J. Kissel, R. McCarthy

At my request, after seeing that the EY BSC 10 vacuum pressure has dropped below 1e-5 [Torr] (see attached trend), Richard has turned on the H1 SUS ETMY ESD at ~2pm PST. I'm continuing to commission the chain, and will post functionality results shortly. 

Also -- 

I've found the ESD linearization force coefficient (H1:SUS-ETMX_L3_ESDOUTF_LIN_FORCE_COEFF) to be -180000 [ct]. I don't understand from where this number came, and I couldn't find any aLOGs explaining it. I've logged into to LLO, their coefficient is -512000 [ct]. There's no aLOG describing their number either, but I know from conversations with Joe Betz in early December 2014 that he installed this number when the LLO linearization was switched from before the EUL2ESD matrix to after. When before the EUL2ESD matrix the coefficient was -128000 = - 512000/4 so we was accounting for the factor of 0.25 in EUL2ESD matrix. I suspect that -128000 [ct] came from the following simple model of longitudinal force, F_{tot} on the optic as a result of the quadrant's signal voltage, V_{S} and the bias voltage V_{B}, (which we know is incomplete now -- see LLO aLOG 14853):
     F_{tot} = a ( V_{s} - V_{B} )^2
     F_{tot} = a ( V_{s}^2 - 2 V_{s} V_{B} + V_{B}^2)
     F_{lin} = 2 a V_{s} V_{B}
where F_{lin} is the linear term in the force model, and a< is the force coefficient that turns whatever units V_{S} and V_{B} are in ([ct^2] or [V_{DAC}^2] or [V_{ESD}^2]) into longitudinal force on the test mass in [N]. I *think* the quantity (2 a V_{B}) was mistakenly treated as simply (V_{B}) which has always been held at 128000 [ct] (or the equivalent of 390 [V] on the ESD bias pattern) and the scale factor (2 a) was ignored. Or something. But I don't know.

So I try to make sense of these numbers below.

Looking at what was intended (see T1400321) and what was eventually analytically shown (see T1400490), we want the quantity 
       F_{ctrl}
       -------
      2 k V_{B}^2
to be dimensionless, where F_{ctrl} is the force on the optic caused by the ESD. Note that comparing John / Matt / Den's notation against Brett / Joe / my notation, k = a. As written in T1400321, F_{ctrl} was assumed to have units of [N], and V_{B} to have units of [V_{esd}], such that k has units of [ N / (V_{esd}^2) ], and it's the number we all know from John's thesis, k = a = 4.2e-10 [N/V^2]. We now know the number is smaller than that because of the effects of (we think) charge (see, e.g. LHO aLOG 12220, and again LLO aLOG 14853).

In the way that the "force coefficient" has been implemented in the front end code -- as an epics variable that comes into the linearization blockas "Gain_Constant_In,"  (see attached) -- I think the number magically works out to be ... close. As implemented, the linearized quadrant's signal voltage is as shown in Eq. 13 of T1400490, except that the EPICs record, we'll call it G, is actually multiplied in
     V_{S} = V_{C} + V_{B}(1 - sqrt{ 2 [ (F_{ctrl} / V_{B}^2) * G  + 1 + (V_{C}/V_{B}) + (V_{C}/V_{B})^{2} * 1/4 ] )}
Note, that we currently have all of the effective charge voltages set to 0 [ct], so the equation just boils down to the expected
     V_{S} = V_{B}(1 - sqrt{ 2 [ (F_{ctrl} / V_{B}^2) * G  + 1] )}
which means that 
     G == 1 / (2 k) or k = 1 / (2 G)
and has fundamental dimensions of [V_{esd}^2 / N]. So let's take this "force coefficient," G = -512000 [ct], and turn into fundamental units:      
     G = 512000 [ct]             {{LLO}}
         * (20 / 2^18)     [V_{dac} / ct] 
         * 40              [V_{esd} / V_{dac}] 
         * 1 / (V_{B} * a) [(1 / V_{esd}) . (V_{esd}^{2} / N)]
     G = 9.5391e9 [V_{ESD}^2 / N]
   ==>
     k = 5.37e-11 [N/V_{ESD}^2]  {{LLO}}
where I've used V_{B} = 400 [V_{esd}] and the canonical a = 4.2e-10 [N/V_{esd}^2] originally from pg 7 of G0900956. That makes LLO's coefficient  assume the actuation strength is a factor of 8 lower from the canonical number. For the LHO number, 
     G = 180000 [ct]             {{LHO}}
         * (20 / 2^18)     [V_{dac} / ct] 
         * 40              [V_{esd} / V_{dac}] 
         * 1 / (V_{B} * a) [(1 / V_{esd}) . (V_{esd}^{2} / N)]
     G = 3.2697e9 [V_{ESD}^2 / N]
   ==>
     k = 1.53e-10 [N/V_{ESD}^2]  {{LHO}}
Which is within a factor of 3 lower, and if the ESD's as weak as we've measured it to be it may be dead on. So maybe whomever stuck in 180000 is much smarter than I.

For now I leave in 180000 [ct], which corresponds to a force coefficient of a = 1.53e-10 [N/V_{ESD}^2].