Re: temperature test

[Andrei Semenov <semenov@jlab.org>]

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