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For example, if a function generator has a minimum load of 1kohm and a maximum load of 50Mohm, if we connect less than 1kohm, the output amplitude will distort.

But what happens to the output waveform if we connect 60Mohm?

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  • \$\begingroup\$ I don't know that I've ever seen that. In theory though, the same rules apply, and the function generator needs some minimum amount of output current in order to develop it's programmed function. A very high load probably would starve the feedback path and result in inaccuracies, but I'm guessing. \$\endgroup\$
    – user39962
    Commented May 10, 2014 at 16:04

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Most of the function generators' output impedance is \$50\Omega\$ and they give rated terminal voltage only if they are terminated with a \$50\Omega\$ load. So any other load connected to function generated gives amplitude different from that of the displayed.

Some function generators support 'High-Z' mode in which displayed voltage is delivered to the load if the load resistance is greater than a minimum rated value. But I guess there is no upper limit for the load resistance in such cases. You will get undistorted output if you connect a \$60M\Omega\$ resistance. Where did you see this?

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  • \$\begingroup\$ for the connection of function generator, is it 50 ohms connect series with the load resistance or parallel connection? for my understanding, the connection is in seris, by using voltage divider rule, if we connect the load resistance higher than 50 ohm, we will get undistorted output. basically what is the purpose of set it at "high Z" mode?if high Z then it is 1Mohm, then means for series connection, we need connect load resistance greater than 1Mohm to get undistorted waveform?so what is the purpose of high Z mode? \$\endgroup\$
    – user40353
    Commented May 11, 2014 at 16:16
  • \$\begingroup\$ 50 ohms is in series with the source. In this mode, it is assumed that the output will be terminated with 50 ohms. So in order to get user set voltage, source produces twice the the terminal voltage desired. Changing mode doesn't affect the output resistance of the signal generator, but rather informs the signal generator, about the circuit in order to modify its output so that the user selected amplitude can be seen between the output terminals. In "High Z" mode, the assumption is that the load is going to be high Z and hence the source voltage is directly set to the desired terminal voltage. \$\endgroup\$
    – nidhin
    Commented May 11, 2014 at 19:00
  • \$\begingroup\$ THanks for your explanation,it clear all my cloud in my head.. between I am designing a function generator, but I couldnt measure the output current,and output impedance, what wrong in it?so I try to connect trial and error, to obtain the minimum resistance value that will obtain undistorted waveform, and the result show that the minimum resistance value is 1kohm. One more question, do function generator have minimum capacitor value?ex: if capacitor value lower than the minimum capacitor value, it will distort the amplitude of waveform? \$\endgroup\$
    – user40353
    Commented May 12, 2014 at 2:41
  • \$\begingroup\$ No. But there will be maximum value for capacitance that a function generator can drive. Logic is simple. Since reactance is inversely proportional to capacitance, if impedance has a minimum value, capacitance should have a maximum value. \$\endgroup\$
    – nidhin
    Commented May 12, 2014 at 12:20

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