081W

At 27V

(27/2)/.0009 = 15000 or 15k.
Power dissipated is 13.5*.0009 = 0.012W

So if I leave the 5.1k resistor in place and change the voltage to 18v the current will double
(18/2)/5100 = 0.0018A
And therefore the power will double. 9*0.0018 = 0.0162W

If I leave the 5.1k resistor in place and change the voltage to 27v the current will triple
(27/2)/5100 = 0.00264A
And therefore the power will triple. 13.5*0.00264 = 0.0357W

All of these power figures are below the .2W T092 limit.

So I guess i can just leave the 5.1k drain resistor in place.
Or use a 5.1k resistor in series with a 10k trimpot and see how the change in current effects the sound.

Leroy

[PRR]

> how the value for Rd should be calculated

You don't "calculate" that. You pencil a rough value to suit what ELSE is in your plans.

How do you calculate the size of your Man-Cave? You don't. 5'x5' is small and limited. 100'x100' is extravagantly expensive. In most real cases you have other constraints: basement is 12' wide, or you have 30 feet of motorcycles to display.

As one rule-o-thumb: pick plate resistor (plate, drain, collector, no huge difference) 2X to 5X smaller than the LOAD. The input to the next stage. Amplifier, tone-stack, volume-pot. Do not forget stray capacitances.

Another, maybe better, guide is: plagiarize plagiarize plagiarize! Just steal a value from a similar plan. Let the other guy do the brainwork and trial/error.

So you see 5.1K on a plan. But maybe you have a bucket of 4.7K. It isn't critical.

> if I want the same current draw at 18v

Then you are raising the output impedance without suggesting a change to the load impedance. Does it matter? What IS the load impedance? What specs do you have to meet?

I would point out that if your simple goal is "MORE!!", then maybe you don't just want more voltage, you may want more current. Fer example: I want more power in my house. I don't want more voltage, that's normally fine. What I want is more appliances without so much lamp-dim. I want more current (actually lower line resistance).

[LeroyP]

Hi PRR

The Load is 100k.

I'm confused now. I'm only new to solid state design. I've been designing valve amps for years and I'm used to using load lines to calculate the required resistor values.
If there is no resistance in the source and the drain value doesn't matter much. How do we determine the current draw?
My goal was not to increase gain, but just increase the level before clipping.

Leroy

[Frank_NH]

MOSFETs flow current through a drain/load resistor as long as the gate voltage is higher than the source voltage by the so-called threshhold voltage (Vt) which for a BS170 MOSFET is about 2.1V (though it can vary from 0.8V - 3V according to the datasheet).  You can use the two bias resistors to set the gate voltage relative to the source so as to ensure the MOSFET is on (i.e. Vgs > Vt).

For the SHO, if you want more clean headroom, I would try increasing the source voltage to 18V and maybe playing around with the gate resistors to get Vd closer to 1/2 Vsupply.  This is easy enough to breadboard - let us know what you find. 

[PRR]

> drain value doesn't matter much. How do we determine the current

Same as tubes.

If you have a 300V supply, do you set the plate idle point at zero V? At 300V? No, you pick some point "about halfway" from zero to 300V. 150V is an obvious choice when you have no clue. When you plot a small triode on the plate curves, you find that the triode doesn't pull-down as good as the plate resistor will pull up. Fender 12AX7 100K 1.5K biases to about 70% of plate supply. This works good because 12AX7 against 100K can only pull-down to like 40%, and the abrupt clipping at that extreme is generally not-nice while the compression in pull-up is generally more musical.

A MOSFET will pull-down really good, so you might pick a point of 50% maybe 40% of supply. Exact point is probably non-critical.

That's a maximum output choice. For max gain: since Gm always rises with current, and gain rises with load resistance, setting drain voltage very low (20%-10%) gives nearly double the current in the same load resistor and increased gain. For FETs the advantage goes about as the square-root, so maybe 1.3X more gain, no big deal. For BJTs Gm is proportional to current so gain is almost double (however BJTs usually give "too much gain" for practical signals so this is not much use).

Knowing B+, bias point, and resistor, you know current.