Re: temperature test

[beni zihlmann <bzihlman@indiana.edu>]

Hi Andrei,

you are correct the factor of 100 is there. but that does not change 
anything
about the conclusion that the power consumption/dissipation in the PMT is
small only 25uWatts per PMT. Indeed the BASE has to provide peak currents
on very short time scales but that does not drive the overall power 
consumption
estimation. 50kHz for a single channel (single block with PMT) is high 
enough. this is
not a hodoscope it covers 4x4cm transverse dimension, in 2x2m wall.
With the use of Cockcroft-Walton bases the HV is generated right in the base
which will reduce the average power consumption in the BASE by a factor 
of 10
compared to regular bases (according to Paul Smith).
I am not worried about the PMT load I am worried that my temperature 
measurement
means nothing.

cheers,
Beni




Andrei Semenov wrote:
> On Thu, 31 Jan 2008, beni zihlmann wrote:
>
>   
>> 1) if I assume a signal height of 1V and as signal shape  a triangle
>>     with base 50ns. I get a peak current of 20mA and the signal results
>>     in 5*10^-12C.
>>     assuming a very high rate of 50kHz this will lead to 0.25*10^-6C.
>>     this charge corresponds to a effective DC current 0.25uA.
>>     Taking the 1V*0.25uA=0.25uWatt. That is what the PMT will consume
>>     as power.
>>     please correct me if this is all wrong.
>>     
>
>
> No, it's not quite correct. The triangle-shaped pulse with a peak
> current of 20 mA and a base of 50 ns has the charge:
>
> Q_pulse = (20 * 10^-3)*(50 * 10^-9) / 2 = 500 * 10^-12 C
>
> (viz., 100 times bigger than in your estimation.)
>
> With 50 kHz rate, the effective mean DC current from PMT will be 25 uA,
> and the power that PMT will consume JUST FOR OUTPUT SIGNALS will be 25
> uWatt .
>
> Again: if you want to have close-to-be-linear PMT response, it's not
> enough to supply just 25uA from the base. The base must provide the peak
> current of 20 mA (or even more taking into account the range of
> amplitudes), and it's hardly possible to transport such a peak current
> (though the long power cables from HV supply to the base) during the PMT
> pulse rise time (viz., 1-2-3 ns). Usually people use capacitors on the
> base for this, but the base must be able to re-charge these capacitors
> without drop of the voltage on PMT dynodes. It means that (without ability
> of the base to supply the significant current in a short time without
> unstability of the output voltages) the pulse shape will be distorted for
> sure => no good time and amplitude resolutions etc. Or the power supply
> (for PMT bases) must be located really close to the bases (and heat the
> volume).
>
> Is the rate of 50 kHz per PMT really considered to be a big rate in this
> experiment? Hm... If so, it's a really (and unusually)  VERY low rate
> conditions, and I'm not quite sure why you are worry about PMT load at
> all...
>
> Andrei
>
>