Would the voltage measured at the 50 ohms still be doubled to the reading on the function generator?
I am new to the concept, from what I know, the 50ohms function generator outputs double the voltage if it is connected to a oscilloscope that has high input impedance. If now we were to use the 50 ohms function generator for a 50 ohms load and then measure the voltage drop on the load, would the resulting voltage still be doubled?
If now we were to use the 50 ohms function generator for a 50 ohms load and then measure the voltage drop on the load, would the resulting voltage still be doubled?
No, the signal generator's 50 Ω output impedance and the 50 Ω load form a 2:1 potential divider and that means that the voltage seen across the load is precisely what the signal generator's display is saying it should be i.e. not double (as seen with a high impedance load such as an oscilloscope).
So, if the signal generator is set to deliver 1 volt RMS then, when connected to the appropriate load resistor (usually 50 Ω) then you will get 1 volt RMS across the load.
The voltage on a
simulate this circuit – Schematic created using CircuitLab
Figure 1. Equivalent circuit.
The source and load resistor form a 2:1 voltage divider.
Measurement of VO without a load will result in a reading of twice the rated output. In other words, the output is only calibrated when running a 50 Ω load.
The signal generator provides 2Display Volts, where Display is the voltage seen on the signal generator's display. Now if the signal generator is terminated with a 50 ohms load the measured voltage across the load is 50/(50+50)(2Display)=Display, assuming a 50 ohms output impedance for the generator. If connected to a high impedance load you get 2Display Volts at the output, which is twice the voltage seen on the display.
