Offset signal from AWG with DC power supply in series, OK?

[Aldo22]

Hi
I'm sure this question has been asked many times before, but I haven't found what I was looking for.

Is it OK to offset a signal from the AWG with a battery-powered DC power supply in series?

I have basically tried it with low voltage and no load.
It seems to work (photo).
But can it be dangerous for these devices if I apply e.g. 20Vpp + 10VDC?
What happens under load (e.g. 50Ω)?

My question: Can I try it out or can something go wrong?

Thank you!

[benj38]

Think about it for a second: the battery will be connected to the output of the signal generator in series with the load resistance. Thus, if the load resistance is low enough, you can certainly damage the generator if you exceed the voltage you are allowed to feed into its output (don't look for a spec of what this maximum voltage is, it is almost never specified by the manufacturer).

The solution is very simple: connect a blocking capacitor in series with the output of the AWG, thus preventing the DC voltage from feeding into it.

The value of the capacitor depends on the frequency of the waveform, and the load resistance: it should present a small enough impedance at this frequency compared to the load.

Two important points to keep in mind:

• The capacitor will obviously also block the average dc value of the generator's output waveform, so if this is not zero you will have to compensate for it by adjusting your DC voltage supply accordingly to supply this as well.
• You want to avoid switching the DC supply into the circuit abruptly since (if this DC voltage is high enough, the capacitor big enough, and the load resistance low enough) the resulting inrush current charging the capacitor that will pass through the generator may damage it. Instead, bring the DC voltage up gradually.

[benj38]

I will add the following clarification to my previous reply:

Note that in that reply I was focusing on protecting the generator from the DC supply, but not the DC supply from the generator.

The reason is that it is quite common for the generator to present to the DC supply a similar or even higher impedance than the load (e.g., both may be 50ohms) and thus, a substantial portion of the DC supply voltage will fall on the generator's output and may jeopardize it.

In contrast, unless one has a good reason to suspect otherwise, it is usually a safe bet to assume that the DC supply will present to the generator a substantially lower impedance than the load, and thus only a small portion of the generator's AC voltage will fall on the DC supply.

[Aldo22]

Thanks for the answers!
That makes sense.
I just wanted to ask about the protection of the battery-powered DC supply, but you answered that already in the second post.