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n-FET Drain to Ground Resistance

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sngecko

Fri Jan 27 2012, 08:21PM

Registered Member #3447 Joined: Fri Nov 26 2010, 11:10PM
Location: North Jersey
Posts: 97

I'm being a bit lazy asking for your input here, but I work weird hours and I can't get into my lab.

Given the circuit below, if this circuit were un-powered (no grid or anode voltage or op-amp supply), would anyone expect resistance at the drain to ground of only ~500 Ohms?

This seems really low to me, but I had to head out of the house before I could tear out this circuit and test it. I think there might be a mistake in how I mounted the FET, given that the IRF830 that I'm using has a flange that is tied to the drain. I have shown, though, that the drain is isolated from the chassis, so I can't figure out why resistance is so low.

Scott.

m4ge123

Fri Jan 27 2012, 09:11PM

Registered Member #4118 Joined: Mon Oct 03 2011, 04:50PM
Location: MD
Posts: 140

Did you get 500 ohms by measuring the MOSFET with a multimeter? If so, you may have connected the probes backwards so that you're measuring the body diode.

sngecko

Fri Jan 27 2012, 09:23PM

Registered Member #3447 Joined: Fri Nov 26 2010, 11:10PM
Location: North Jersey
Posts: 97

By the Power of Gray Skull! I may have done just that. I will have to recheck it later...

On the other hand, the forward voltage drop is supposed to be only 1.6V for the IRF 830's diode.

Note: It turns out that I measured it correctly. What this thread may not be aware of is that I recently accidentally discharged a >10kV, 2uF capacitor through this circuit... So, it's no surprise that these FETs are toast.

Proud Mary

Fri Jan 27 2012, 09:32PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Have you considered what might happen in the case of gate drive failure, Scott? Your MOSFET would almost - but not quite - stop conducting, and you would have a potential divider consisting of the laser tube, the triode valve, the MOSFET, and R across your high voltage line. Could the voltage drop across drain and source exceed the maximum permitted for the device - 500V, I see from the data sheet?

Faults proliferate, and it can sometimes be difficult to identify the first cause from which all the others flow.

sngecko

Fri Jan 27 2012, 09:56PM

Registered Member #3447 Joined: Fri Nov 26 2010, 11:10PM
Location: North Jersey
Posts: 97

I do typically start at near 0V on the gate. You're exactly right, Proud Mary.

As Steve Connor pointed out a while back, I did consider the clamped situation. I calculated a maximum Vds of ~150V given the 3-500Z's charts. I explain my reasoning at Note 3 on my Current Regulator page.

Proud Mary

Fri Jan 27 2012, 11:00PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

This is what you say in Note 3:

2) Now, weâ€™re going to try to make the plate as tantalizing as possible to the electrons, but deny them to flow freely by clamping the FET. If Vp-g equals 20000V and 15V is on the grid, this puts us way off the chart. We can approximate the near-linear 1ma line with Vf-g = .006452 x Vp-g + 2.5V. In this case, if we were to put 20000V on the plate and allow 1ma through the FET, there would be ~130V on the FET drain. Of course we have to add the 15V on the grid, for a total of ~145V on the FET drain. The IRF830 laughs at this, with its 500V max voltage rating. This is assuming a 1mA current, which is by no means negligible. But I think we can assume that if it were to drop to near-zero, we should still be within voltage tolerances. When I receive the tubes, Iâ€™ll test the actual no-load cathode voltage.

Now, if we remove the grid drive, then the drain current will fall to ~25 Î¼A, which then becomes the maximum current that can flow through the whole potential divider series string of laser tube, triode valve, MOSFET and R.

The laser tube starts to fire, so a big portion of the 20kV supply will appear as a pulse on the triode anode. The triode conducts, so its equivalent resistance becomes quite low, but still it can only pass the 25 Î¼A limited by the switched-off MOSFET. What do you think the drain-source voltage would be then?

What would happen in your laser tube if ionisation were to start (and its resistance start to fall) but then stop again after some nanoseconds when the voltage across it collapsed? It would stop conducting and the voltage build up on it again at the rate allowed by the 25 Î¼A. Could this account for the oscillatory behaviour? Which components would determine the time constant?

Added afterthought: If you were to replace your MOSFET and R with a single resistance, across which, for example only, 100V was dropped during conduction of the triode, and you were to tie the grid to Earth by a high resistance 'grid leak' then you would have a control grid voltage of -100V, that is, within the normal range of grid bias for this valve. But you have tied the grid to +15V, removing the fail-safe design feature for this particular valve of having a low quiescent current at zero bias. The electrons will naturally think they are looking at an anode, and some of them will jump ship rather than go all the way to the intended anode, as we saw before. In normal use of 3-500Z as an RF power amplifier, grid current will only flow on positive half cycles, but in your case it will flow all the time. The impedance of the valve will be low, not much of your 20kV will be dropped across it, and your 500V drain-source limit will be exceeded.

My instinct is that you should have fixed negative bias on the grid.

sngecko

Sat Jan 28 2012, 01:11AM

Registered Member #3447 Joined: Fri Nov 26 2010, 11:10PM
Location: North Jersey
Posts: 97

As I read your post for the first time, I was reminded of something from The Hitchhiker's Guide to the Galaxy. As a bowl of petunias is falling through the atmosphere, it thought, "Oh no. Not again." It was speculated, as I try to follow your excellent explanation, that if I understood exactly what you were saying, I would know a lot more about the universe than I do now.

Anyway, it is certain that the grid drive failed at some point. I have not yet deduced whether it failed before or after the big shock or before or after the initial (doomed) tube lighting. At any rate, I had started the process by clamping even that 6.12mA grid current, so the rest of your analysis should hold at least until the current is allowed to increase past a point where the discharge can be sustained.

I wish I was better prepared to answer your questions... Here goes:

(1) What might the drive voltage be as soon as the discharge tube begins conducting with insufficient current to sustain itself? I can only answer by trying to measure it in some way. So much relies on characteristics of the 3-500ZG that are hard to find in the literature.

(2) I think your explanation of oscillation makes sense insofar as I understand it. I noticed that the pitch of the oscillation increased as I increased the current. This makes pseudo-sense to me, as the new current level was allowing the tube to start more frequently. I don't think I understand why there is a "minimum sustainable current," though. Is it a result of the ionized gas having a lower resistance?; thus, if the current is sufficiently restricted, there is insufficient power to maintain the ionized gas channel?

(3) As for the time constant, I imagine that there are any number of structural capacitances in series with resistive elements or coupled to the chassis in some way.

So, in retrospect, I should have attempted to start the tube with sufficient gate drive on the FET to permit 6.12mA (current from grid) + 6.5mA (optimum current for the laser) to flow (12.62mA) in order to prevent the oscillation from beginning in the first place?

Afterthought Response: The +15V arrangement derives directly from a paper: Michael J. Posakony, â€œA High Voltage Current Regulator for Laser Gas Discharge Tubes,â€ The Review of Scientific Instruments Vol. 43, No. 2, pp. 270-273 (February 1972).

Proud Mary

Sat Jan 28 2012, 02:03AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

As a practical proposal, if you were to substitute the laser tube for a resistance equivalent to the tube's at its optimum working current, then you can adjust the triode grid bias to whatever value and sign is needed for it to draw that amount of anode current. Given the variable impedance in the cathode line, this procedure can only point you in the right direction, rather than provide a final answer.

I have noticed one small error in the circuit diagram, where we see that a tantalum capacitor shown decoupling the grid of one triode is not shown on the other, so there may be other errors too. Who can say?

sngecko

Sat Jan 28 2012, 03:21AM

Registered Member #3447 Joined: Fri Nov 26 2010, 11:10PM
Location: North Jersey
Posts: 97

You'll notice that the tab above the tantalum decoupling cap connects directly to the other grid bias, indicating that both grid bias voltages are driven from the same +15V supply. Thus, the single decoupling cap.

Proud Mary

Sat Jan 28 2012, 07:08AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

gotcha!

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