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30th March 2017, 08:12 #1
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To find the source resistance of the signal generator
Can I use the following to find out the internal source resistance of the signal generator?
I use a DC voltage source driving the signal generantor and 10K ohm resistor in series.
Base on the current and voltage drop across the 10K ohm then I can find out how much the source resistance of the signal generator.

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30th March 2017, 09:13 #2
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Re: To find the source resistance of the signal generator
The loop is a bit misdrawn, and the scheme would require
you to know the generator's internal voltage amplitude
or the 10K resistor's current (which is not shown to be
measured, nor the added series resistance of doing that).
If your signal generator is the normal 50 ohms then what
you measure at the 10Kohm resistor owes more to the
resistor tolerance than the generator impedance. A "10K"
1% resistor could be 9900 to 10100 ohms; an ideal 10000
ohm resistor would make the total loop resistance 10050.
You'd need to know the "10K" resistor's value to much
better accuracy. Or, use a test load that is closer to your
best guess for Zout so the tolerance matters less (or not
at all).
Using two different resistor values is better. Like open
circuit amplitude, and loaded amplitude, do the math.
Presuming your signal generator isn't the kind that blows
its output driver up when you drive it into an open load....

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30th March 2017, 10:01 #3
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30th March 2017, 22:51 #4
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Re: To find the source resistance of the signal generator
Since most sign gens are meant for 50 ohms,
I'd use something in that range. For example
a ~40 ohm resistor and a ~60 ohm resistor
and solve the two equations for Rout.
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