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Re: De-Q'ed glow diode

Posted by VoltSecond (M) on September 4, 2002 at 18:02:43

In Reply to: De-Qed glow diode posted by Mark Kelly on September 3, 2002 at 22:35:05:

Yes. We can add a cap with no series resistance for good high frequency performance and add a second cap with a series resistance to keep the Q under control.

For the glow tube lets assume 100 ohms for the ESR and 4 mH inductance. Lets also assume the other components around the glow tube don't add any damping or change the effective inductance of the glow tube.

Knowing Zo = sqrt(L/C) or Z^2 = L/C

Using C = L/Zo^2 and we'll use 100 ohms for Z.

100 ohms with 4 mH becomes C = 0.4 uF.

With Qseries = Zo/Rseries we'll get a Q of 1 (fairly well damped.)

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If the ESR in the glow tube goes away at high frequencies for the same reason the glow tube looks inductive at high frequencies, we can add an external RC damping network.

If we leave C = 0.4 uF and use another 0.4 uF in the damping network the best Rdamp for this for minimum Zo peaking is:

n = C_damp/ C_circuit

Rd_opt_Zo = Ro * sqrt[ (4 + 3 * n) * ( 2+ n)/ { 2 * n ^2 * ( 4 + n) } ]

Ro = 100 ohms

n = 0.4 uF/ 0.4 uF = 1

Rd_optimum = 100 ohms * SQRT((4 + 3*1)*(2+1)/(2 * 1^2 * (4+1))) = 145 ohms (use 150 ohms)

If we use C damp as 1 uF.

n = 1.0 uF/0.4 uF = 2.5

Rd_optimum = 100 ohms * sqrt((4 + 3*2.5 )*( 2+2.5 )/( 2 * 2.5^2 * (4+2.5) ) ) = 79.8 ohms (use 82 ohms)

Because the damping cap has series resistance, it does not count fully to the maximum allowed capacitance across the glow tube. My estimate is if the max ESR of the glow tube is 200 ohms and the damping resistor is 50 ohms, we count 200/(200+50) of the damping cap towards the maximum parallel capacitance.

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If the glow tube is fed with a 0.5H inductor, for my first guess, I would assume the inductance at the glow tube intersection to be the inductances in parallel or 1/(1/0.5H + 1/4mH) = 3.97 mH. This is not significantly different than the 4 mH of the glow tube itself.

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