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I have a Lock-In amplifier that can generate sinusoidal signals up to 1.5V peak. There is an output port and input port, and I can obviously fix the frequency or amplitude of the signal I am sending out from the output port.

This device also has a switchable input port impedance and an output load impedance too. The input port impedance can switch between 50 ohms and 1M ohms. The output impedance can switch between 50 ohms and "HiZ"?

This is the description I got for the output impedance being turned on or off:

"Select the load impedance between 50Ω and HiZ. The impedance of the output is always 50Ω. For a load impedance of 50Ω, the displayed voltage is half the output voltage to reflect the voltage seen at the load."

I'm still learning some AC circuiting, so I was hoping if someone could explain me or point me in the right direction to try understand the point of this better.

Thanks a lot for your help

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The internal layout of the output voltage measurement instrument and the output impedance, RS.

The impedance of the output is always 50Ω. For a load impedance of 50Ω, the displayed voltage is half the output voltage to reflect the voltage seen at the load.

I suspect that they are indicating that the setup is as shown in Figure 1. The generator has a 50 Ω source impedance and is designed to drive a 50 Ω load. In that situation the voltage at the load will be half that of the internal signal source due to the 2:1 voltage divider. The voltmeter has been calibrated to read half the actual internal voltage.


From the comments:

So if I send a sine wave with Vpk at 1 V, the actual output from this will be a Vpk of 0.5 V?

Try it.

  1. Set the frequency down low enough so that it's in spec for your multimeter's AC True RMS voltage range - typically 50 to 400 Hz.
  2. Measure the output voltage on a 50 Ω load.
  3. Record the built-in meter's voltage.
  4. Now see if you can reconcile the two. The built-in meter may be using pk-pk which will be \$ 2 \sqrt 2 \$ times the RMS value.
  5. Now measure the open-circuit voltage. It should be double reading you got in (3) above.
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  • \$\begingroup\$ So if I send a sine wave with Vpk at 1V, the actual output from this will be a Vpk of 0.5V? \$\endgroup\$
    – digeridoo
    Nov 30 '19 at 16:20
  • \$\begingroup\$ If the load is also 50 ohms, yes. If it is high impedance, no. \$\endgroup\$ Nov 30 '19 at 16:54
  • \$\begingroup\$ @digeridoo: See the update. \$\endgroup\$
    – Transistor
    Nov 30 '19 at 17:29
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A lot of RF measurement and generation equipment uses a 50 Ω input or output impedance. The reasons for this are that any cable has some inductance and capacitance and therefore some characteristic impedance. It can also be shown that if the output impedance of the generator, the characteristic impedance of the cable and input impedance of the load are all the same transmission line effects are minimised. You also get maximum power transfer if the input impedance of the load matches the output impedance of the generator.

50 Ω has been chosen as an industry standard for this. Why this impedance is another story and beyond the scope of this question.

For the input impedance the generator will switch in a resistance to GND when 50 Ω is selected.

The output impedance is always 50 Ω and the switch is just for convenience. When set to 'HiZ' it displays what you would see with for example an oscilloscope which typically has a 1 MΩ input impedance while the '50 Ω' setting would be for something like a spectrum analyser which will have a 50 Ω input impedance.

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  • \$\begingroup\$ No need for the MathJAX, Warren. SE supports HTML entities such as &Omega, μ, ± (plus-minus) and lots of others in the posts (but not in the comments). Also SI standard recommends a space between the numbers and the symbols. \$\endgroup\$
    – Transistor
    Nov 30 '19 at 17:21
  • \$\begingroup\$ @Transistor, Thanks I've just learned something, html formatting does look better. I was not aware I could use it on this site. \$\endgroup\$ Nov 30 '19 at 17:27

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