james89es@...

I thank everyone who responded. Here's my next thought:

terminated cable from my 547's calibrate-out jack to the 1A1's
input. With a signal of 100 v, the terminator would draw 2 amps ?!?
(I=E/R)

It seems to me, a humble hobbiest, that 50 ohms would intefere with
test circuits in general - compared with a 1meg scope probe or a
10meg DMM probe.

What am I failing to understand? (alot, I'm sure) :-)

-Jim

Stan or Patricia Griffiths <w7ni@...>

james89es@yahoo.com wrote:

I thank everyone who responded. Here's my next thought:

terminated cable from my 547's calibrate-out jack to the 1A1's
input. With a signal of 100 v, the terminator would draw 2 amps ?!?
(I=E/R)
Not really because the 547 calibrator is not capable of putting out 2 amps.
It has a rather high output impedance at 100 volts so the voltage will be way
below 100 volts. At 0.2 volts and lower, however, the 547 Calibrator has a
50 ohm output impedance which means if you terminate it in 50 ohms, half of
the 0.2 volts will be dropped across the internal 50 ohm impedance and the
other half will be dropped across the 50 ohm terminator. In other words, if
you terminate the 547 Calibrator in 50 ohm when it is set to an output of 0.2
volts or less, it puts out exactly HALF the voltage marked on the front
panel.

It seems to me, a humble hobbiest, that 50 ohms would intefere with
test circuits in general - compared with a 1meg scope probe or a
10meg DMM probe.
In general, 50 ohms does load the circuits too much that you usually want to
measure. Most very high frequency circuits are also low impedance circuits
so they can stand 50 ohms of loading. This is why high frequency sampling
scopes, which are generally 50 ohms input impedance, are still useful for
making circuit measurements. Some high frequency circuits cannot stand 50
ohms of loading either so that is why the FET probe was invented . . . high
frequencies and high impedance at the same time. (They won't handle much
voltage, however . . . there's no free lunch.)

Stan
w7ni@easystreet.com

What am I failing to understand? (alot, I'm sure) :-)

-Jim

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