This under voltage circuit may be used as an add-on to an existing power supply or standalone. The circuit draws power
from an existing power supply, which may be up to 36 Vdc. Once voltage drops below a preset level the output swings high. A
CA3140 MOSFET op-amp is used as a comparator, the output can drive loads of up to 10mA so can drive a low current LED
directly. Additional buffering is required for higher loads.
The circuit is a standard comparator with the non-inverting input set to a reference voltage. Once the voltage at the inverting
input drops below a preset amount the output swings high. Although not shown, it is easy to connect the output to a LED or relay
or drive additional circuitry.
With the values used above the under voltage is set at 10 Volt. A practical use would be to provide an alarm or shut off power
from a 12 Volt car battery and prevent deep discharge. The reference voltage is set by zener diode D1 to 6.2 Vdc. The comparator
will swing high when the voltage on pin 2 (inverting input) reaches approximately 6.1 Vdc. The trip voltage "Vtrip" is set by
the circuits supply voltage, Vs and the potential divider R1 and R2.
| Vtrip =
|| Vs R1
|| << Vref
| R1 + R2
This circuit uses four parameters: Vs, the supply voltage, Vtrip the limit at which the circuit operates, Vref, the reference
voltage and finally the ration of the potential divider R1 and R2. On designing the circuit you must know three of the parameters.
You will know the supply voltage, Vs and the value you want you want the circuit to operate at, Vtrip. The reference voltage
Vref must always be a few volts less than the lowest voltage that the supply will drop to.
In this example, I have the designed the circuit to operate on a car battery, nominal voltage 12. I want the trip to operate at
10 Vdc, i.e. Vtrip=10. I have set the reference voltage to 6.2 Volt but could have chosen a 5.1V zener diode or other lower value.
The comparator will switch when the non-inverting input drops 0.1V lower than reference voltage. Because I want my circuit to
activate at 10V the supply voltage Vs will have to drop to this value. The voltage at pin2, will be the ratio of Vs R1 / (R1+R2)
R1 and R2 can be any values, but I have chosen R1 to be 22k. A formula can now be derived.
Re-arranging the terms for R2 the formula becomes:
In my example I have chosen R1 = 22k and Vs = 10 so:
Simulating Under Voltage
| R2 =
|| 10 * 22k
|| - 22k = 14k
Trying this circuit with LTspice , I used a piecewise linear source for Vs, which simply changes voltage from 12 Volt to
8 Volts in 4 seconds. In real life, depending on the load a car battery may take several hours or even days to drop
to this value. In any case you can see the comparator output switch from near zero volts to about 8 Volts. Bear in mind
that the supply voltage decays in time so any succeeding circuit must be designed to allow for this.
For anyone wishing to experiment further, the LTspice version is available below: