I have two non contact voltage detectors (Klein and Greenlee) ..both are rated to detect AC voltage with a sensitivity range from 50 to 1000 V. I used these detectors on 9 different cell phone power adapters with only the USB cable attached to the adapter (so no current is flowing). On 5 of the power adapters it states that the output is DC at about 5 V with a range from 0.3 to 1.0 amps. Everything I have researched says that these non contact detectors will only detect AC voltage; not DC. On four of these adapters I detect voltage and on one I do not. The results are the same for both detectors and are repeatable. Can anyone explain why these detectors that are rated for AC use at 50 to 1000V apparently can detect voltage on a DC supply at about 5 V?? …and why voltage is not detected on one (1) of the power adapters??
Electronic – Why does a non contact AC voltage detector detect voltage on a DC line that is not flowing
voltage-detector
Related Solutions
This device works on capacitive coupling, not inductive coupling, as the inductor you have labeled "R2 lead" suggests. Capacitive coupling has the advantage of working equally well regardless of how much current is flowing in the wire. You probably want to know that wire is live even though the light is off at the moment, so this is good.
This probably also explains why the R2 lead is bent. This serves to increase the effective surface area of that plate of the capacitor, increasing the capacitance. My tester (Klein Tools NCVT-1) has a solid, flat probe in the tip.
Rather, R2's lead is one plate of the capacitor, and your hand is the other:
simulate this circuit – Schematic created using CircuitLab
Recall that capacitance is the ratio of charge to voltage:
$$ C = \frac{Q}{V} $$
But you can re-arrange that as:
$$ V = \frac{Q}{C} $$
We don't know what \$Q\$ is, but we know that it's roughly constant, owing to the very high input impedance of U1A.
Further, we know that the capacitance is a function of the separation between the plates, so as your hand moves towards or away from the probe, C is changing. So if Q isn't changing fast enough to be significant, and C changing rapidly as you move your hand (one plate) closer or farther away from the other plate (the resistor lead), then \$V\$ must change. Beep!
You can in fact get this effect by moving the probe rapidly towards or away from any conductive object. I have many times observed my probe beeping briefly when I bring it near a circuit I have disconnected at the circuit breaker.
To validate this explanation, try this experiment: tap the probe against your hand, rapidly, and vigorously, until it's beeping frequently. Then, reduce your taps to still somewhat rapid, but gentle contacts, or even not-quite contacts, with your hand. You should find it to be quite sensitive. Now, hold it touching your hand for a moment. Ten seconds is sufficient with my tester. Now try the same gentle pattern of contacts. You should find it much less sensitive.
Here's what I expect is occurring: when you go through the initial round of smacking, you increase the charge on the capacitor by injecting a lot of noise through triboelectric effects, wildly varying capacitance to your hand, RF currents picked up by your body, coupled through that capacitance, microphonic effects, jarring the battery, and so on. Point is, when you are done, \$Q\$ is very likely non-zero. This increases the voltage swing as you vary the capacitance between the probe and your hand, and thus, the sensitivity.
When you hold the probe near your hand, the charge separation you initially established has a chance to equalize through leakage paths past the capacitor. Since we are talking about a very high impedance, and a low charge, there need not be much leakage to be significant.
I tried to reproduce your drop tests, but could not, reliably. I didn't want to sacrifice my tester to solder the contacts, so I tested with an unmodified tester. If I smack the probe against my hand, I can get it to beep. But if I smack the opposite end of the device, and I'm careful to not be holding it in any way to be close to the probe end, it does not beep. I suppose battery contacts bouncing could still explain this behavior, and by smacking the other end I'm just not jarring the battery as effectively. However, battery jarring does not explain the experiment I previously described. Perhaps both mechanisms are significant.
Since it appears that the voltage across the optoisolator input is low enough for the controller to interpret it as a low, just put it directly in series with the switch:
simulate this circuit – Schematic created using CircuitLab
Since the door controller is already limiting the current, you don't need any additional resistors in series with the LED.
If the current sourced by the controller was more than the optoisolator could handle, then a slightly more complicated circuit using a transistor would be required.
Best Answer
They are not detecting the DC voltage of the power supply. They are detecting the AC that is present due to leakage current in the power supply.
Every power supply will have leakage, and when the power supply is not grounded, this leakage current has no low resistance return path, and will create a voltage.
This can be detected by non contact AC voltage detectors.
Switch mode power supplies have this by design (C1), or by parasitic effect of the transformer inside.
The leakage current is very low, below a few hundred uA or even less. You might have noticed a "buzz" feeling touching some appliance, that is the result of this leakge.
Large appliances can leak a higher current, and this might give a tingle.