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Tesla coil output voltage

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LAN_403

steelq

Sun Dec 21 2008, 11:31PM

Registered Member #1709 Joined: Sun Sept 21 2008, 11:44AM
Location:
Posts: 2

Hi

I need formula for tesla coil output voltage.

Is my formula ok?:

C1- capacitance of primary resonant circuit
C2- capacitance of secondary circuit

Vout=Vin*sqrt(C1/C2)

Avi

Mon Dec 22 2008, 01:27AM

Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410

wouldn't you need to factor in resonant rise and coupling factor?

LithiumLord

Mon Dec 22 2008, 02:15AM

Registered Member #1739 Joined: Fri Oct 03 2008, 10:05AM
Location: Moscow, Russia
Posts: 261

Hmm, this actually can be quite true for an SGTC - before the sparkgap fires off the energy stored in the tank cap is CU^2/2, which, if using a losless sparkgap approximation and having no corona loss in the model, is all transferred into the secondary at the point of the first notch, and therefore the formula is correct. However in reality there is pretty much of additional losses and of course you just can't miss out the corona loss as it's the actual way a coil gives off it's output power.

Linas

Fri Dec 26 2008, 08:31AM

Registered Member #1143 Joined: Sun Nov 25 2007, 04:55PM
Location: Vilnius, Lithuania
Posts: 721

but does it true, that DRSSTC pulsed power can be 100KW ??
if i have 1200Apeak, and 300V in bridge, so it means that power have to be 360KW ??
of course if voltage are same phase as current, ( but as i know, in L current shifted pi/2 than voltage )
so how calculate DRSSTC pulsed power ?? (and output voltage)

Arcstarter

Fri Dec 26 2008, 09:44AM

Registered Member #1225 Joined: Sat Jan 12 2008, 01:24AM
Location: Beaumont, Texas, USA
Posts: 2253

I believe watt is energy over an amount of time. One watt is equal to one joule of energy per second. Power is the rate at which energy is used or generated.

The pulse time in a drsstc is low, which means less energy used. If it was always on at the same current(of course that would not be oscillating at all)and voltage of the pulse, it would be more watts.

I am not too sure at what i am getting at. I am just saying that a watt is the power over time, so a pulse from a discharge of a capacitor would be about the same as the watts put in, even though the voltage is the same but the current in the pulse is MUCH greater than the charging current. Another example would be, say you have a capacitor that hold 1000 joules. That would be able to source 1000watts for one second. A capacitor that holds 100,000joules of energy would be able to source 100,000 watts for one second. That would be 10,000 watts for 10 seconds, 1000watts for 100 seconds, and so on.

Best example. Say you have a 3900uf 400vdc capacitor. That is 312joules. It would take about 312joules to charge up (almost, you must remember that all capacitors leak, at least a tiny bit). That means if you had the current, it would take 312 watts to charge it in one second. So 312 watts for one second is what the charger must be capable of. That means if you were charging it to 400vdc, and you wanted it to charge in one second, it would require 1.2820512820512820512820512820513 amps, for one second.

Or, say you wanted to charge that very same capacitor in 10 seconds. Again, 312joules, which is 312 watts per second. Divide 312watts by 10 and you get 31.2. So that means to charge the capacitor in 10 seconds, it would use 31.2watts for 10 seconds. Divide 31.2 by 400 volts and you get the current, which is 0.078 amps. That means 0.078 amps is what it would pull for 10 seconds to get a full charge.

I might be wrong, it is 3:40 in the morning here. Seems logical.

HV Enthusiast

Fri Dec 26 2008, 02:43PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

Linas wrote ...

but does it true, that DRSSTC pulsed power can be 100KW ??
if i have 1200Apeak, and 300V in bridge, so it means that power have to be 360KW ??
of course if voltage are same phase as current, ( but as i know, in L current shifted pi/2 than voltage )
so how calculate DRSSTC pulsed power ?? (and output voltage)

Thats simply instaneous or peak power. 360kW. Even a static discharge to your light switch as you walk across the carpet is probably on the orders of tens of kW peak power. But only for nanoseconds.

Whats more important is what your average power is, which is on the order of several hundred watts to perhaps a few kW for most DRSSTCs.

Nicko

Fri Dec 26 2008, 04:55PM

Registered Member #1334 Joined: Tue Feb 19 2008, 04:37PM
Location: Nr. London, UK
Posts: 615

What's (excuse the pun) the conversion efficiency? If I measure the RMS voltage & current at my variac, for a simple SSTC, what conversion efficiency can I expect in order to calculate output power?

Nick

MOT_man

Fri Dec 26 2008, 10:30PM

Registered Member #1127 Joined: Mon Nov 19 2007, 12:08AM
Location:
Posts: 139

Output voltage on Tesla secondary =

input voltage via transformer peak (sqrd root) X
secondary inductance (mH)/ primary inductance (uH)

This gives you peak output in volts AC - or DC depending on the system you are running.

So I give my my example - I have a 12 kV NST that has a primary gap setting of 0.75" - that gives my 16500 V of peak voltage in gap due to capacitance/ring up.
My inductance on the secondary measures out at 130.2 mH 1990 turns and my primary comes in at 20.2 uH for 10 turns. So here is what I do. I convert my secondary and primary over to henries. I have 0.1302 H (sec) / 0.0000202 H (pri) = 6445.54
16500 V (sqrd root) = 128.4
128.4 X 6445.4 = 827.589 kV peak at current setting.

When I run my 14.4 kV PT - my gap setting could be open at 1" - which gives 22 kV charging voltage
Secondary output therefore would be 22 kV (sqrd root) = 148.32 X (sec/pri relationship factor) 6445.54
148.32 X 6445.54 = 956 kV peak output.
However - you must also understand that the closer in the tap moves due to higher the capacitance the higher the output voltage Why? Well - the inductance of the primary drops even further. Therefore lets take the measurement at turn 5 -
Secondary is still 0.1302 H (130.2 mH sec) but the primary is .0000101 (10.1 uH primary) = 12891.089
1912 kV or 1.912 MV using 22 kV peak input voltage from a 14.4 kV PT.

Using my 12 kV NST unit set at turn 5 - I'd be getting 1655.8 kV or 1.6558 MV.

So there you have it.

...

Sat Dec 27 2008, 01:45AM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

These numbers (going by capacitance or inductance) would only be even close to being true for a tesla coil that doesn't break out, has no resistance, etc. In reality you are mainly limited by the impedance of the streamers, which is very complex and depends heavily on things like frequency, the length of the pulse, the repetition rate, etc, etc.

Also, for reference, the Vout=Vin*sqrt(c1/c2) is the same thing as Vout=Vin*sqrt(l2/l1) since at resonance C1/l1=C2/l2. But that assumes that the coil does not break out, and that energy is conserved, neither of which are very good assumptions

MOT_man

Sun Dec 28 2008, 06:56AM

Registered Member #1127 Joined: Mon Nov 19 2007, 12:08AM
Location:
Posts: 139

Point taken. Remember that a Tesla Coil is "on" only for a short duration. This formula is only counting into effect the absolute peak output in V. Frequency and pulse length effect the physical size of the streamer due to current. I have yet to see a volt meter accurately measure the output of a Tesla Coil - I doubt one exists.

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