Adapters drawing power when nothing plugged in

So, the usual wall-wart or power brick adapters that convert AC to DC, use a teensy amount of power when they are not plugged into a device on the DC side.

My question is, what about DC-DC converters? I’ve built a 12V battery bank and have mounted 12V accessory sockets (cig lighter style sockets) to use certain auto accessories, and it’s just convenient to use those plugs because they have inline fuses within.
I’m going to put in a little USB adapter that sits in the socket and puts out 5V in USB socket form. So this adapter is a 12VDC to 5VDC adapter.

I’m wondering if anyone has any info on whether a DC-DC adapter as such would draw any power when nothing is plugged into the USB socket.

Yes, it will draw current from the 12V source, any common regulator topology will. How much is anyone's guess. The only way to know if it matters to your battery charge level is to measure it in operation. Most likely it will be inconsequential, but you won't know 'til you know...

INTP:
I’m wondering if anyone has any info on whether a DC-DC adapter as such would draw any power when nothing is plugged into the USB socket.

If some voltage conversion is being done by a power adapter, then there will usually be at least some amount of power being used up by the system… associated with the conversion activity.

You could use a DC/DC converter with a spec sheet.
This one only draws 200uA idle current.
Leo..

Southpark:
If some voltage conversion is being done by a power adapter, then there will usually be at least some amount of power being used up by the system… associated with the conversion activity.

I get the impression that you’ve glossed over the part about no device drawing current through the converter. I know there are losses during use, I’m wondering about the idle consumption of a converter fed power on its input but nothing connected to output.
Namely, I was wondering/hoping that no connection on output would mean net open circuit for 0 power consumption.

INTP:
Namely, I was wondering/hoping that no connection on output would mean net open circuit for 0 power consumption.

Think about it - If the unit draws no residual power, how will it 'know' when an external device has been connected and hence 'waken up' For any device to be permanently active (to provide output-device-connected detection) it must draw power, even if only a few milliwatts.

That logic is not. It's as if you said, 'how does a battery know when to provide power to a light'.

A plain resistor is a 'DC-DC converter' in a sense. It doesn't need a wakeup signal.

So, think about it. There are plenty of cases where completing a circuit is all the wakeup involved. Your sweeping statement isn't as categorically obvious as you try to portray.

But thanks for taking the time to respond, nonetheless.

But you were referring to a 'real' DC to DC convertor which is an active device. Whilst a resistor etc is a passive device it demands no current if it's 'output' doesn't form a completed circuit. You could use a resistor or 7 volt zener to drop from 12 to 5 but only at a single current value. If you add ANY form of 'smart' (active) voltage monitor it will have some burden (even if only milliwatts) on the supply side to drive its internal circuitry.

And so TIL that 'real' DC to DC converters are 'active devices', if your phrasing is correct.

But still- 'knowing' when something is connected isn't the part that's requiring power, as far as I understand it. Plugging something in is a pretty drastic change to the state of affairs. It doesn't take some little AI to decide if something was in fact plugged in.
Just like the AC-DC converter wall wart situation, it is not a power draw caused by needing to be alert to the presence of something getting plugged in. The idle draw is just due to the nature of the components, which I imagine the transformer is the culprit. DC-DC doesn't have the same transformer, which is why I'm on this pondering that I'm pondering.

All DC to DC converters are a form of switch mode power supply, so even with no load on the output, the power supply has to keep switching something on and off to maintain the required regulated output voltage.
In any DC - DC power supply when switching, power is lost in the switching fet or transistor,
power is lost in the catch diode, and power is lost in the inductor.
Power is also lost in the output voltage divider which combined with the internal referance
determines what the output voltage is.
The internal voltage referance itself has to consume power to work.
The amount of power lost is usually related to the maximum output power the converter can supply, so the higher the output power, the more energy is lost, even with no load.

mauried:
All DC to DC converters are a form of switch mode power supply, so even with no load on the output, the power supply has to keep switching something on and off to maintain the required regulated output voltage.

My understanding was always that there's no output voltage on a smps until there's a load, although I recall reading that's not the case on more recent ones. (I use an atx on my bench, and had to stick a big resistor on the output since Arduinos and leds and other tiny loads didn't wake it up.)

All I know just have power. How else would it know when to switch on? And what is "a load"? 1A 1mA? 1uA? 1nA?

juma_yetu:
My understanding was always that there's no output voltage on a smps until there's a load

How could you know ?

Is there a 'no load' method of measuring a voltage ?

And as has been suggested, the very act of sensing a load, must introduce a load, no matter how small.

INTP:
I get the impression that you've glossed over the part about no device drawing current through the converter. I know there are losses during use, I'm wondering about the idle consumption of a converter fed power on its input but nothing connected to output.
Namely, I was wondering/hoping that no connection on output would mean net open circuit for 0 power consumption.

I focused on your question about - does your DC-DC converter draw any power if nothing is plugged into the 5V (your USB socket) output port.

So the answer was --- yes... the active DC-DC converter still draws power if you don't plug anything into that 5V socket.

Conventional switching power supplies require/required sufficient load at all times - to prevent instability (oscillations) that could destroy the power supply itself, which meant some kind of internal load was required to avoid this kind of thing.

@srnet and septillion: When I converted my atx smps for bench use, a gazillion posts all over the web said to stick a resistor over the output. I went with 10K 10W: without a load like that, there is actually no voltage. I have no idea how that works, but it gels with the exact same problem I had many years before. That was a proprietary PC architecture (French brand Normerel), and when I took the ps out of the case but with the mains into the now-removed ps, there was no output voltage. In that case, I just stuck it back in the case and the motherboard drew enough to turn it on.

As I say, I have no idea what the electronics behind that is, but it is. Although as I also said, I have read that more modern smps devices do not suffer from that limitation.

INTP:
A plain resistor is a 'DC-DC converter' in a sense.

If you're talking about a resistive voltage divider..... then just remember that the output voltage of the voltage divider will usually be influenced by a current-drawing component connected to that output terminal (of the divider). The divider doesn't have any features for regulating (or trying to regulate) the output voltage.

@juma_yetu, you confuse two things :wink:

First, the resistor. The purpose of that is to add some load because (especially older) PSU's find it a bit hard to deliver a stable voltage without load aka to make it stable. Not because there is no voltage at all.

Second, a disconnected PSU. It probably uses the same as a ATX supply, you need to pull a wire HIGH or LOW to activate it. As for a ATX, POWER_ON to GND. But to make that possible, internal it always derives a voltage. In the case of a ATX that's the 5V_STANDBY.

ATX power supplies wont run without a suitable load, simply because they cant regulate the
5V rail properly, and as an ATX power supply has both overvoltage cut outs on both the 12V and the 5V rail, the power supply will try to start, the 5V rail will very briefly go above 5V and the PS then shuts down.
Only the auxiallary 5V supply will keep running.