Answer to Question #4338 Submitted to "Ask the Experts"

Q

I have a follow-up to the answer to  Question #2255, Category: Instrumentation and Measurements. The response included: "This creates an electrical pulse each time these radiations interact with the wall or gas in the tube. However, when the avalanche arrives at the anode, the voltage drops, and the G-M tube is not active to additional radiation detection. This is the 'dead time.' My question is: Why does the voltage drop and what causes it to return to normal? One suggestion has been made that all the fill gas is ionized every time radiation enters the instrument, although I haven't seen it stated in literature anywhere that all the gas is ionized. If that is true, then once the ionization stops, the flow of electrons stop, and so the current flow stops. Another explanation is that all the radiation energy transferred to the gas is dissipated via the ionization so the voltage drops.

A

As you know, when the Geiger discharge is initiated, the avalanche process grows until a large number of ion pairs have been created, each ion pair consisting of a free electron (not attached to any gas molecule or residual molecule) and a positive ion produced when the electron is stripped from the gas molecule. These positive ions move much more slowly than do the electrons so that the electrons migrate to the positively charged anode (central electrode) and are collected before the positive ions have barely moved. The large number of positive ions formed in the discharge form a kind of dense charge sheath around the anode so that instead of the anode seeing simply the negatively charged cathode it now sees a highly positively charged field of ions so that the electric field strength between the positive ions and the positive anode is considerably less than what it would be if the positive ions were not present. The avalanche process occurs within very close proximity to the anode and, when the electric field strength close to the anode drops below a critical value, it will no longer be sufficient to cause the ion avalanche to propagate, and the propagating ionization process (referred to as gas multiplication) ceases. The electric field strength is proportional to applied voltage so a decrease in field strength is equivalent to an apparent reduction in voltage.

Keep in mind that in a typical G-M circuit, the cathode is at ground potential, and the anode is at high positive voltage; this positive potential is provided by an appropriate power supply that is isolated from the collecting anode by a rather large electrical resistor, thus providing a means by which the voltage at the anode can drop following a discharge event. The avalanche process is stopped by the inability of the reduced electric field close to the anode to promote the ionization rather than by the ionization process affecting all the gas. (Again, remember that the avalanche is taking place only within a very small part of the gas volume very close to the anode.) Indeed, in most G-M detectors, if one increases the applied voltage, the size of the output pulses will increase somewhat as more gas is ionized at the higher voltage.

Once a Geiger event has occurred, another one cannot take place until the positive ions have been removed to a sufficient degree. The positive ions are removed by drifting relatively slowly in the electric field toward the cathode. As the positive charges drift away from the anode the charge density (charge per unit volume) decreases and electric field strength (and voltage) gradually increases; positive charges that reach the cathode are neutralized. The electric field strength increases until, eventually, another gas multiplication avalanche process may take place. I hope this is helpful to you.

George Chabot, PhD, CHP