Become a Site Supporter and Never see Ads again!

Author Topic: battery box resistor versus gain question  (Read 1820 times)

0 Members and 1 Guest are viewing this topic.

Offline udovdh

  • Trade Count: (0)
  • Taperssection Member
  • ***
  • Posts: 986
battery box resistor versus gain question
« on: January 03, 2006, 10:57:23 AM »
Hello,

In my simple battery box the resistor controls the gain that the FET on the mic capsule generates.
More resistance is more gain.
If I lower the resistance, I get less gain.
Currently the resistor is at factory recommended value.
I need less gain, as I wrote elsewhere, for loud situations. (-6 dB would be nice)
I can lower the resistor a little without problems.
The manufacturer, though, cannot say if the lowering of the resistance versus gain is a linear process.

Does anyone here have some kowledge about what type of behaviour, curve I can expect?
I am getting closer to the minimum resistor value, but have been at 22 times the minimum value.
What form has the gain/resistance graph?
« Last Edit: January 03, 2006, 11:00:40 AM by udovdh »

Offline lordbelial

  • Trade Count: (4)
  • Taperssection Member
  • ***
  • Posts: 533
  • Gender: Male
  • Barcelona got tapers!
Re: battery box resistor versus gain question
« Reply #1 on: January 03, 2006, 12:28:34 PM »
Hello,

In my simple battery box the resistor controls the gain that the FET on the mic capsule generates.
More resistance is more gain.
If I lower the resistance, I get less gain.
Currently the resistor is at factory recommended value.
I need less gain, as I wrote elsewhere, for loud situations. (-6 dB would be nice)
I can lower the resistor a little without problems.
The manufacturer, though, cannot say if the lowering of the resistance versus gain is a linear process.

Does anyone here have some kowledge about what type of behaviour, curve I can expect?
I am getting closer to the minimum resistor value, but have been at 22 times the minimum value.
What form has the gain/resistance graph?

doh...

Please, could you post some schemas?

I can't easily understand what you're trying to say without taking a look to the schema you're using in yout BB design...

Thnks
Actual Gear:

stealth  - AT943 (c,o,sc,h) > ST-9100 > Edirol R09HR/I-River IHP-116(CFMod)
Ultrastealth  - DPA 4061 > ST-9100 > Edirol R09HR/I-River IHP-116(CFMod)
Open - BSC1-K1/K2/K3/K4 > Segue Dogstars > Marantz PMD671 busman t-mod

Playback: PC > M-Audio Fast Track Pro > KRK RP6 actives

My shows on the archive: http://www.archive.org/bookmarks/tapemaniac
Member of  Busman Audio team

Offline udovdh

  • Trade Count: (0)
  • Taperssection Member
  • ***
  • Posts: 986
Re: battery box resistor versus gain question
« Reply #2 on: January 03, 2006, 01:23:43 PM »
A simple battery box.

9V battery via resistor to mic capsule.
Where the resistor reaches the mic capsule, a capacitor is attached as well to filter the DC component and get only the audio component for the recorder.
Ground of the battery is ground for the mic and line out.

 +--[ R ]------- +9V
 |
 |   C
 +--||----- audio
 |
 |
|O
 |
 +----------------- ground

Offline graemecogger

  • Trade Count: (0)
  • Taperssection Newbie
  • *
  • Posts: 38
  • Gender: Male
    • Graeme's Galleries
Re: battery box resistor versus gain question
« Reply #3 on: January 03, 2006, 03:25:22 PM »
The effect will depend on the input impedance of the pre-amp you are connecting to.  Ignoring bass rolloff and other frequency dependent effects:

if Rb is the battery box resistance and Ri is the pre-amp input impedance, the voltage gain is proportional to

Rb*Ri / (Rb + Ri)

If you need to work in dB, remember that -6dB is half the voltage gain and -20dB is 1/10 the voltage gain.

To reduce the level by 6dB, the new battery box resistance should be (where Rb is the original battery box resistance):

Rb_new = Rb*Ri / (Rb + 2*Ri)

 

RSS | Mobile
Page created in 0.065 seconds with 32 queries.
© 2002-2024 Taperssection.com
Powered by SMF