Capsule bias through FET?

Recently I repaired a Neumann KMR82i microphone.
I noticed that the polarisation voltage is put on the capsule through a gate-drain junction of a FET, in fact a BF245B.
(Source wire was simpy cut off.)
What could be the advantage of using a FET here, instead of the usual high value (1 G.ohm or so) resistor?

Cheap available high resistance?

Just trying to be innovative by engneer? 
Lower cost?
Did you measured voltage before and after extra FET?

The FET went short circuit, that was the whole problem!

RuudNL said:

The FET went short circuit, that was the whole problem!

Click to expand...

I would replace it and measure just from curiosity

The microphone already went back to the owner.
I have put in a 1 G.ohm resistor and the microphone works perfectly now.
According to the owner: more output than before.
(Which he liked.  Because it is a 'shotgun' microphone usually used on a distance, high output is a benefit.)

That's  why i was thing about voltage measurement - i think they just used it to drop the voltage, but this is just theory, didn't have it on my bench.

one of my first  diy mic's I used a fet instead of a High R resistor and a red LED for bias of the tube

...of course I didn't know what I was doing, an old radio engineer from east berlin's Eterna Schallplatte gave me the schematics.

RuudNL said:

I have put in a 1 G.ohm resistor and the microphone works perfectly now.
According to the owner: more output than before.

Click to expand...

Does this mean, that FET resistance is even higher than 1G ?
I could not find anything about resistance in datasheet for BF245 ...

I think they relied on the leak current to the gate.
When the microphone arrived, I measured 45 V before and after the FET.
Of course this is strange, because you would expect a voltage drop after the biasing element, because of the high resistance.

measuring high value resistors/fets is not so easy. I think the way to do is by building  a voltage devider with the fet an another resistor with a known value 100m for instance, than you can calculate the resistance  from the voltage drop.

But one thing is certain: if you measure 46 Volts on both sides of the FET (used to bias the capsule), there is something wrong!
The load of the voltmeter should cause a significant voltage drop if the FET is supposed to have a high resistance.
So: I didn't want to take any risk, and replaced it with a 1 G.ohm resistor. That cured the problem.

RuudNL said:

But one thing is certain: if you measure 46 Volts on both sides of the FET (used to bias the capsule), there is something wrong!
The load of the voltmeter should cause a significant voltage drop if the FET is supposed to have a high resistance.
So: I didn't want to take any risk, and replaced it with a 1 G.ohm resistor. That cured the problem.

Click to expand...

Yeap, usual with typical digital multimeter it should show around 3.6V after high impedance when it is 500M-1G range.
Maybe these FET doesn't had such high impedance, maybe it wasn't main goal here.
Note that some mikes have 1M (like schoeps) or 4.7M (like C414C) for polarisation voltage node.

That FET leaks 5nA (max) at 20V gate-source bias, which looks like 4 GOhm (min).  Given that source/drain doping and poly dielectric thickness is pretty tightly controlled, it's entirely possible that at the time it was the most 'reproducible' high-value resistance they could find. 

They probably also had a large quantity of them at a good price, so why not?

Two patents to do a search for

US8331584B2

US6,453,048B1