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I'm trying to replicate the Nixie clock designed by Kevin Lee (https://0x7d.com/2017/nixie-tube-clock/). It uses a NXP PCA9685 PWM controller and a MSD42WT1G BJT to drive each tube cathode separately.

Here an extract of the scheme enter image description here

The PWM signal is 5V and the voltage drop on the tube (when a cathode is conducting) should be around 133V. I don't know the "HV DC" anode voltage.

My questions are:
1) When PWM out is high, the BJT is in forward active or in saturation mode?
2) If it's in saturation (as I think), why R1 is placed between emitter and GND, and not between tube and collector?
3) Here the author says

For example, a 1K ohm resistor and Vb = 5V will result in Ve = ~4.4V and Ic = ~4.4mA regardless of Vc

but this should be possible only if in active mode, right? So the BJT is in active mode? If yes, why?
4) Assuming Vce negligible, the collector current is (HV DC - 133)/1000?
5) Is this the best way to arrange the circuit?
6) If the transistor is used only as high voltage switch, is it better a BJT or a MOSFET?

Thank you for your help!

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  • \$\begingroup\$ I don't use tubes, but I would assume the tube voltage during conduction would be lower than the strike voltage, thus leaving Q1 in forward active mode (as a current source). \$\endgroup\$ – sstobbe Mar 21 at 15:45
  • \$\begingroup\$ The Nixie Tube becomes ionized, and the current must be limited. \$\endgroup\$ – analogsystemsrf Mar 21 at 15:50
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I'm going to throw out a HV DC voltage of 170V for any calculations. Basically, the large emitter resistor acts as negative feedback, limiting how much current the transistor will try to drive through the Nixie tube.

  1. What is the mode of operation of the transistor? It depends (on the load). If the load of the Nixie cathode is "low", say 22kΩ, then the transistor will be in forward active mode. In forward active mode, the emitter resistor creates negative feedback to the common-emitter amplifier, limiting the current to about 4.4mA, as the author suggests. However, if the load of the Nixie cathode is "high", maybe 220kΩ, then the transistor will be saturated, and the current through the cathode will be less than 4.4mA.

  2. The emitter resistor provides negative feedback, limiting the maximum current that the transistor will attempt to pull. The typical way that I've seen current limiting implemented, with an anode resistor, would use a much higher value than 1kΩ.

  3. The BJT is most likely in active mode. Again, the emitter resistor is providing negative feedback. If more current flows through the collector, then it will also flow through the emitter resistor. A higher current through the emitter resistor will make the emitter voltage higher, reducing the voltage at the base, which will "turn off" the transistor.

  4. Assuming Vce is negligible, Ice will be limited by the emitter resistor to about 4.4 mA.
  5. I think it's probably fine. There is a certain amount of safety from using an anode resistor, as it will always work, and it simplifies the the cathode side electronics slightly. If the NPN fails short for any reason, then damage to the Nixie tube may occur (1kΩ is not large enough without the NPN). The real benefit I see is that multiple cathodes can be illuminated simultaneously, and each cathode current will be roughly the same (maybe equal brightness). If the goal is to do fancy PWM transitions between digits, this circuit may be a better option.
  6. The way this circuit is designed, I would only use a BJT. This topology does not work nearly as well with MOSFETs when the same (logic-level) inputs are provided.

Added: The reason this works better with an NPN compared to an N-channel MOSFET is that the NPN provides tighter current regulation to a fixed value with less tendency to overshoot. The base current helps with correcting load current overshoot/undershoot when turning on, and during load changes. The actual load current is also more predictable with the NPN, as the base voltage variation (0.6-0.8V) will be negligable in determining the current, while the Vth of the MOSFET will have more variation due to temperature and manufacturing variance. Also worth mentioning, if you ever do saturate the NPN, it may behave poorly, letting in a lot more current then expected. I've included a schematic that's set up for a brief simulation below:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I had to think about what you meant by feedback, but then I realized if the voltage drop on the emitter resistor is high, the B-E voltage will be less and less base current so less conductance, so less voltage drop on the emitter resistor. Simple, but clever. \$\endgroup\$ – zeta-band Mar 21 at 18:12
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    \$\begingroup\$ @zeta-band This technique is called emitter degeneration (or source degeneration if used with FETs). It's a very common form of negative feedback used in things like amplifiers and current sources. \$\endgroup\$ – Hearth Mar 21 at 19:32
  • \$\begingroup\$ @W5VO can you explain better why a mosfet doesn't work well in a similar configuration? Do you mean doesn't work with the same circuit, and so the circuit has to be modified, or doesn't work at all? \$\endgroup\$ – Marco Mar 22 at 13:09

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