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I was looking at this circuit, and there was a resistor connected to ground after the output of a signal generator(50 ohm on the top left) . The explanation was that adding it allows the signal generator to see a constant load. Note: On the right is an op-amp.

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Can someone explain what that means or why its added ?

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  • \$\begingroup\$ If your edge speed is 5X or 10X the electrical length of your input wiring or cable, you can skip the 50 ohms. Except the attenuator will not be accurate. \$\endgroup\$ – analogsystemsrf Jul 4 '17 at 17:39
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Short answer :

Because the signal generator was designed to work with a 50 ohm load.

The reasons behind this constraint dives into the complex world of transmission lines, impedance matching and a lot of complicated mathematics.

What you need to know is that without this 50 ohm load resistor, your signal generator will not behave as expected.

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That explanation to see a constant load is not a very good explanation. I think that is an anwer given by someone who does not understand this very well. What the person probably meant is that it makes the load more predictable which is true but also needed.

The opamp's input is a virtual ground so without the 50 ohm resistor the input impedance would be 220uF in series with 1 kohm. That is a pretty constant load depending on the frequency range.

The real reason why the 50 ohms needs to be there is to terminate the Attenuator properly. If it is for example a 10 dB attenuator, it will only attenuate by 10 dB when it is terminated with 50 ohms.

The same is true for the signal generator, it also needs a 50 ohms load which is should be provided by the attenuator. If you do not load the signal generator with a 50 ohm load, it will deliver 100 mV when you set it to deliver 50 mV. Only when terminated properly (with 50 ohms) will it really deliver 50 mV.

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