Yeah the resistors are 3.7KV 10Mohm .5 watt. I was going to make chains of 5 to get a good total voltage rating.
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----- Reply message -----
From: "Jim Lux" <jimlux@xxxxxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Subject: [TCML] Voltage/power dissapation drop across HV DC current limiting resistors.
Date: Mon, Sep 28, 2015 9:27 AM
On 9/28/15 5:18 AM, Jim Lux wrote:
> On 9/27/15 6:51 PM, Matthew Sweeney wrote:
>> No that is exactly correct. Another user mentioned that I should be
>> careful
>> using NST for VDG seemingly because there could be short down in the
>> inside
>> of the VDG to the HV brush, allowing potentially lethal currents up to
>> the
>> collector. I've not been able to confirm this exact reason - maybe
>> someone
>> else can chime in on this.
>>
>> I was mostly only concerned about the maximum wattage I'd need for
>> resistors, which I realized I can actually easily calculate by assuming a
>> short to ground. This gives me the maximum possible current flow through
>> the resistor network and thus I can calculate maximum wattage.
>>
>
> Your problem won't be the power handling of the resistors, but their
> voltage handling.
>
> In any case, a typical 15kV, 30mA NST essentially has an inductive
> impedance of 15E3/30E-3 ohms or 500 kOhms. That is, with the output
> shorted (0 volts), it will source 30mA: with a 500k load, it will source
> 15mA at 7.5kV (which is 7.5E3*15E-3 watts = 112.5 Watts) and with no
> load, it sources 0mA at 30kV.
>
> The 250k load is the maximum power dissipation point (Thevenin, and all
> that).
Ooops.. 500kohm load..
>
> In your case, you're worrying about limiting the current if there's a
> spark or short? What's your actual current limit? Microamps is pretty
> small. Milliamps is also small, but more practical.
>
> If you're just interested in limiting current, you could put the
> secondary of another NST in series (leaving the primary disconnected).
> That would limit the current to half.
If you want to limit to, say, 1mA, then you need about 15 Megohms in
series, and it's going to dissipate (1E-3)^2 * 15E6 or 15 Watts total.
If you put 15 resistors, each 1 Meg, in series, then each one will
dissipate 1 watt, and each one will have to stand-off 1 kV, which is
quite reasonable. 2W resistors are readily available, although I'd make
sure there's plenty of air circulation.
if you want to limit to 100uA, then use 10 Meg resistors. Then the
power dissipation will go down substantially: 0.1 watt for each resistor.
>
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