Why is gain lower with same Rd/Rs ratio, but lower resistor values?

Thread Starter

ericknabe

Joined Mar 3, 2020
11
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
 

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Thread Starter

ericknabe

Joined Mar 3, 2020
11
I read somewhere that when using a source resistance of over 100 ohms the equation could reliably be simplified to Av ≈ -Rd/Rs. Was I then mislead?
And then, when factoring in for gm, does it make sense- or is it accurate, rather- to say that if your goal is to have the same sensitivity before clipping, then lowering (or raising) both the drain and source resistors in proportion to the power supply will accomplish that goal in each case?
 

LvW

Joined Jun 13, 2013
1,022
I read somewhere that when using a source resistance of over 100 ohms the equation could reliably be simplified to Av ≈ -Rd/Rs. Was I then mislead?
And then, when factoring in for gm, does it make sense- or is it accurate, rather- to say that if your goal is to have the same sensitivity before clipping, then lowering (or raising) both the drain and source resistors in proportion to the power supply will accomplish that goal in each case?
At first - what you have read "somewhere" is wrong.
Secondly, your drawing showes thar Rs is bypassed with a large capacitor. Hence, the gain is simply: A=-gm*Rd||Rload.
 

Thread Starter

ericknabe

Joined Mar 3, 2020
11
Secondly, your drawing showes thar Rs is bypassed with a large capacitor. Hence, the gain is simply: A=-gm*Rd||Rload.
I forgot to mention I omitted the bypass capacitor.

I think that with gm factored in, then it seems like halving the values of Rd and Rs also halve the gain which explains why I’m getting the same amount of clipping by doing that on a 9v supply vs the original values on the 18v supply. Correct?
 

Audioguru again

Joined Oct 21, 2019
1,501
An old mechanical buzzer produces harmonic distortion like that when its pulse width is not 50%.
Also, 54 years ago the vacuum tubes amplifier was also clipping like crazy.
 
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