AC coupling with no common ground - Page 1
Forum section: electronics / beginners

author: m.elsayed
author_rank: Contributor
timestamp: August 09, 2025, 01:46:31 am
content: Background: I am trying to simulate injecting a small digital signal(3-5VDC 10-50 kHz) to a high voltage AC signal (220VAC 50Hz)My understanding is that connecting a capacitor between the signals will block DC voltage from passing and allows only AC, and since the PWM signal has much higher frequency it will dominate and inject itself to the larger AC signal.And since there is no common ground, the small signal circuit does not really see 220V potential nor the other way around.My question is how to analyze and simulate such circuit when there is no common ground? and how this even works?Any pointers on this exact topic from a book or a video would be much appreciated.In the attached schematics, I basically switch between the two "grounds" then run the simulation, but I really have no idea if this even makes sense.P.S. I am not really trying to implement such a circuit since I do not really know what I am doing, and this is just a mere curiosity.

author: ledtester
author_rank: Super Contributor
timestamp: August 09, 2025, 04:09:35 am
content: Quote from: m.elsayed on August 09, 2025, 01:46:31 amBut this is assuming that current can flow through the capacitor.The signal generator V2 won't be able to push any electrons through the capacitor because there is no return path.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 09, 2025, 06:12:42 am
content: Also draw the return wire in the simulation, and add a capacitor both in the "signal" wire and in the "GND / return" wire. With only one capacitor, there is still a galvanic connection between the input and output. Then you can connect one of the sides to "GND" (or "zero" in spice) and for the rest you can create 230V and 5Vac sources and combine them at will.I'm not exactly sure whether those sources need a DC path to the zero node. If they need it, you can add an extra Giga Ohm resistor just to make the spice simulation happy. (ngSpice needs this sometimes, other spice simulators add this implicitly when they think it's needed.

author: m.elsayed
author_rank: Contributor
timestamp: August 09, 2025, 10:58:21 am
content: Quote from: ledtester on August 09, 2025, 04:09:35 amThank you for your response, is this one of the situations where it works in theory and the simulator but not in real life?My understanding --which might be completely wrong-- is that when there is an AC signal the capacitor will charge to the voltage potential it sees on the side that is emitting the signal (relative to this side ground).So in this case the capacitor will charge to 3-5V(relative to the circuit on the right ground), then discharge the 3-5V to the carrying signal on the left.I know that a transformer is the right answer, but still curious whether this in practice will work or not.Quote from: Doctorandus_P on August 09, 2025, 06:12:42 amThank you, I had a feeling that there is nothing to stop the current from passing through to the small signal side --at least in practice--, so this is a great insight.

author: PGPG
author_rank: Frequent Contributor
timestamp: August 09, 2025, 07:46:42 pm
content: Quote from: m.elsayed on August 09, 2025, 01:46:31 amUntil you speak about normal current/voltage signals and not about radio-transmissions you should always assume that each current have to flow in closed circuit and each voltage to be understand as signal have to be measured against some other point in circuit.If you have battery powered source generating 5V pulses at its output (when referred to its battery (-)) and connect this output to anything then no current will be flowing from your output to anything as this current have no way to return back.Your generator output instead being injected in the node you expect will have no influence at that node, but this device battery (-) will be jumping up and down as nothing protects it against such jumping. The only way to inject your output signal is to make something to make your sentence "And since there is no common ground" be false.

author: m.elsayed
author_rank: Contributor
timestamp: August 09, 2025, 08:54:06 pm
content: Quote from: PGPG on August 09, 2025, 07:46:42 pmThank you, no, radio wave is out of the question, the signal is 30kHz at most.I think this makes it much clearer to me, for current to flow via capacitive coupling there must be a return path.I will run another simulation but this time will focus on the grounds(-) and observe how they behave.But I am struggling to understand why would it work in the simulation? in the circuit above I have no common ground and I simply switch between the two grounds when probing and I can see the signal is injected just fine.

author: MariuszD
author_rank: Regular Contributor
timestamp: August 10, 2025, 05:28:03 am
content: Quote from: m.elsayed on August 09, 2025, 08:54:06 pmWithout a return path, there will be no injected signal. In the simulation, the current flowing from V2 will be zero. The voltage V2 will be visible only on one terminal, the one not connected.In the simulation, the circuits are ideal, their operation is simple, an open circuit cannot work.In the real circuit, it will be some very small signal because the return path will close due to parasitic capacitances. Capacitances will be in the range of a few picofarads. 5pF at 30kHz has a reactance of 1 megaohm. A few microamperes will flow into the grid. The grid at 30kHz has high impedance. As a result of this current, a voltage of several hundred microvolts will appear.You can't learn anything about the operation of a real circuit dependent on parasitic parameters from this simulation.Neon line tester works thanks to these parasitic capacities.https://www.electrothinks.com/2020/03/Neon-line-tester-circuit-working-explanation.html

author: Zero999
author_rank: Super Contributor
timestamp: August 10, 2025, 11:08:00 am
content: Capacitors will work. Two wires are of course needed: live as well as neutral.If the DC power supply is considered to be SELV then the capacitor needs to be a low vale (look up what mains leakage currents are permitted by the safety standards and be sure not to exceed them) and Y1-rated (look this up if you don't know what it means). The PCB must have sufficient creepage and clearances, again look it up.

author: mtwieg
author_rank: Frequent Contributor
timestamp: August 10, 2025, 01:21:02 pm
content: Quote from: m.elsayed on August 09, 2025, 10:58:21 amQuoteWhat specifically do you mean by "works"?Quote"I can see the signal is injected" meaning what, specifically?As others have pointed out, there's no return path in your circuit. Therefor zero current will flow anywhere. That means there will be zero voltage across R1 or C1. They effectively have no effect on anything, regardless of what values they have. So at node SIG you will see just the 220V AC from V1, and at node G you will see the sum of V1 and V2. But is that a "working" circuit?

author: inse
author_rank: Super Contributor
timestamp: August 10, 2025, 01:29:53 pm
content: I‘d advise to stay away from 230V until the principles of electricity are understood

author: Seekonk
author_rank: Super Contributor
timestamp: August 10, 2025, 11:10:39 pm
content: I might suggest using a small common mode choke as a transformer. They work pretty good in that range and you will have great isolation. You still need a capacitor on the line side. I can't see getting much signal thru a 100K resistor.