Is this a correct way to do a low voltage cutoff for motorcycle use? (Zener diode + relay)
You want to cut off power to a USB charger when the voltage from your battery/regulator falls below 14V.

Is this a good way to do it?

Your design will, in fact cut off power to Vcc when the input voltage falls below a certain level. However:

the Zener diode should have a 13.3 or 13.4V Zener voltage if you want it to trip at 14V. Your circuit actually trips at the Zener voltage plus approximately 0.6-0.7V.

your circuit may &quot;oscillate&quot; when the voltage is near 14V. The circuit has no hysteresis. If near 14V and the voltage sags, the relay will open. This might allow the voltage to rise just enough to turn the relay back on. Which might cause the voltage to sag etc.

When your battery / regulator voltage falls below 14V, your USB charger (Vcc) will be completely disconnected. However, a small current will still flow through your circuit. Although the current is small, if the motorcycle is unused for an extended period of time, this current could drain the battery.


The G5LE-1-DC12 relay requires 33.3mA to operate, and has a coil resistance of 360\$\Omega\$
The SS8050 is able to draw 1.5A current, which is quite sufficient for the relay, and has less than 1 uA of leakage current when off. It has a current gain between 40 and 300.
I am unable to find a specification for the forward &quot;off&quot; current of a 2EZ14 Zener diode. This is unfortunate, because whatever current leaks through the Zener diode, and is not diverted by R5 will cause a leakage through Q1 between 40 and 300 times greater.
Decreasing the value of R5, may help to decrease current into Q1 when the &quot;line&quot; voltage is below 14V. However, it will (obviously) increase the the leakage through D2 and R5. I believe the sweet spot occurs when the two leakage currents (through Q1 and R5) are equal, (but I haven't checked that).