HID lamp arcs are actually fairly low voltage once they are ignited.
The ignitor must generate very high voltages to strike the arc, but maintaining the arc is the responsibility of the ballast.
The ballast produces a high-frequency output at relatively low voltage with a controlled current. It is a constant-current source; that's how it works, since the resistance of an arc is almost zero. It is easier to maintain a stable arc with a high-frequency AC current than with a DC one, thus, the use of a high-frequency oscillator in the ballast for a HID lamp. (This is the same reason, by the way, that high frequency power supplies are used in inert-gas-welding systems; with a DC welder you must touch the work to initiate the arc - this is not necesasry with a high-frequency energy source.)
Basic electrical laws (Ohm's Law) says that E = IR. Since "R" is very low, if you apply a lot of voltage then I (current) will be extremely high!
This will cause the bulb to burn up or even (in an extreme) to explode due to vaporization of the electrodes.
So with an arc lamp what you do is use a charge-train device (a series of diodes and capacitors) on a constant-current power supply to "pump" an output capacitor for striking. With each "half-cycle" into the capacitors the current is "pumped" into the final HV capacitor by a series of diodes that allow flow into the capacitor but deny the reverse-flow back out. Since the supply is a constant-current device, and voltage is allowed to "float", the voltage will build until it is sufficient to strike the arc. This is how the lamp "ignites." (It is also why, if you ever turn on a ballast without a lamp inserted in the socket, or with a defecitve lamp, you can burn up the ballast - the voltage will rise until the insulation in the ballast fails and it arcs internally!)
Once it has struck the arc provides a VERY low resistance path for energy to flow. Since the striking circuit can provide almost no current (but a very high voltage), it does not contribute to the operation once the lamp strikes and the direct connection to the ballast's circuitry takes over. The ballast's job is to provide a high-frequency controlled-current electricity feed to the arc. The voltage across an operating HID lamp is actually quite low; the current is regulated to prevent the lamp (and power supply) from melting down.
If the current control circuitry in the ballast fails, it can fail "wide open" and will lead to a runaway. The fuse is there to prevent this failure, which could lead to the lamp exploding, as it will open before current can rise to dangerous levels.
One potential problem with any electronic circuit of this kind is that as supply voltage falls the current required to maintain the total wattage rises. This causes the circuit to run hotter, since heat losses depend on current flow, not voltage. With a sufficiently low supply voltage that can deliver lots of current (and NiMH batteries fall into this category) you could conceivably get a thermal failure of some part of the regulating circuit in the ballast.
This MAY be what is happening, but without tearing the ballast apart I can't say with any kind of certainty.
It is interesting to note though that Welch-Allen, who makes nearly ALL of the HID lamp/ballast combinations used in various "small" applications (including dive lights), has TWO separate ballasts for their 10w lights. One is designed for NiMH batteries, the other for Alkalines.
The ignitor must generate very high voltages to strike the arc, but maintaining the arc is the responsibility of the ballast.
The ballast produces a high-frequency output at relatively low voltage with a controlled current. It is a constant-current source; that's how it works, since the resistance of an arc is almost zero. It is easier to maintain a stable arc with a high-frequency AC current than with a DC one, thus, the use of a high-frequency oscillator in the ballast for a HID lamp. (This is the same reason, by the way, that high frequency power supplies are used in inert-gas-welding systems; with a DC welder you must touch the work to initiate the arc - this is not necesasry with a high-frequency energy source.)
Basic electrical laws (Ohm's Law) says that E = IR. Since "R" is very low, if you apply a lot of voltage then I (current) will be extremely high!
This will cause the bulb to burn up or even (in an extreme) to explode due to vaporization of the electrodes.
So with an arc lamp what you do is use a charge-train device (a series of diodes and capacitors) on a constant-current power supply to "pump" an output capacitor for striking. With each "half-cycle" into the capacitors the current is "pumped" into the final HV capacitor by a series of diodes that allow flow into the capacitor but deny the reverse-flow back out. Since the supply is a constant-current device, and voltage is allowed to "float", the voltage will build until it is sufficient to strike the arc. This is how the lamp "ignites." (It is also why, if you ever turn on a ballast without a lamp inserted in the socket, or with a defecitve lamp, you can burn up the ballast - the voltage will rise until the insulation in the ballast fails and it arcs internally!)
Once it has struck the arc provides a VERY low resistance path for energy to flow. Since the striking circuit can provide almost no current (but a very high voltage), it does not contribute to the operation once the lamp strikes and the direct connection to the ballast's circuitry takes over. The ballast's job is to provide a high-frequency controlled-current electricity feed to the arc. The voltage across an operating HID lamp is actually quite low; the current is regulated to prevent the lamp (and power supply) from melting down.
If the current control circuitry in the ballast fails, it can fail "wide open" and will lead to a runaway. The fuse is there to prevent this failure, which could lead to the lamp exploding, as it will open before current can rise to dangerous levels.
One potential problem with any electronic circuit of this kind is that as supply voltage falls the current required to maintain the total wattage rises. This causes the circuit to run hotter, since heat losses depend on current flow, not voltage. With a sufficiently low supply voltage that can deliver lots of current (and NiMH batteries fall into this category) you could conceivably get a thermal failure of some part of the regulating circuit in the ballast.
This MAY be what is happening, but without tearing the ballast apart I can't say with any kind of certainty.
It is interesting to note though that Welch-Allen, who makes nearly ALL of the HID lamp/ballast combinations used in various "small" applications (including dive lights), has TWO separate ballasts for their 10w lights. One is designed for NiMH batteries, the other for Alkalines.