If in doubt with your calculation, try this online calculator http://www.mcamafia.de/nixie/ncp_en/ncp.htm Very simple, and does the job quickly
On Nov 21, 11:25 pm, John Rehwinkel <[email protected]> wrote: > > 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.
