Hi Mark, First, you're using a diode to replace the voltage drop of a ballast. The effective voltage is easy to calculate, you don't have to worry about peak voltages, DC, etc. Since a silicon diode is very efficient you can regard it as open circuit for half a cycle when it's not conducting and a short (less than 1 volt drop) for the other half cycle when it's conducting. The power to the heater is normal for the conducting half and zero for the other so the average power is halved. Since the power is equal to the square of the voltage it has to be root 2 (0.707) for half power, that's equivalent to approximately 85 volts for 120 volts input.<BR>However, if you substitute the ballast with a diode there will be more current surge at turn on. The ballast resistor will limit the current surge, it's due to the voltage divided by the total circuit resistance. Since the diode has a very low effective resistance the current is limited just by the heaters; if they're 1/4 resistance when cold the current will be 4 times. With the ballast, it will dilute the effect, even with zero ohms for the heater the ballast will provide some resistance to limit the current. In practice it might limit to 2 or 3 times normal at startup, depending on the total heater string voltage and hence how much ballast resistance is used. The thermistor will give a soft start. You're right about the diode running cool, there's not the power wasted that's dissipated in the ballast. Tube heaters are a lot more tolerant to surges than light bulbs since their filaments normally run close to melting heat. When tubes are run from a transformer winding in parallel they will also surge, the transformer has a low source resistance. However there's no risk of some tubes heating faster than others and the voltage dividing unevenly between them while stabilizing in a series connection.<BR>Don Black.<P>------------------<BR>

I certainly agree with Don. A diode is no replacement for a ballast tube. The diode has no current limiting, and drops no voltage as the ballast tube does. Ouch. Those poor tubes. As for the thermisors, my intention was to put the thermistor in series with the line. A 3A unit will work- as I just tried on one of my sets. If you want to put it in series with the filaments, choose a lower value. These Keystone Current Limiters come in all ratings.<P>Scott<P><P>------------------<BR>

<BLOCKQUOTE><font>quote:</font><HR>Originally posted by Don Black:<BR>When the ENIAC computer was operating tube reliability was a major problem just from the sheer number (18,000) used. Any failure stopped it until repaired.<BR>Don Black.<P>[/B]<HR></BLOCKQUOTE><P>Hi Don,<P>I have to disagree in part with your statement. True a tube failure could shut down the computer but they had a clever way of preventing this. I have a friend who used to work at SAGE as an IBM tech. He explained this all to me once. The MTBF with so many vacuum tubes was too low to allow this computer and those like it to operate more than maybe 20 minutes with all tubes functioning. They actually had an error checking diagnostic program that cycled the power supplies to high and low limits and identified at what voltage excursion the individual circuits would fail. As some cicuit failure points came close to the nominal operating voltage those circuits were flagged for the maintenance crews to repair/replace before they actually failed. Very clever. This diagnostic routine was run on a daily basis separate from a customer's program.<P>The term MTBF actualy was first coined when so many pieces of technology were first assembled in this one machine. Most people think it stands for 'Mean Time Between Failures', but it actually meant 'Many Tubes Blow Frequently'. <IMG SRC="http://www.antiqueradios.com/forums/wink.gif"><P>Rob<P>------------------<BR>