Is there a way to make an inverter that only draws power when a load is connected?

Question

I’m working on a wearable battery project and I think it’s best to use AC power for the voltage and current adjustments, as well as stable transmission over long distances. Because it’s battery powered it needs to be as low energy as possible.

For that reason, I was wondering if it was possible to have the inverter be “always on” but only draw the amount of power needed to run the connected devices. Furthermore, I imagine the inverter itself would draw some power, so I assume it would need to draw power to run itself only when a device is drawing power. Would anything to that effect be possible?

I’m picturing an arrangement wherein a device, for example a fluorescent light tube (I won’t be using one for the project but it’s a common AC load device), would be connected to the inverter output directly. The circuit suddenly becoming closed would allow current to flow and the inverter would begin to operate normally. This would happen without drawing a fixed current, but rather just enough to run the light and the inverter. Ideally all of that would happen without digital logic components being involved.

Answer 1

If the output of an inverter is not connected to anything, there is no current flow though it. What you are asking for is actually the natural state of any power device; no load connected = no current flow. There is however a "burden" load on your battery to power up the internal control architecture of the inverter assembly itself, which would be necessary if you want the inverter ready for action as soon as a load is connected. On something "wearable" this can be a few watts to a few dozen watts, we have no way of knowing how you are designing it.

Answer 2

The circuit suddenly becoming closed would allow current to flow and the inverter would begin to operate normally.

Normally current doesn't flow through an open circuit. You don't have to do anything special to get that result.

This would happen without drawing a fixed current, but rather just enough to run the light and the inverter. Ideally all of that would happen without digital logic components being involved.

An inverter is typically a voltage source, and as such it doesn't push out a fixed current or wattage into the load. Rather a high resistance load will draw only a little current, and a low resistance load will draw more current according to Ohms Law I = V / R. You naturally get the behavior from all voltage sources weather it be a AA battery or an AC outlet in your home.

For that reason, I was wondering if it was possible to have the inverter be “always on” but only draw the amount of power needed to run the connected devices.

Unless the inverter design is very inefficient it's normally going to draw more power from the source if the load is consuming more power. This is typical behavior for any well designed power converter.

Furthermore, I imagine the inverter itself would draw some power, so I assume it would need to draw power to run itself only when a device is drawing power. Would anything to that effect be possible?

That's sort of a paradox. Before the inverter is on the device has no power supply, therefore the device can't draw power until the inverter is already on. So the inverter can't use the fact that the device is drawing power as the turn on criteria. The only way to find out if the device will draw power is to give it some and see what happens.

Because it’s battery powered it needs to be as low energy as possible.

Having the inverter on is going to burn a little power all the time regardless of if you have a device connected. If you want to eliminate that wasted power then you can just put a jumper into the connector that powers everything up when the cable is plugged in. As you can see from the picture, the power to the inverter (IN_POS) is disconnected from the inverter when the connector on the external device cable is unplugged.

Answer 3

I’m picturing an arrangement wherein a device, for example a fluorescent light tube, would be connected to the inverter output directly. The circuit suddenly becoming closed would allow current to flow and the inverter would begin to operate normally.

The problem is that a circuit closed through a flourescent light tube is very hard to detect, because no current will flow at all until you apply a high voltage.

In theory, one could design an inverter with a control circuit which would occasionally "pulse" it to "poll" if there is a tube connected. If current flow is seen, it would stay active. If it isn't, the source would go back to sleep for some period of time until checking again. Really careful design and choice of polling interval could possible give this usefully long standby life. But you'd have to get deep into the implementation details of a DANGEROUSLY HIGH VOLTAGE CIRCUIT, so this is inadvisable.

Really it would be better off either change to a lower voltage lighting technology, or else devising some other means of load detection.

For example, consider how this would be inactivated. Would you remove the tube itself? That could be dangerous if it leaves sockets exposed. If you unplug a wiring harness, then in addition to designing that harness to be intrinsically safe, you could design it with a low voltage connection detect circuit.