I'm using a function generator that has a 50ohm internal resistance. Does this affect any circuit that I may connect it to and if so, how would it affect the results?

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    \$\begingroup\$ The output of the function generator with internal resistance [R] will drop its output voltage by [D] according to how much current [I] your circuit draws from it, according to ohm's law: [D = IR]. \$\endgroup\$ Commented Apr 30, 2014 at 3:03
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    \$\begingroup\$ This might be helpful, I suspect. Why your function generator outputs twice the programmed voltage \$\endgroup\$ Commented Apr 30, 2014 at 3:35

1 Answer 1


The 50 Ohm internal resistance changes the behavior from that of an ideal signal source is one simple manner: It behaves just like adding a 50 Ohm resistance in series with the voltage source, completing the circuit through your load:


simulate this circuit – Schematic created using CircuitLab

As the function generator is specified with an internal resistance of 50 Ohms, its signal specifications would be for a 50 Ohm load.

For example, for a 5 Volt signal output setting, and using a matched 50 Ohm load for X above, the current through the loop will be: I = V / R = 5 / 100 = 50 mA. This is the nominal, specified case. As load 'X' is changed, the current drawn will change, and thus the voltage across the load changes correspondingly.

  • \$\begingroup\$ So for a load it appears you get a voltage divider \$\endgroup\$
    – codedude
    Commented Apr 30, 2014 at 4:51
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    \$\begingroup\$ Yes. To be meticulously correct, the internal 'resistance' is an impedance, which will vary with the signal frequency. The voltage divider phenomenon varies correspondingly. For a well designed function generator, the variation is negligible across the specified operating frequency range ('flat response'), but with lower end devices, or ones that have not been calibrated for ages, the variation can be significant. \$\endgroup\$ Commented Apr 30, 2014 at 5:05

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