> > Try disconnecting the diode on your breadboard and applying a 12V DC > > to the output node (where you expect 170V) and see what voltage you > > measure at the base of Q2. Do this with the oscillator unpowered (no > > 12V where it usually goes, just where it usually doesn't). You should > > see about (0.7/170)*12 = 49.4mV. If its much higher than that you are > > likely getting too great of an output voltage and it's burning out you > > FET. > > Measured 54.7mV. I guess that is fine.
54.7mV is good, it means that you'll come in somewhere under 170V and not way over. So your FET will be safe from that perspective. Another point that someone in the thread had made was to ensure that your diode wasn't a general purpose diode but something quick (not sure if signal diode is the right term here). If it is too slow the drain voltage on the FET can still get quite high before the diode has a chance to forward bias and conduct any appreciable current. > I can't measure the voltage between the base and the emitter on Q2 > yet. Don't want to put in another FET and burn it like the other 3. I > want it to be the last thing. When you do end up finally powering up with your FET back in the circuit you'll probably want to have a look at the CTRL pin (collector of Q2) to make sure that it does go low on occasion. Otherwise your oscillator is always be going. I would expect that the oscillator would stop for a bit and then restart. Like I said. I haven't built one of these yet but I plan too very shortly and I'm really interested in knowing what ends up being the root of your problem. Steve -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To post to this group, send an email to neonixie-l@googlegroups.com. To unsubscribe from this group, send email to neonixie-l+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/neonixie-l?hl=en-GB.
