FeeelElec FY6900 question/issue

[Prizmatic]

Just received one of these cheap waveform generators and would appreciate other owners input (possible daft question warning). According to the user manual this is a 50 ohm output device, yet when I connect it to my oscilloscope (analogue Philips PM 3070, with a 50 ohm pass-through terminator), it displays half the p/p voltage than it is set for (ie: 5v set on generator, 2.5v displayed on scope). Conversely without 50 ohm terminator the displayed voltages on the scope are the same as the generator is set for. My ageing scope has no modern magic like input impedance detection or even a manual setting for it So is this normal behaviour for the FY6900, or have I been sold a lemon?
Thanks.

[toli]

Not familiar with this unit, but in most function generators you need to set this up in the menu to high-z or 50ohm according to expected load, then it matches.

[Prizmatic]

I've not found any high-z setting in the menu's, no mention of it in the manual.

[Grandchuck]

It is not a lemon (unless there are other issues).  It has been set up for calibrated output for Hi-Z loads.  Only a true voltage source (zero internal impedance) has constant voltage output.  It looks like a nice signal source for the cost.

[Prizmatic]

That is what I don't get. If its 50 ohm output impedance as advertised surely it should be calibrated for 50 ohm loads? Hopefully someone who owns a FY6900 will get back to me on this.

[bdunham7]

It is working exactly as advertised.  The displayed voltage is the open circuit (Hi-Z) voltage and if the output is connected to anything with a lower impedance, the voltage will drop accordingly.  Knowing that the output impedance is 50 ohms allows you to do that calculation if you know the input impedance of the device receiving the signal.

[Prizmatic]

Thanks for the responses. So basically the generator does not measure the loaded output voltage directly?

[bdunham7]

Thanks for the responses. So basically the generator does not does not measure the loaded output voltage directly?

It doesn't 'measure' it exactly, but it uses a DAC and op-amp to accurately output the voltage, so I suppose in a way the op-amp circuitry is measuring the output voltage--but this happens before the output impedance.  So think of an accurate voltage source with a 50R resistor afterward, a la Thevenin if you like.

[Prizmatic]

Got it. Ok, not what I expected but question answered. Thanks for bringing me up to speed on it.

[radiolistener]

That is what I don't get. If its 50 ohm output impedance as advertised surely it should be calibrated for 50 ohm loads?

No, if it's 50 Ω output, it means that it's voltage will be divided by two when loaded to a 50 Ω load.

Just because:



where Z1 is a generator output impedance and Z2 is a load impedance.

for Z1 = Z2 = 50 Ω: Vout = V1 * 50 / (50+50) = V1 * 0.5

when you connect 75 Ω load to the generator, the voltage will be different:

Vout = V1 * 75 / (50+75) = V1 * 0.6

So basically the generator does not measure the loaded output voltage directly?

No, it doesn't measure the voltage.
For what reason it needs to measure it? This is not multimeter, this is generator.