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For a project I am working on I need to take a DC voltage source, around 12V, with at least 1A of current, and I need to add it to a generated DC pulse train which will be from -4 to 4V and will be very low current, say less than 20 mA. So I need the output to be from 8V to 16V DC pulse and I need the current to be from 0.98 to 1.02A.

Someone told me about a summer, but you can imagine googling "summer" does not bring up much that is not related to the season of summer. Googling voltage summer tends to bring up summing amplifiers using op-amps which only handle very small current. And remember I do not want any current or voltage gain. I have my two inputs and I want to simply sum their voltage and currents together.

Does anyone have any ideas for how to accomplish this? Any help would really be appreciated.

In case you are wondering, my 12V, 1A voltage source will be a switching converter from mains supply (110V, 60Hz to 12V DC)

My -4V to 4V pulse train will be from an op amp output setup as a comparator.

Thank you for any help you can give.

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  • \$\begingroup\$ If it is a high speed signal maybe capacitive coupling. Otherwise, you really need some sort of an amplifier that can handle 1 amp. \$\endgroup\$
    – MadHatter
    Commented Nov 5, 2015 at 3:52
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    \$\begingroup\$ This actually doesn't make sense, not the way you say it. The problem is that if the -4V to 4V pulse train only has 20 mA of oomph to it, it will basically have zero effect on the voltage of the 12V supply. I think, but I am not sure, that what you want is to have a power supply whose voltage can be modulated by a -4V to +4V control signal. This can certainly be done, but it is not an "off the shelf" type of thing. I think you should explain a lot more about why you are doing this and what you are trying to accomplish. \$\endgroup\$
    – user57037
    Commented Nov 5, 2015 at 4:08
  • \$\begingroup\$ Oh, provide a link to the 12V source you are looking at. \$\endgroup\$
    – user57037
    Commented Nov 5, 2015 at 4:09
  • \$\begingroup\$ Output current is determined by the load. What is your circuit driving, and why do you want the current to vary from 0.98 to 1.02A? \$\endgroup\$
    – Bruce Abbott
    Commented Nov 5, 2015 at 5:21
  • \$\begingroup\$ Hello, please disregard the question. I was told by my supervisor that the way this is accomplished is by summing the two together. It turns out that was completely wrong and it is done using an op-amp as a comparator fed to a push/pull transistor amplifier. The final product is -15V to +15V and is done entirely by switching the gates of the push/pull transistors. Thanks for answering, but turns out I was on the wrong track the whole time. \$\endgroup\$
    – humanistscience
    Commented Nov 5, 2015 at 21:06

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you need what called a clamper circuit, you can add any signal (pulsed) to a dc voltage

enter image description here

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