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Re: Bypass Caps

Hi Kevin,

> Does anyone have a good source for small (.001uf 20kvac) bypass caps?
> I have blown up too many neons, and I do not want to blow another.

There are many surpluse houses that carry Sprague "doorknob" caps.  Most of
them charge a little more than what you can get them for at a "hamfest".
One of the places that carry them is C & H.  Their phone number is 800-325-9465.
They carry plent of 30 KV and 40 KV units.

>   How does one figure the proper bypass caps to use?

You asked for it.  It is a long explaination.  Not knowing your familiarity
with these calulations, I will explain it as well as I can.

In figuring the correct value ofthe bypass cap, both the operating frequency and
the 60 Hz supply must be considered.  While you want to pass as much of the
operating frequency (RF) as you can, you want to pass as little of the 60 Hz as
you can.  So, you end up with a trade-off (surprize).  As the cap value gets
larger and passes more RF, the leakage of the 60 Hz supply also goes up.  With a
pole transformer (or many neons) 20 mA of 60 Hz leakage is no big deal.  But with
just a couple of neons, that much leakage represents a large portion of the available
power to the tank circuit.  So, you must first decide just how large of a leakage you
are willing to live with.  Let's just say you want no more than 5 mA of lost supply
current.  Let's also suppose you are using 15 Kv transformers.  This means you will have
7.5 Kv per side, to ground reference.  Now calculate the minimim reactance you can live
with.

R=E/I, R=7,500/.005, R=1.5 Megaohms, minimum reactance needed to keep leakage below 5 mA

So, looking at the formula for the reactance of a capacitor, to calculate the value needed

                    1
Xc = --------------------------------
     2pi x (capacitance) x (frequency)

                      1  
1,500,000 = ------------------------- = 1.77 pF
           2pi x capacitance x 60 Hz

Now, I would actually select a value of available capacitors to use.  There are
30 KV -at- 500 pF available.  So, lets use 3 of these per side.  That gives us 1500 pf.
So using the above formula, we get very close to the 1.5 Megaohm value of reactance.

But what about the reactance at operating frequency?  Keep in mind that the lower the
frequency, the higher the reactance and vice-versa.  Let's suppose 200 KHz is the
operating frequency.  So, using the above formula substituting in 200,000 Hz for the
60 Hz, we get 500 Ohms.

But what about actual RF attenuation when the choke is placed into the circuit?  For
a series choke/parallel capacitor low-pass filter arrangement, the basic formula for
attenuation is:

Voltage out           Capacitor reactance             
----------- = ------------------------------------- 
Voltage in    Choke reactance - Capacitor reactance

or it can be stated,

Vo     Xc
-- = -------
Vi   Xl - Xc

So, what attenuation do you want?  75%?  90%?  95%?  This you have to decide.  Let's say
we want 90%.  So, the Vo/Vi portion of the formula becomes .1 (1500/15,000 = .1) for 90%.

So, substituting in the Xc value we figured and the 90% attanuation, we should arrive at
a choke reactance value needed.

       500
.1 = --------
     Xl - 500

Xl = 5500 Ohms

Now we need to know what the value of the choke is in Henrys.  The formula for inductive
reactance is:

Xl =  2pi x (inductor value) x (frequency)

Value needed at operating frequency (still assuming 200 KHz).

500 = 2pi x (inductor value) x 200,000
Inductor value will be 4.3 mH

The voltage drop of the supply caused by this choke should also be considered.  So, plugging
this value back into the above formula, we get reactance at 60 Hz:

Xl = 2pi x .0043 x 60 = 1.6 Ohms

Without even calculating the voltage drop at 7500 volts, we know this is OK.

So, we come up with a 1.5 pF capacitor and a 4.3 mH choke, for 90% attenuation at 200 KHz
with a 5 mA loss of supply to ground using a 15 KV supply.

By increasing the the choke value and reducing the capacitance value, we can keep the 90%
attenuation while reducing the leakage of the supply.  At least that is how it seems at
first glance.  Ed Harris pointed out a very important detail yesterday.  What happens to
the reactance of the transformer itself as the frequency increases?  After all, that
is what is in parallel with the capacitor.  Ed gave 2 figures he arrived at by actual
test for a neon transformer that would apply here.

100kHz                  4.2kohm 
300kHz                  1.8kohm 

So interpolating from the above values for 200 KHz, we get 3 kohms.  So, with the above
capacitor, we will get 85.7% of the remaining RF across the cap and the remaining 14.3%
across the tranformer.  If we decrease the value of the capacitor, its reactance goes up.
It won't take much to make the reactance of the cap equal the reactance of the transformer.
When you get to this point, each will be passing 50% of the remaining RF.  You can see
where this goes.  If you get a capacitor that is too small, the transformer becomes the
easier path for the RF, instead of the capacitor as planned.

So, where does all of this lead?  Considerng the value of the capacitor it becomes clear
that a value needs to be selected that provides a much lower reactance than the transformer
at operating frequency while still not leaking too much current from the transformer(s) to
ground.  You will have to accept a fair amount of leakage if you want to protect the
transformer.  That is all too clear.

The choke value needs to be as high as possible for high attenuation.  At 60 Hz, the
reactance will be so low for a choke below 10 mH that the loss can be ignored.  You will
be working with choke values below 10 mH, most likely.

A few last points on the chokes.  Constructing them should be done on a fairly large
powdered iron core.  It is important that core saturation by magnetic flux be avoided.
The cores available from Hosfelt electronics for $1.49 each, part number T-250-B, are a
good choice.  You will be in no danger of saturating those units, if a hundred turns or
so of wire are used.  Care must be taken when you construct these chokes that you don't
burn them out between layers or turns.  This is a matter of experimentation.  I recently
posted information on how I wound my chokes with these cores.  Air core chokes are
worthless in these circuits due to core saturation and provide no protection to speak of.

OK.  I'm done.  I hope this answered your question about bypass capacitor value selection. 

Scott Myers