Author Topic: Need help with DC biasing an AC voltage source  (Read 1550 times)

Offline bobfrancis1980

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Need help with DC biasing an AC voltage source
« on: November 12, 2023, 09:45:52 PM »
I have been trying to use ChatGPT to understand how one would DC bias an AC voltage source but I still don't understand the various explanations it gave on how the two voltage sources would be physically connected to combine them and what the circuit would look like.

For example, say I wanted to combine a DC voltage source of +100V with an AC voltage source of 100V so that the output oscillates between +200V and 0V. How would I do this?

Offline klugesmith

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Re: Need help with DC biasing an AC voltage source
« Reply #1 on: November 12, 2023, 10:15:40 PM »
Good to know we can't be replaced by ChatGPT.
In circuit models, a voltage source is a two-terminal component.
Instantaneous voltages add when two sources are connected in series.

Note: to do what you asked, we need an AC source with voltage amplitude of 100 V.
It would generally be identified as a 70.7 volt AC source, going by its RMS voltage.

An exception relevant on HV forum is the nameplate on X-ray transformers, where the term "kVp" follows the peak voltage in kV.
It's traditional but kind of sloppy, since kVp is not a unit of measurement.  Same goes for stating wall voltage as 120 VRMS. 
RMS is generally implicit when speaking of AC voltages and currents, including RF transmitter and receiver and random noise signals, and capacitor ratings.

What would a typical portable "True RMS" voltmeter indicate if connected to the combined waveform in picture above?
In DC-volts mode it would say 100 V, having averaged out the AC.
In AC-volts mode it would say 70.7 V, having switched to AC coupled input.
To get the RMS value of composite waveform (swinging 0 to 200), user can add the squares of DC and AC components, then take square root of sum.
DC source alone produces 10 watts of heat in the load resistor.
AC source alone produces 5 watts of heat in the load resistor.
The two sources in series produce 15 watts.
It works that way because the DC and AC components are orthogonal.   It would not work if we combined two DC or two AC sources.
« Last Edit: November 12, 2023, 10:35:15 PM by klugesmith »

Offline Twospoons

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Re: Need help with DC biasing an AC voltage source
« Reply #2 on: November 12, 2023, 10:48:53 PM »
I have been trying to use ChatGPT to understand how ...

ChatGPT is not a reliable source of information.

Offline bobfrancis1980

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Re: Need help with DC biasing an AC voltage source
« Reply #3 on: November 13, 2023, 02:08:03 AM »
Thank you very much for the response and graphics. From what I understood you would have the AC voltage source in series with the DC source to create the DC offset.

Thanks again :)

Bob

Offline TizianoBll

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Re: Need help with DC biasing an AC voltage source
« Reply #4 on: November 13, 2023, 06:39:50 PM »
Hi,
One easy way to do that is to use this circuit (See photo). It's just a single diode rectification but, instead of connecting to the smoothing capacitor you connect to the diode.
The disadvantage of this circuit is that at 50Hz it won't be able to deliver high currents.

*The capacitor acts as a DC source (since it gets charged to 100VDC by the 100Vpk ac source and the diode).
« Last Edit: November 14, 2023, 08:33:14 PM by TizianoBll »

Offline klugesmith

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Re: Need help with DC biasing an AC voltage source
« Reply #5 on: November 13, 2023, 10:53:02 PM »
Series capacitor is a most excellent suggestion. Thanks for using a circuit simulator to demonstrate.

The same "voltage doubler" circuit topology is found in practically all microwave ovens.

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Re: Need help with DC biasing an AC voltage source
« Reply #5 on: November 13, 2023, 10:53:02 PM »

 


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