I doubled the values of R4, R5 and R6. Is this the right approach? I thought this would keep the current and the gain the same as the 9v version but double the headroom.
The current in the MOSFET is set by the voltage across the gate-source terminals. In turn, this is set by the voltage on the gate and the resistance on the source to ground.
You've doubled the gate voltage by doubling the supply voltage and leaving the resistances - and more importantly the ratio - of R1 and R2 the same. This ratio needs to be changed to get the current down.
This is all fairly well (I think
) explained in "Designing MOSFET Boosters", here: http://geofex.com/Article_Folders/mosboost/mosboost.htm
That's why I wrote the article - for people who want to know why instead of painting by the numbers.
The bottom line is that if you want the same current but higher voltage, leave the source resistor the same, and correct the ratio of R1 and R2 to get the gate voltage back down to where it was with the lower supply voltage.
At that point, you have a choice. If you leave R4 at the same value, with the same old current it will drop the same voltage and you will not have changed headroom at all. If you double R4, it will move the drain voltage lower, back towards the middle of the new, higher supply voltage. By adjusting the value of R4, you can put it in the middle of the available voltage (supply minus source resistor voltage) and get the most headroom from the new supply voltage. This will probably double the gain as well.
The drain current is set by the voltage across the source resistor. That's set by the gate voltage and the gate threshold voltage, which varies from MOSFET to MOSFET. The DC position of the drain voltage is set by the drain current and the value of the drain resistor, that being subtracted from the power supply. And that determines headroom.
Notice that an opamp running from +/- 18V - as most of the common ones will - will yet again double headroom, and give you all the gain, and more, of a MOSFET.