Re: HV Probe and DMM Input Impedance

n4buq
 

I do have older VTVMs (HP 410B, 410C, etc.) so I could use it with those as well. I have an older HP probe (459A) designed to be used with those (has a different resistor) but it's missing the spring-loaded end-cap as well as has a different resistor value. I took a look at that one yesterday and noticed the resistor has a rather sticky film on it. I wonder if that was originally there to reduce the effect of arcing?

If the resistor in my "new" probe didn't have fingerprints on it before, it does now. I'll be sure to wipe down. Thanks for the tip!

Thanks,
Barry - N4BUQ

----- Original Message -----
From: "Michael A. Terrell" <@michaelaterrell>
To: TekScopes@groups.io
Sent: Monday, October 28, 2019 9:42:01 AM
Subject: Re: [TekScopes] HV Probe and DMM Input Impedance

That probe was made for a common, TV shop grade VTVM, with a 10M input
impedance, for AC measurements. In DC mode, a 1M resistor at the probe tip
gave a total of 11M impedance. This was done to reduce the input
capacitance which would detune RF stages, or quench an oscillator. This was
critical while servicing the high impedances encountered on tube based
electonics.

Add 1090M to 10M, and you get 1,100M wich is 100 times the impedance with
the HV probe. Be sure there are no fingerprints on the body of that
resistor, or it may arc over under HV use.

On Mon, Oct 28, 2019 at 9:36 AM n4buq <n4buq@...> wrote:

This last weekend, I found an EICO HV probe in very nice condition at an
estate sale. Opening it up, it contiains a 1.09G ohm resistor in series
with the tip and cable. I connected it to the input of my Fluke 27 DMM and
measured a low voltage source (all I had handy at that moment). From what
I could tell, the probe gives me a 100x scale factor (e.g 10VDC measured
0.1VDC). While I may need to measure some higher voltages to confirm
whether this is really accurate, it appears to be at least somewhat
accurate.

I have a question, though, regarding the theory of the way this works.
The Fluke has a 10M ohm input resistance which, if I'm thinking about this
correctly, makes the measuring circuit 1100M ohms of which 10M ohm is the
meter and the remaining resistance in the voltage divider network is the
probe's resistor; however, I'm having trouble with the math.

Intuitively, (for me, at least), to obtain a 1/100 divider, I would think
that ideally the probe resistance should be 0.990M ohms with the meter
providing the remainin 10M ohms. But I find it odd that the resistor has
that odd value which makes it seem like it was almost intended to work with
a 10M device.

If I'm not mistaken, those probes were intended to be used with a
particular device (meter) that provided the proper readings but not sure
about that either (not finding a lot of info on this probe).

Am I off base here? I know that some of the HV probes designed to work
with the Fluke are designed to connect differently and I think the meter is
used in mA mode with them but not sure about that.

Is it a false expectation that the meter give me a 1/100 reading when used
with the probe in that manner? Is it also possible that the 1090M ohm is
giving me a "close enough" with that low voltage test and the difference
would become more measurable with higher voltages?

Sorry - this should be simple but, for some reason, I can't make it make
sense to me at the moment.

Thanks,
Barry - N4BUQ





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