Supply voltage and drain/source resistors

[EricKnabe]

So I'm trying to do a runoff groove fetzer valve type of thing. I'm making a jcm800 pedal clone type of thing. Im using eMOSFETs rather than JFETs so that I can bias them to the proper Q point somewhat independently of my already chosen drain and source resistors (with which I set the gain.) But obviously, with a much lower supply voltage using the same resistance on the drain and source provides too much overdrive because of the lower output swing. But reducing the drain resistor (and therefore the gain) by the same ratio that I've reduced the supply voltage by actually gives a completely clean tone. Which clearly isn't accurate to the original design. But strangely if I keep the ratio the same- which should give the same gain (right?)- as the original, but I lower both the drain and source resistor (say by a factor of 10 if the original supply voltage was 10x the new supply voltage) then the amount of overdrive is actually right on the money. Not too much and also not too clean.

ASo to rephrase, if the original supply voltage was 180V, with a drain resistor of 100k and a source resistor of 1k5, then at 18 volts it's over-overdriving. If I reduce the gain by a factor of 10, (10k drain resistor) then the circuit will not overdrive at all. If I reduce both of them to 10k and 150 respectively, then the gain should be the same as before, given by the ratio of Rd/Rs, yet it does not clip the 18V power rail nearly as much, and in fact is just right for what I'm trying to do.

I would normally be fine to simply accept that and design around it, but I actually would like to know WHY and what is going on, so I can understand the principles and concepts behind why that is working. Thanks!

[antonis]

1. MOSFETs aren't Tubes..
2. They also are non-liner devices..

They exhibit a Vgs slope related to V_DS, which has to do with "channel modulation parameter" (width/lengh plus oxide capacitance & other nasty micro-things..) and results into variable item's output impedance..
That impedance is considered in parallel with any other impedance on item's output (Drain resistor // Volume pot // Load..) and it's "modulated" by Drain/Source channel current..
So, Drain equivalent resistance for gain calculation isn't the sole value of Drain resistor..

That said, let's go to Source resistance..
You're are confused a bit on actuall gain calculation, 'cause Source resistor is effectively by-passed via 22μF capacitor..
It doesn't stand R_D/R_S but g_m X R_D for Gain calculation..
So, altering Source resistor value only counts on bias point and not on Gain calculation..
(actually, it does count 'cause alters a bit g_m value but in a much lesser affection..)

[EricKnabe]

Oops I probably should have mentioned I omitted the bypass cap. But I should also mention I never meant to suggest mosfets are tubes. I'm actually going to completely rephrase my question in a way that might be easier to understand:

So I've created this schematic based on an 18 volt supply but cascaded a few stages to generate clipping in a guitar preamplifier. If I lower the supply voltage to 9 volts, it's quieter, but also clips more due to having less voltage swing available. If I lower the gain in proportion to the supply voltage by decreasing the drain resistor by half, then the signal is actually too clean, overdriving much less than it does at 18 volts. However if I keep the gain the same with the same Rd/Rs ratio but drop the value of both resistors by half, then the same amount of clipping is generated as with the 18 volt supply.

my question is simply: why?
Thanks. Keep in mind I've omitted the bypass capacitor on the source.

[antonis]

But I should also mention I never meant to suggest mosfets are tubes.

Feeding that specific circuit with 180V brings me in mind TUBES.. 

>If I lower the gain in proportion to the supply voltage by decreasing the drain resistor by half, then the signal is actually too clean, overdriving much less than it does at 18 volts.<

Can't see any reason to implicate Gain with Supply voltage..

[EricKnabe]

I'm actually referring to a reply made by PRR to a question I had a long time ago when I first started out on this little project. He said you'd have to drop the gain proportionally to the supply voltage drop for the same input sensitivity before overdriving the circuit. So I tried dropping the drain resistor by half for an 18 to 9 volt power supply drop, but as I said the resulting signal was too clean. Yet when I dropped both the drain AND source resistors by the same proportion to the the power supply drop, then the clipping was at the same amount as with the original 18volt supply.
The reason I originally used 180 volts as an example is because this is a "tube emulation" type of circuit.

In any case, I'm just learning as I go and that is why you see me so often in this forum asking questions. I really want to understand why some things work and some things don't. In this case I want to know why I need to keep the ratio of Rd/Rs the same but with proportionally lower values to get the same clipping before overloading the circuit.

[antonis]

Now I see..

Paul told you (or to somebody else..) to alter the Gain according to headroom availiable..

You proceeded in altering Gain by only altering Drain resistor value..

But Gain = g_m X R_D..
g_m isn't a constant parameter.. Its value varies according to 2I_DQ /(V_GS - V_TH)

You can see now that Gain didn't alter by only the proportion of Drain resistor altering..

[EricKnabe]

So if I solve for gm using a 10k or 100k drain resistor (and completely omit the source resistance) I still get the same gain for both. But if I change the drain resistance, then the value of Vgs will also change, correct?
So that must be what I'm doing wrong then. Am I getting there now?

[antonis]

Do not omit Source resistor unless you realy want a DC grounded CS amp..
(bias stability issues..)

I posted you a link about MosFet amp on another thread:https://www.diystompboxes.com/smfforum/index.php?topic=124383.0, which, I presume, you didn't read thoroughly..

[EricKnabe]

Maybe I misspoke. I did not mean omitting source resistor from the circuit, I mean I was using the gain equation for the bypassed resistor, and still getting the same value for both a 10k and 100k resistor. Then I used the equation that includes the source resistor: Av= -gm*Rd/gm*Rs+1 and even then I still got the same answer whether I used 10k and 150, or 100k and 1.5k.
So I'm guessing that my mistake was that by decreasing the values of Rd and Rs I'm also changing the value of Vgs and therefore the value of gm. Is that correct? And if so, does that mean I can reliably get the same amount of clipping at different supply voltages by using resistor values that vary proportionally to changes in the supply voltage? (Hopefully that makes sense)

P.S. I have read that article many times, in fact I posted the thread you originally linked it in because I couldn't find the answer in that article, but admittedly I probably did not understand what I was reading at all.

[antonis]

You maybe are right, so let's face it the practical way..

Choose Supply voltage and Drain working current..
Drain resistor will result according to desirable Q-point bias..
Choose, from manufacturer datasheet, V_GS appropriate for above desired Drain current..
For chosen V_GS and Drain current, choose Source resistor..
(its value will not affect gain as far as bypass cap is large enough..)
Go back to datasheet and find(estimate) g_m (g_fs) for particular Drain current and V_DS..
Calculate Gain via g_m X R_D..

You might result into some conflicting values (like desired Gain vs Drain resistor) but you can still look on the bright side of life.. 

[EricKnabe]

I've taken your advice, designed a few stages. I've made calculations based on the correct equations. But now I've noticed another similar peculiar phenomenon. If I drop the supply by 1/10th as I'd been doing, (in a simulation for purely theoretical and learning reasons) and I create a stage with 1/10th the gain as the original stage. With the same value for Rd, I solve for Rs per the equation gm*Rd/gm*rs+1, however it's much too clean. But if I take the answer given for Rs and then I drop its value by 1/10th, then it's exactly as dirty as the original.

So I was led to believe that if you drop the supply voltage by 1/10th then you need 1/10th the gain. But apparently you actually need slightly more. What is going on here?

[antonis]

If I drop the supply by 1/10th as I'd been doing, (in a simulation for purely theoretical and learning reasons) and I create a stage with 1/10th the gain as the original stage. With the same value for Rd, I solve for Rs per the equation gm*Rd/gm*rs+1, however it's much too clean. But if I take the answer given for Rs and then I drop its value by 1/10th, then it's exactly as dirty as the original.

Let's take it once more..
(hoping for the final one..)

R_S MUST be taken into account for Q-point setting ONLY..!!
As far as it's bypassed, DO NOT take into account its value for Gain calculation..!!
(gm X Rd/(1 + gm X Rs) only stands for Source "directly" seen BOTH by DC AND AC..)

Into 4^thof "my" steps above is included the well stated (I hope for it.. ):
>For chosen V_GS and Drain current, choose Source resistor..
(its value will not affect gain as far as bypass cap is large enough..)<

So, decide for Source resistor value, set it and verify for proper V_GS and FORGET IT..!!

P.S.
Capital letters indicate attention points only..
(and definately not any sign of yelling or dismay/frustration..)

[EricKnabe]

I am not using a bypass capacitor.

[antonis]

Ahaaa..!!!

So, we are wearing our fingers, keyborads & mouses for wrong schematics & infos.. 

Plz, post both High & Low supply schematics with items values..