> Thanks, John. I'm a little confused though. I thought that the firing > voltage on the INS-1 was 65V and the current was .5mA. Do I not need to > worry about reducing the voltage going to the INS-1 down from ~160-180V as > long as I address the current issue? Also, is "maintaining" voltage in your > reply the same thing as firing voltage.
Right, if you calculate your resistor for the correct current, it will drop the required voltage (Ohm's law and all that). Firing (or "striking") voltage is generally higher than the maintaining voltage. The firing voltage is the voltage required to initiate the discharge, after which the voltage drops some, to the "maintaining" voltage. For the INS-1, the striking voltage is specified as not less than 65V and not more than 90V. The maintaining voltage is specified as not more than 55V. > In other words, let's say my PSU voltage is 180V. Would my calculation be: > > (180V-65V) / 0.0005A = 230k resistor? You have the right idea, but you want to use the 55V figure, as the steady state value should be that or less. (180V - 55V) / 0.0005A = 250k resistor. With the 230k resistor calculated from the striking voltage, you'd get 0.54mA (if the tube's maintaining voltage is 55V), which is probably just fine too. It's not critical. > Also, in terms of the Power rating needed on the resistor, can you tell me if > this is the right calculation. I want to make sure I am not missing anything > now to avoid burning out a component later. I wasn't sure if I should be > using the current that I am looking for (.5mA) or the current that is flowing > into the resistor from the PSU for this calculation. > > P= I^2 x R = .0005A^2 x 230000 ohms = 0.0575 watts. Assuming this is > correct, does that mean I can use anything 1/8W (.125) or better on the > resistor. The same current flows through the resistor and the indicator, the 0.5mA you mentioned. Even though the power supply is capable of supplying more current, the resistor limits it to this value. Your calculation is correct, any reasonable resistor will have sufficient power dissipation capability. However, there's another subtle factor that's often overlooked, and that is resistor voltage capability. Some small resistors can't hold off a lot of voltage, so you might want to use a quarter watt or half watt unit for its voltage capability, even though you won't be dissipating much power. You can also solve the calculation the other way, volts times amps: (180V - 55V) x 0.0005A = 0.0625 watts (a larger figure due to the difference between striking and maintaining voltages). - Cheers, John -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To post to this group, send an email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/neonixie-l?hl=en-GB.
