What does it mean for a signal to be 5V with respect to one leg of an AC input?

This power supply provides 5V with respect to one leg of the 24 VAC input. I'm trying to understand what that means because this provides 1.2 amps and I only need 20 mA for what I'm trying to do. It seems like hot and neutral are going to each end of the regulator. Does this mean a high signal would be in phase and amplitude with 24 VAC and a low signal would be the 24 VAC peak minus 5V? Can I do something similar to this with a LM7805?

• It's a bit hard to tell with the schematic snippet. There's a ground reference on your voltage regulator. Where is the ground on the 24 V AC? – Transistor Sep 20 '16 at 17:06
• There isn't a ground symbol for the 24 VAC on the entire schematic because the 24 VAC supply is being provided by another controller that has a 110 to 24 transformer. – Justin Manuel Sep 20 '16 at 17:19
• without seeing that source in your schematic, (and particularly, what is the neutral leg connected to, assuming 24VAC is the hot leg) I'm not sure how much we can tell you about it. – The Photon Sep 20 '16 at 17:27
• "There isn't a ground symbol for the 24 VAC on the entire schematic because..." Either the other leg of the 24VAC gets to Ground somehow, or it's not connected at all and therefore useless. Please show the entire circuit including both legs of the 24VAC. – Bruce Abbott Sep 20 '16 at 17:31
• halckemy.s3.amazonaws.com/uploads/pdf_file/file/133912/… – Justin Manuel Sep 20 '16 at 17:32

I think the rather unconventional SMPS arrangement produces a -5V supply with respect to one side of the (floating) 24VAC input.

In other words, the 'ground' is -5V with respect to J4 pin 2. Presumably this matches up with the inside of the device and allows communication using the 5V logic translator.

It does look kind of dodgy- but the Schottky diode should protect against obvious issues.

An LM7905 would do something similar, but its not rated for enough voltage and would be very inefficient.

• I think you are correct because when I connect the ground on my scope probe to J4 pin 5 and the probe to J4 pin 2, The scope says that it's resting at -5 volts. When I run some logic through it, the voltage toggles between 0V and -5V. – Justin Manuel Sep 20 '16 at 19:32
• And the 24VAC transformer definitely goes to J4:2? Then this is it. – Spehro Pefhany Sep 20 '16 at 19:34
• So one can connect the input and output of a regulator to each leg of a transformer to produce a negative DC voltage? What is the purpose of C2 and C3? Why aren't they connected to virtual ground like C4? – Justin Manuel Sep 20 '16 at 19:45
• It's more like the supply of ~+33V (half-wave rectified and filtered 24VAC) is being connected between the output of the regulator and the input rather than more conventionally between the ground and the input. That's why the input filter caps C2/C3 are in an unconventional arrangement. This chip is being run uncomfortably close to the absolute maximum 40V input (it sees 38V nominally). I don't like it for that reason alone. It could easily see 45V if the transformer is lightly loaded. – Spehro Pefhany Sep 20 '16 at 19:59

First of all, there is no Hot and Neutral. Based on the fact that you are asking that the DC voltage is relative of one of the legs, I assume that pins 1 an 5 of J4 are connected to a transformer's secondary.

To call one node ground, you don't need to connect it to anything special (e.g. earth.) You just call it ground and that's the end of the story. Then if you put a DC regulator, the resulting DC voltage is going to be reffered to that leg of the transformer. In this case the leg that is connected to pin 5.

Your schematic can be simplified this way:

simulate this circuit – Schematic created using CircuitLab

In the schematic, the ground symbol is connected to J4:5. However, if the pin is not connected to earth also, the voltage of J4:5 with respect to earth is unknown. However, J4:2 is 5VDC with respect to J4:5.

As you may see, I omitted the ground symbol because it's only used to name the node connected to J4:5.

Yes, you can use a LM7805, but the setup shown on the diagram is much more efficient.

• OP says pins 1 and 2 of J4 are the 24V input. Also note the input filter capacitors are connected to the power supply 5V output. What you've drawn is sensible but it doesn't match the schematics. – Spehro Pefhany Sep 20 '16 at 19:11
• I meant to simplify; however, added the capacitors. The point is that the voltage of J4:2 with respect to J4:5 is 5VDC. – Krauss Sep 20 '16 at 19:22
• But is the transformer connected to J4:2 as the OP says it is, or is it J4:5 as you show? – Spehro Pefhany Sep 20 '16 at 19:24
• Please note that J4:1 is connected to the input of the regulator and J4:2 is connected to the output. You can't connect the transformer to the regulator one leg to the input and one leg to the output. Also note that D1 is a half wave rectifier. The regulator is reffered to J4:5 thus that pin should be the other leg of the transformer. – Krauss Sep 20 '16 at 19:34
• Generally speaking, a DC offset to an AC supply is useful if you want to, for instance, switch the "hot" side of the supply with a MOSFET. It can be used to provide the necessary voltage to the gate. – Ian Bland Sep 20 '16 at 20:15