Re: HV Oscilloscope probe (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Sat, 1 Sep 2007 10:38:38 -1000
From: Chester Lowrey <hilo90mhz@xxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: HV Oscilloscope probe (fwd)

You have it almost right, but its the current that rises, not voltage,
because the impedance of that capacitor is going down... So in other words
the internal components start dissipating massive amounts of energy at high
frequency and overheat...

This is why some homebuilt HV probes use lots of ceramic capacitors in
parallel/series, not because you cant get one capacitor with the right
capacitance, but because you need the higher power dissipation/surface area
provided by all those capacitors.



On 9/1/07, Tesla list <tesla@xxxxxxxxxx> wrote:
>
>
> ---------- Forwarded message ----------
> Date: Sat, 01 Sep 2007 07:26:52 +0200
> From: Herwig Roscher <herwig.roscher@xxxxxx>
> To: Tesla list <tesla@xxxxxxxxxx>
> Subject: Re: HV Oscilloscope probe (fwd)
>
>
> Hi Jim,
>
> > Probably because it forms a capacitive divider.. the 4pF starts to be
> > pretty low impedance (132 ohms at 300 MHz), and working into the
> > usual 1Meg paralleled with 15-20 pF, it's only a 10:1 divider.
> > See, e.g., http://www.highvoltageprobes.com/faq.html for a discussion
> > of HV scope probes.
> - Thank you for your explanation. So the kind of the divider
> schematic probably causes the voltage across the components to rise
> with frequency and is the reason for the specified decline of applied
> voltage with increasing frequency.
>
>
> Bye             Herwig
>
> ----------------------------------------
> Greed is the root of all evil !
> ----------------------------------------
>
>
>
>