[Patrick Turner]

While perusing the Net via Google to find a simple way of measuring
high value resistance of perhaps 1 giga ohms or more,
I stumbled on the subject of high voltage measurements at
http://www.realhamradio.com/High_Vol...own_Tester.htm

Then I surfed onto
http://www.somis.org/D-amplifiers1.html

I quote from their site
.................................................. ......


"""""Tubes vs. FETs
When the first RF power FETs were introduced, it was commonly thought
that FETs would eventually replace bipolar transistors and gridded
electron tubes in HF power amplifiers. Since RF power FETs work better
at 50V than they do at 12V, FETs have not replaced bipolar transistors
in 12V mobile applications. Another difficulty with FETs is cost. A pair
of FETs that can produce 1200W PEP at 29MHz cost about six times more
than an electron tube, or tubes, that can do the same job. The FETs'
input power requirements are 50V at 50A, i.e., 2500W--so there's
considerable heat to dispose of. Meeting the cooling requirement is not
nearly as easy as it is with tubes because tubes operate quite happily
at surface temperatures that destroy silicon devices.

In low power applications at room-temperature, solid-state devices can
last 100 years. However, at the junction temperatures encountered in
high power applications, the P and N doped layers slowly diffuse into
each other--thereby steadily eroding the device's amplifying ability. A
relatively-large rigorous cooling system is needed to achieve a
reasonable operating life from high power solid-state RF amplifying
devices.

Another difficulty with solid-state high power RF amplifiers is their
power supply requirements. Tubes are quite tolerant of moderate
variations in their anode supply voltage. Transistors, however, are
fatally sensitive to over-voltage. It is much easier to build a 3000V,
0.8A unregulated supply for a tube than to build a regulated 50V, 50A
power supply with over-voltage, over-current and over-temperature
protection circuitry for high-power solid-state devices.

The bottom-line is that 1500W, HF, gridded electron tube amplifiers are
more efficient, more forgiving, easier to cool, more compact, weigh
less, are more tolerant of high SWR and are less costly than 1500W HF
semiconductor amplifiers. For instance, a pair of legal-limit FETs cost
about $800 from Motorola®. The efficiency is about 10% less that what
one can achieve with gridded electron tubes."""""

So my observation makes me ask, "could building a high power class AB2
audio amp of say 1,500 watts
be easier, cheaper than a 1,500 watt SS amp?"

The answer may be fertilizer to some minds here. ( either thought
provoking, or just BS )





Why would I want to measure high value resistances?

Well, last Thursday I was in Sydney where I spent 4 hours with a guy who
had been
tinkering with ESL speakers during some of his spare time for the last
20 years,
in between efforts with a wide range of SS and tube amps.

He had a pair of his ESL set up powered with 30 watt class A SET with a
pair of parallel
6C33C russian triodes.

These amps easily drove his unique design of ESL, about which I shall
devote a special post
on later..

But I heard tight bass, and crisp defined transients, finely detailed
brass and massed strings,
and a sense of intimicy and connection with the recorded event.
It wasn't tiring; none of the brightness of many other ESL.

He had a HV voltage insulation tester which could swing a meter even
when you breathed on it,
and it could tell you if a piece of paper had high or low resistance
when layed across the terminals.
But no calibrated readout.
But he was also able to say why calibration wasn't necessary; if R for
the membrane coating was above a giga-ohm,
all was well, and not a "low resistance 100Mohms".

I have done a good long search on Google under "multimeter schematic"
and drawn a blank.

It possible to apply 9V from a batter to two terminals, and have a
voltage divider to
work the input to a darlington pair of signal bjts to give some
indication which works a meter.
My friend's was like this but again, no calibration.

Does anyone have a schematic which will read say 0-30M, 30M to 100M,
100M to 300M, 300M to 1,000M, &above 1,000M ?

The proven workable schematic should be based around applying a low
voltage of say 9V across the DUT,
and the measurement based on the amplification of small dc current flow.

Patrick Turner.

[Chris Hornbeck]

On Sun, 10 Jun 2007 14:24:06 GMT, Patrick Turner
wrote:

It possible to apply 9V from a batter to two terminals, and have a
voltage divider to
work the input to a darlington pair of signal bjts to give some
indication which works a meter.
My friend's was like this but again, no calibration.

Does anyone have a schematic which will read say 0-30M, 30M to 100M,
100M to 300M, 300M to 1,000M, &above 1,000M ?

The proven workable schematic should be based around applying a low
voltage of say 9V across the DUT,
and the measurement based on the amplification of small dc current flow.

How about a 9 volt battery and a collection of shunts
across a DVM, the resistance under test and the DVM/shunt
forming a simple voltage divider?

Assuming a 10 MegOhm DVM with a usable sensitivity of
10mVolts full scale, required shunts would be (very
roughly; values would need to be trimmed for scale
accuracy) :

100MegOhm FS = 100K Ohms shunt

1000MegOhm FS = 1Meg Ohm shunt

etc. I hope I haven't made my usual three orders of magnitude
error somewhere. Arf.

Much thanks, as always,

Chris Hornbeck

[Patrick Turner]

Chris Hornbeck wrote:

On Sun, 10 Jun 2007 14:24:06 GMT, Patrick Turner
wrote:

It possible to apply 9V from a batter to two terminals, and have a
voltage divider to
work the input to a darlington pair of signal bjts to give some
indication which works a meter.
My friend's was like this but again, no calibration.

Does anyone have a schematic which will read say 0-30M, 30M to 100M,
100M to 300M, 300M to 1,000M, &above 1,000M ?

The proven workable schematic should be based around applying a low
voltage of say 9V across the DUT,
and the measurement based on the amplification of small dc current flow.

How about a 9 volt battery and a collection of shunts
across a DVM, the resistance under test and the DVM/shunt
forming a simple voltage divider?

Assuming a 10 MegOhm DVM with a usable sensitivity of
10mVolts full scale, required shunts would be (very
roughly; values would need to be trimmed for scale
accuracy) :

100MegOhm FS = 100K Ohms shunt

1000MegOhm FS = 1Meg Ohm shunt

etc. I hope I haven't made my usual three orders of magnitude
error somewhere. Arf.

Much thanks, as always,

Chris Hornbeck

OK, let us suppose we have 1,000M, ( 1 giga-ohm ) to measure.

My Fluke has a 0-300mV dc millivolt range which does not work anymore,
but has LCD display plus digits and auto ranging for mV expressed down
to
0.001mV.
The 1M shunt would make the meter have about 800k of input resistance,
so 1.2M would give a round figure of 1M.

So the series R between 9V and 0V with meter and 1,000M to measure =
1,001M,
and I should have I = 9V / 1,001,000k which = 9uA,
thus producing 9mV dc across the 1M of the meter input R.
1mV would indicate R = 10Gohms
1.0Vdc would indicate quite low resistance

This will move around if the connection isn't constant, so a cap of
say 0.033uF across the meter would remove fast V changes.

Have I answered my own question?

A j-fet source follower input with following opamp with gain = say 100
so 1Gohm gives a 1V reading, with a less sensitive meter. A diode
limiter would be needed.

Assuming the above arrangement is OK, I can make a probe for measuring
resistance of the membrane coating for ESLs using a pair of coins as
electrodes soldered
to two threaded rods mounted in a piece of clean polycarbonate plastic
so hand effects are eliminated.

If I used 100V dc for the resistance test, the same gear can be used
without fear of meter
overload, or fusing the coating on a membrane easily.

BTW, my Fluke is very vague about R 1ohm, and using a voltage source
and shunt R of
say 10.00ohms would tell me current in 1ohm, and the voltage across the
tested R
is divided by the current and R more accurately found.

0.01 ohms with 1Amp flow gives 10mV across R, and read as 10V across the
meter.
R = 0.01V / (10V / 10ohms)
0.1 ohms with 1A gives 0.1V, and 10V across the meter
1ohm gives I = 10V / 11ohms = 0.909 A = 9.1 V approx, and 0.91V across
the 1ohm,
so R is calculated at 0.91V / ( 9.1 / 10 ) = 1.0 ohms.

More simply R tested = 10 x VR / Vmeter


My temporary dc bench supply fitted with a 10V regulator should
do the trick without smoke generation.
For low wattage resistors, R shunt on the meter could be 100ohms.

So 0.01ohms gives R = 100 x 0.001V / 10V ohms.

Patrick Turner.