5
\$\begingroup\$

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?

\$\endgroup\$
5
\$\begingroup\$

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:

schematic

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ So for a load it appears you get a voltage divider \$\endgroup\$ – codedude Apr 30 '14 at 4:51
  • 2
    \$\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\$ – Anindo Ghosh Apr 30 '14 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.