measure the secondary voltage of a tesla coil.

[paulj]

Good evening, I was wondering if with the minimum of material, we could measure the voltage of the secondary tesla coil?
I contacted another forum which suggested that I try first with a small model, a capacitive divider, and an oscilloscope, it's just to measured a peak only.

Regards, Paul

[davekni]

Tesla coil voltage measurement is a fun challenge, but not at all easy to get accurate.  A capacitive divider can work until arcs form.  Once arcs form, capacitance from the arc changes the divide ratio.  If the sensing is on the opposite side from a breakout point, it might remain somewhat close.

My preferred approximate measurement technique is measuring secondary coil current and frequency on a scope.  (Frequency changes as arcs form, so needs measuring.)  With a known secondary inductance, voltage can be calculated from current and impedance (I*2*PI*F*L).  This approximation discounts two factors.  First: Secondary coil capacitance adds current passing through only part of the coil.  Second: Primary coil current induces secondary current that is not due to secondary voltage.  This latter effect can be removed by subtracting the appropriate (for your coil) amount of measured primary current from measured secondary current.  I make an extra tap on primary CT for this purpose.  The first issue of coil capacitance is harder to deal with.  As arcs grow, some of the coil capacitance is now to the arc instead of just to top-load and ground.

With a reasonably-large top load, I think voltage by secondary current can get with +-10%.  This is based on playing at "low" voltages where my 15kV probe can measure the top directly.

[paulj]

thank you for your feedback, I suspected that it would be complex. For other members don't hesitate to give your opinion it would be very interesting.

[Duane B]

I think that measuring the base current of the secondary is an excellent way to determine the secondary voltage. In my experience with a base driven resonator, the secondary voltage is directly proportional to the base current, and that includes when the coil is making streamers or not. Every half cycle of Fr the total energy in the inductance is converted to the total energy in the capacitance, and vice versa. That's why the current and voltage are directly proportional to each other. This is somewhat easy to visualize when there is no streamers loading the coil. A little more difficult to visualize with streamers. Maybe I'm wrong though? Can someone please post envelope waveforms of the base current and output voltage of a DRSSTC? Maybe direct me to a site where they are already shown? The actual voltage and current are not important, just if the waveforms are directly proportional to each other or not. Output voltage could be measured with a scope probe and a small wire several feet from the coil.

Edit: Secondary voltage and base current are very closely related. Dave's response is the best. It would be nice to get some more discussion going on about this topic.

[Uspring]

In JavaTC the "Equivalent Energy Inductance (Lee)" can be used to calculate the top load voltage. The docs state that "Lee is meaningful when considering the forward transfer impedance to the resonator. Vtop/Ibase = sqrt(Lee/Cee) = Zft." The effect of secondary coil capacitance, that Davekni mentioned, is included in the JavaTC calculation.

The primary induces additional voltage in the secondary:

V = 2*pi*f*M*Ipri, where M is the mutual inductance.

When adding this voltage to the voltage caused by Ibase, one has to observe the phase relationship between the voltages.

[Duane B]

I prefer this equation: Vtop = Ibase/2*pi*f*C

[Duane B]

OK, here is some data I just collected on one of my small coils.

Secondary L = 173.8 mH
Secondary C = 20.32 pF
Frequency = 84.690 kHz
Temperature = 72 F
Humidity = 33% (outside, don't know what it was inside)
Top load is a 12" by 3" spun aluminum toroid with no breakout.

I measure a peak current of 2.4 amps in the base before sparks break out on the side of the toroid.

Using this equation, Vtop = Ibase/2*pi*F*C , I get a peak voltage of 221.960 kV

The temperature and humidity is given for a reference only and not used in my calculation.
I could find one reference on the Internet from Antonio Carlos M. deQueroz about spark volyage levels from toroids.
He indicates a 11.8" by 3.9" toroid will reach a peake of 282 kV before sparking.

Does anyone have data for peak voltage before a spark breaks out on a 12" by 3" spun aluminum toroid?
I'm sure secondary diameter, length, and other factors may effect it.

[Mads Barnkob]

I wrote about some known DIY methods for that in another thread: https://highvoltageforum.net/index.php?topic=473.msg2959#msg2959

[Duane B]

Mads, thanks for the links you provided. Very informative stuff!

I notice that Terry Fritz's equation for Vtop is the same as Davekni's equation above. (1)
Those equations are essentially the same as mine. However, the reason I use the capacitive reactance in my equation is because, unlike the other equations with inductive reactance, you do not have to deal with the voltage (or current) induced into the secondary by the primary. Determining peak voltage is more complex via the inductor because the excitation source is combined with it.

Hydron's measurements of current into and out of the top load toroid is very interesting. As are the other topics dealing with current into the spark. I would have been more interested in comparing the current into the toroid with the current in the secondary base. Peak voltage occurs before spark breakout, so current in the spark is meaningless for determining peak voltage.

(1) Note that all of these equations essentially rely on the secondary system as being a series lumped circuit. That is good, because the ideal Tesla coil is a simple series lumped RLC circuit composed of the proper components!

[davekni]

I notice that Terry Fritz's equation for Vtop is the same as Davekni's equation above. (1)
Those equations are essentially the same as mine. However, the reason I use the capacitive reactance in my equation is because, unlike the other equations with inductive reactance, you do not have to deal with the voltage (or current) induced into the secondary by the primary. Determining peak voltage is more complex via the inductor because the excitation source is combined with it.

Yes, for voltage before breakout, capacitance is simpler, little confounding with primary current as you mentioned.  After breakout, inductance is much more accurate.