While reading some experimental setup of a paper, I've came across with a statement saying,

"Since a DC power supply is not capable of finely regulating the discharge current at very low currents( in the mA range), a separate current-limiting low-inductance resistor was added to enable the stable operation of the magnetron at very low currents while using the power supply in the voltage mode."

To me who is not familiar with electronics, this sentence wasn't clear to figure out how "a current-limiting low-inductance resistor" is related to the stable operation. They didn't provide the circuit, but I guess it is in series, considering 'current-limiting'.

P/S is on the voltage mode(to keep the current the set value of 100 mA), and the current is from a gas discharge probably with some oscillating nature.

I got the meaning of 'current-limiting' which is a general role of resistor, however, can't understand how stable operation" is achieved with the described resistor.

I found LR circuit example , where time constant is given by L/R, so the low inductance will reduce response time. But, it seems controversy to me because if the circuit reacts quick for discharge oscillation, the inductance wouldn't suppress the effect of discharge oscillation. Then, doesn't make sense of the description 'to enable the stable operation.' And the example of LR circuit is kind of situation for DC circuit, and the page was explaining the current time history when a switch is closed.

I assume 'low-inductance' may somehow related, but can't clearly figure out alone.

I will appreciate if anyone could help me to understand.

** Edit - schematic which I assume is added.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Plasma has a negative resistance characteristic. Together with an inductor this makes an oscillator. The non-inductive resistor in series makes the total resistance positive. \$\endgroup\$
    – τεκ
    Commented Apr 9, 2018 at 14:03
  • \$\begingroup\$ Helium Neon laser tubes usually have or incorporate a "ballast" resistor almost always have such a resistor. The placement is also often indicated as close to the tube electrically. \$\endgroup\$
    – KalleMP
    Commented Apr 9, 2018 at 14:13
  • \$\begingroup\$ @τεκ Thank you for your comment. I looked up 'the negative resistance' and got to know the meaning. But, I couldn't understand "The non-inductive resistor in series makes the total resistance positive". Would you be able to explain me a bit more about it? \$\endgroup\$
    – Hwi
    Commented Apr 9, 2018 at 14:28
  • \$\begingroup\$ @KalleMP Thank you for your comment. As far as I understood, the 'ballast' resistor is being used in the plasma tube in series to limit any over current. So, I think the description in text saying the 'current-limiting' meant that role as like the ballast resistor does. Then, is "low-inductance" anyhow related to the 'ballast' resistor..? \$\endgroup\$
    – Hwi
    Commented Apr 9, 2018 at 14:36
  • 1
    \$\begingroup\$ To prevent free oscillation you need total resistance to be positive. If discharge has -5000 Ohm you want say external ballast at 10000 Ohm to keep the system sane. \$\endgroup\$
    – KalleMP
    Commented Apr 9, 2018 at 15:01

1 Answer 1


Without a circuit schematic, it's difficult to directly address the problem, but specifically calling out a low-inductance resistor (which are their own class of resistors) suggests that there is a capacitor or capacitance involved, not explicitly mentioned. This in turn suggests that the circuit is prone to ringing or oscillation, which are generally undesirable.

Ringing and oscillation are caused by the interplay of inductance and capacitance in a circuit, and if the conditions are right, your circuit might just go crazy.

The resistance of the low-inductance resistor presumably helps dissipate some of that energy that would otherwise go into the oscillations, while the low inductance would make the circuit less prone to oscillation in the first place.

  • \$\begingroup\$ Thank you for your answer. I've tried to put the schematic I assume.. And, may I ask a bit more detail about your last explanation " the low inductance would make the circuit less prone to oscillation in the first place."? \$\endgroup\$
    – Hwi
    Commented Apr 9, 2018 at 13:38
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    \$\begingroup\$ If a circuit has 0 inductance, it won't oscillate (since oscillations are caused by caps and inductors trading energy back and forth, though inductors can be mimicked by certain active circuits). Minimizing the inductance helps get you to that 0 inductance condition. \$\endgroup\$
    – Bort
    Commented Apr 9, 2018 at 13:45

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