# Adding small AC signal to a DC current using inductive coupling

Is it possible to add a small AC current on to a DC current in a wire using inductive coupling from an external source. Or if there is another way, I want to add AC ripples to a DC current.

simulate this circuit – Schematic created using CircuitLab

• Yes you can add ac ripple to a DC current. Sine wave level must be adjustable. L1 must handle frequencies of interest including number of turns and core material. – Sparky256 Jul 24 '19 at 23:41

First, there will be just one current in the circuit you show. The current into the coil must be the same as the current out of the cell.

You can do pretty much exactly what you have drawn. Use a transformer with the secondary winding connected as shown, in series with your dc source. You can use a function generator or other sine wave source connected to the primary winding of the transformer.

You will need to select a transformer designed to work at the ripple frequency of interest. If you want your ripple to be at the mains frequency this should be easy.

• I like this answer, and stole the idea for another previous question. – hacktastical Jul 25 '19 at 0:15
• Elliot Thank you for your answer. I have also edited the same current at input and output, thanks for pointing that out. – Ishaan Bassi Jul 25 '19 at 0:56
• Unfortunately, it’s not that simple. In many cases, this doesn’t work as cleanly as this answer describes. – Bob Jacobsen Jul 25 '19 at 11:23

Yes you can do this. The transformer must work at the frequencies of interest and you must keep the DC current through the secondary well below the rated peak current or core saturation will cause unwanted distortion of the ripple.

The answer is, it depends on what’s determining the DC current. Any AC current will also flow though those circuit elements.

For example, if the DC current is being driven by a high quality current source, that source will adapt to cancel the effect of the transformer.

A more extreme example: take the case of $$\I_{DC} = 0\$$ because it’s an open circuit. You get no AC current. But $$\I_{DC} = 0\$$ because it’s a short circuit (a loop of wire) would give you the full AC from the transformer

In many cases, you’ll get only partial cancellation: there will still be AC induced, but less than you would get if the transformer was just driving a short circuit.