The formula for resistive damping is:

h=(E^2)/(2*M*Wo*Rd),

where

   E is the output in volt-seconds/meter,
   h is the damping coefficient (0.5/Q),
   M is the effective pendulum mass in kilograms,
   Wo is the natural frequency of the pendulum in radians/sec, and
   Rd is the total shunt resistance, Rcoil plus Rshunt.

The resulting sensor sensitivity is

  Eeff= E * (Rshunt/(Rshunt + Rcoil)

Most geophones can be efficiently damped because they have high output, low
mass, and low coil resistance. Most amateur sensors lack these properties.
My sensors are efficiently shunt resistance damped. See my web pages at
http://sites.google.com/site/bobmclure90

Bob

On Thu, Dec 16, 2010 at 3:11 PM, Ted Channel  wrote:

>  This is kind of the same subject..................I understand you can
> dampen a coil with a resistor,
>
> Does anyone use this method in place of a Lehman style magnet plate?
> Do you give up anything (voltage) by using a resistor on the coil?
> If it works, why do we still use magnet plates to damp?
>
> Thanks, Ted
>
> ----- Original Message -----
> *From:* Christopher Chapman 
> *To:* psnlist@..............
> *Sent:* Thursday, December 16, 2010 10:00 AM
> *Subject:* Re: Geophone Damping Mass Slew rate
>
>      I have been playing around with a dismantled SPZ geophone damping and
> find if you short the leads to the coil and then drop the mass
> the mass will descend slowly, I guess from meeting the generator current
> being dissipated by the internal resistance.
>
> Questions arise from this.
>
> 1. if there were no internal resistance (superconductor)   then when the
> leads are shorted the mass would fall
>   a bit then stay forever hovering ?
>
>     Correct
>
> 2.the geophone is acting just like a rotating electrical generator
> armature developing torque under electrical loads ?
>
>     Correct
>
> 3. It seems to me that seismic noise rarely hits the resonant frequency
> and
>   you might do better to not increase the damping over what already
>   is there in the mechanical and physical sense ?
>
>     No. The critical damping allows you to get an output voltage flat with
> frequency above the resonant frequency.
>
> 4. Does the rate at which the geophone mass drop under heavy damping
> represent some new fundamental Eigen frequency?
>
>     No. The resonance is determined by the mass of the armature and the
> spring constant.
>
> 5. Can anyone provide me with high school math models which represent themechanical and
> electrical behaviors of the geophone?
>   High school math being trig and algebra minus the calculus?
>
>     The theory is freely available on the Internet.
>
> an Eigen freq not contained in EQ signals then do no damping at all?
> This [little damping] should work for weak EQ signals and not close strong
> ones ?
>
>       An undamped geophone has a single frequency peaked response -
> definitely NOT what you want!
>
>      Regards,
>
>      Chris Chapman
>
>
The formula for re=
sistive damping is:

h=3D(E^2)/(2*M*Wo*Rd),

where

=A0=
=A0 E is the output in volt-seconds/meter,
=A0=A0 h is the damping coeff=
icient (0.5/Q),


=A0=A0 M is the effective pendulum mass in kilograms,
=A0=A0 Wo is the n=
atural frequency of the pendulum in radians/sec, and
=A0=A0 Rd is the to=
tal shunt resistance, Rcoil plus Rshunt.

The resulting sensor sensit=
ivity is



=A0 Eeff=3D E * (Rshunt/(Rshunt + Rcoil)

Most geophones can be e=
fficiently damped because they have high output, low mass, and low coil res=
istance. Most amateur sensors lack these properties. My sensors are efficie=
ntly shunt resistance damped. See my web pages at http://sites.google.com/site/bobmclure90


Bob

On Thu, Dec 16, 201=
0 at 3:11 PM, Ted Channel <tchannel@............> wrote:









This is kind of the same subject......=
.............I=20
understand you can dampen a coil with a resistor,=A0 
=A0
Does anyone use this method in place o=
f a Lehman=20
style magnet plate?
Do you give up anything (voltage) by u=
sing a=20
resistor on the coil?
If it works, why do we still use magne=
t plates to=20
damp?
=A0
Thanks, Ted

  
----- Original Message ----- 
  
From:=20
  Christopher=20
  Chapman 
  
To: psnlist@web=
tronics.com 
  
Sent: Thursday, December 16, 2010=
 10:00=20
  AM
  
Subject: Re: Geophone Damping Mas=
s Slew=20
  rate
  


  

  

  

  

  

  

  

  

  

  

  
dismantled SPZ geophone damping and find=A0i=
f you short=20
  the leads to the coil and then drop the mass=A0the=20
  mass will descend slowly, I guess from meeting=A0the generator current=20
  being dissipated by the=
=A0internal=20
  resistance.=A0
=A0
Questions arise from=20
  this.=A0
=A0
1. if there were no internal resistance=20
  (superconductor)=A0=A0 then when the leads are shorted the mass would=20
  fall=A0
=A0 a bit then stay forever hovering=20
  ?=A0
=A0
=A0=A0=A0 Correct
  
gener=
ator=A0
armature developing=20
  torque under electrical loads ?=A0
=A0
=A0=A0=A0 C=
orrect
  
resona=
nt=20
  frequency and=A0
=A0 you might do better to no=
t=20
  increase the damping over what already=A0
=A0 is=20
  there in the mechanical and physical sense ?

  
=A0=A0=A0 No. The critical damping allows you to get=A0a=
n=20
  output voltage flat with frequency above the resonant=20
  frequency.=A0
=A0
4. Does the rate at which the geophone=20
  mass drop under heavy damping=A0=
represent some new fundamental Eigen frequency?

  
=A0=A0=A0=A0No. The resonance is determined by the m=
ass of=20
  the armature and the spring constant. =A0
  
=A0
  
anyone provide me with high school math=A0models which represent the mec=
hanical=20
  and=A0electrical behaviors of the geophone?=A0
=A0 High school math being trig and=20
  algebra=A0minus the calculus?
  
=A0
  
=A0=A0=A0 The theory is freely available on the=20
  Internet.
  
an Eigen freq not contained in E=
Q signals=20
  then do no damping at all?=A0
This [little dampin=
g] should=20
  work for weak EQ signals and not close=20
  strong ones ?=A0

  
=A0
  
=A0=A0=A0=A0=A0 An undamped geophone has a single=
=20
  frequency peaked response - definitely NOT what you want!=A0 
  
=A0=A0=A0 Regards,
<=
/font>



  
=A0
  
=A0=A0=A0=A0 Chris=20
  Chapman